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Soal Limit Fungsi Aljabar

Limit fungsi aljabar merupakan dasar matematika untuk mempelajari Integral Fungsi, Limit Fungsi Tak Hingga, Limit Fungsi Trigonometri, dan Fungsi Turunan (diferensial).

Penerapan Limit Fungsi Aljabar dalam kehidupan sehari-hari memang sulit untuk kita lihat secara langsung tapi sebenarnya sangat erat kaitannya dengan kehidupan. Hanya saja kita belum tahu lebih jauh, contoh sederhananya adalah pembuatan tanggal kedaluarsa makanan dan menghitung kecepatan suatu kendaraan.

Atau lebih sederhana lagi saat mengukur berat badan, ketika kita melihat hasilnya pada timbangan akan terlihat 62,5 kg. Sebenarnya hasil 62,5 kg ini bukanlah hasil pengukuran yang paling akurat tetapi dapat mewakili hasil pengukuran, karena berat kita mendekati 62,5 kg. Kata “mendekati” merupakan salah satu kata kunci dalam mempelajari limit fungsi.

Cara menyelesaikan soal-soal limit sebenarnya tidaklah sulit apabila kita paham aturan-aturan pada limit fungsi aljabar. Untuk memudahkan kamu memahaminya silahkan pahami step by step pada pembahasan soal limit fungsi aljabar di bawah ini.

Tapi sebelum itu, pahami catatan sederhana tentang limit fungsi agar memudahkanmu dalam menyelesaikan masalah limit fungsi aljabar atau limit fungsi trigonometri.

Berdasarkan definisi limit, Jika nilai Limit Kiri = Limit Kanan = L secara simbol dituliskan $\lim\limits_{x \to a^{+}}f(x)=\lim\limits_{x \to a^{-}}f(x)=L$ maka nilai $\lim\limits_{x \to a}f(x)=L$.

Nilai limit fungsi $\lim\limits_{x \to a}f(x)$ dapat ditentukan dengan cara mensubstitusi nilai $x=a$ ke fungsi $f(x)$.

Langkah untuk menyelesaikan limit fungsi baik itu limit fungsi aljabar atau trigonometri dan limit tak hingga, langkah awalnya adalah substitusi langsung. Apabila setelah subtitusi mendapatkan hasil bentuk tak tentu seperti $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $0 \times \infty$, $\infty – \infty$, $0^{0}$, atau $\infty^{\infty}$ maka harus melakukan manipulasi aljabar dengan cara memfaktorkan, mengalikan dengan akar sekawan, atau dengan manipulasi aljabar lainnya selama tidak menyalahi aturan bermatematika.

Cara Menyelesaikan Limit Fungsi Aljabar

1. Pemfaktoran

Bentuk-bentuk pemfaktoran yang sering digunakan antara lain:

2. Mengalikan Akar Sekawan

Berikut beberapa bentuk akar sekawan dari fungsinya:

3. Teorema Limit Fungsi

Beberapa teorema limit fungsi bisa kita gunakan dalam menyelesaikan Soal Limit Fungsi. Andaikan $n$ bilangan bulat positif, $k$ konstanta, dan $f$ dan $f$ dan $g$ adalah fungsi yang mempunyai limit di $c$. Maka berlaku:

4. Menggunakan Turunan (Aturan L’Hospital)

Aturan L’Hostpital adalah cara alternatif untuk menyelesaikan limit fungsi. Untuk menggunakan aturan ini, kita perlu mempelajari Turunan Fungsi terlebih dahulu, jika belum maka tidak dianjurkan untuk menggunakannya.

5. Suatu fungsi $f$ dikatakan kontinu dititik $a$ jika $\lim\limits_{x \to a} f(x) = f(a)$

Definisi di atas menjelaskan bahwa sebuah fungsi $f$ dikatakan kontinu dititik $a$ jika memenuhi ketiga syarat berikut:

Jika salah satu saja dari syarat di atas tidak dipenuhi, Maka $f$ dikatakan tidak kontinu di $x=a$.

Soal dan Pembahasan Limit Fungsi Aljabar SMA

Soal Limit Fungsi Aljabar berikut ini merupakan hasil sadur dari soal UN (Ujian Nasional), soal Ujian Mandiri, soal SBMPTN (Seleksi Bersama Masuk Perguruan Tinggi Negeri), soal Ujian Sekolah, dan soal simulasi dari berbagai bimbingan belajar.

1. Soal UNBK SMA IPA 2018

Diketahui $f(x)=\begin{cases}3x-p,\ x\leq 2 \\ 2x+1,\ x \gt 2 \end{cases}$ Agar $\lim\limits_{x \to 2}f(x)$ mempunyai nilai, maka $p=…$

$\begin{align} (A)\ & -2 \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 2 \end{align}$

Pembahasan:

Limit Kanan

Limit Kiri

Limit Kiri = Limit Kanan

6 – p = 5
6 – 5 = p
p = 1

Jadi jawabannya yang benar adalah D. 1

2. Soal UN SMA IPS 2018

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}-3x+2}{x-1} =\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 2 \end{align}$

Pembahasan:

Coba kerjakan langsung dengan substitusi,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-3x+2}{x-1} \\ & = \dfrac{2^{2}-3(2)+2}{2-1} \\ & = \dfrac{4-6+2}{2-1} \\ & = \dfrac{ 0 }{1}= 0 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 0$

3. Soal UN IPS 2018

Nilai $\lim\limits_{x \to 2} \dfrac{2x^{2}-x-6}{3x^{2}-5x-2} =\cdots$

$\begin{align} (A)\ & -1 \\ (B)\ & 0 \\ (C)\ & \dfrac{1}{5} \\ (D)\ & 1 \\ (E)\ & 7 \end{align}$

Pembahasan:

Langkah pertama adalah dengan cara memfaktorkan,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{2x^{2}-x-6}{3x^{2}-5x-2} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 2x+3 \right)\left( x-2 \right)}{\left( 3x+1 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 2x+3 \right)}{\left( 3x+1 \right)} \\ & = \dfrac{2(2)+3 }{3(2)+1}=\dfrac{7}{7}=1 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 1$

Cara alternatif (Aturan L’Hospital).
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{2x^{2}-x-6}{3x^{2}-5x-2} \\ & = \lim\limits_{x \to 2} \dfrac{4x-1}{6x-5} \\ & = \dfrac{4(2)-1}{6(2)-5} = \dfrac{7}{7}=1 \end{align} $

4. Soal UNBK SMA IPS 2019

$\lim\limits_{x \to 3} \dfrac{x^{2}-9}{2x^{2}-7x+3} =\cdots$

$\begin{align} (A)\ & \dfrac{1}{2} \\ (B)\ & \dfrac{5}{6} \\ (C)\ & \dfrac{6}{7} \\ (D)\ & \dfrac{7}{6} \\ (E)\ & \dfrac{6}{5} \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{x^{2}-9}{2x^{2}-7x+3} \\ & = \lim\limits_{x \to 3} \dfrac{(x+3)(x-3)}{(2x-1)(x-3)} \\ & = \lim\limits_{x \to 3} \dfrac{(x+3) }{(2x-1) } \\ & = \dfrac{(3+3) }{(2(3)-1) } = \dfrac{6}{5} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ \dfrac{6}{5}$

Cara alternatif menggunakan Aturan L’Hospital.
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{x^{2}-9}{2x^{2}-7x+3} \\ & = \lim\limits_{x \to 3} \dfrac{2x-0}{4x-7+0} \\ & = \dfrac{2(3)-0}{4(3)-7+0} = \dfrac{6}{5} \end{align} $

5. Soal EBTANAS SMA IPS 1995

Nilai dari $\lim\limits_{x \to 0} \dfrac{6x^{5}-4x}{2x^{4}+x} =\cdots$

$\begin{align} (A)\ & -4 \\ (B)\ & -2 \\ (C)\ & 0 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{x \to 0} \dfrac{6x^{5}-4x}{2x^{4}+x} \\ & = \lim\limits_{x \to 0} \dfrac{\left( x \right) \left( 6x^{4}-4 \right)}{\left( x \right) \left( 2x^{3}+1 \right)} \\ & = \lim\limits_{x \to 0} \dfrac{ \left( 6x^{4}-4 \right)}{\left( 2x^{3}+1 \right)} \\ & = \dfrac{ \left( 6(0)^{4}-4 \right)}{\left( 2(0)^{3}+1 \right)} \\ & = \dfrac{ -4 }{1}=-4 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -4$

Cara alternatif menggunakan Aturan L’Hospital.
$ \begin{align} & \lim\limits_{x \to 0} \dfrac{6x^{5}-4x}{2x^{4}+x} \\ & = \lim\limits_{x \to 0} \dfrac{30x^{4}-4}{8x^{3}+1} \\ & = \dfrac{30(0)^{4}-4}{8(0)^{3}+1} \\ & = \dfrac{-4}{1}=-4 \end{align} $

6. Soal EBTANAS SMA IPS 1996

Nilai $\lim\limits_{x \to 5} \dfrac{x^{2}-x-20}{x-5} =\cdots$

$\begin{align} (A)\ & 9 \\ (B)\ & 5 \\ (C)\ & 4 \\ (D)\ & -4 \\ (E)\ & -9 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 5} \dfrac{x^{2}-x-20}{x-5} \\ & = \lim\limits_{x \to 5} \dfrac{\left( x-5 \right) \left( x+4 \right)}{\left( x-5 \right)} \\ & = \lim\limits_{x \to 5} \dfrac{ \left( x+4 \right)}{\left( 1 \right)} \\ & = \dfrac{ 5+4 }{1}=9 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ 9$

Peringatan!
Cara alternatif menggunakan turunan fungsi (Aturan L’Hospital).
$ \begin{align} & \lim\limits_{x \to 5} \dfrac{x^{2}-x-20}{x-5} \\ & = \lim\limits_{x \to 5} \dfrac{2x-1}{1} \\ & = \dfrac{2(5)-1}{1} = 9 \end{align} $

7. Soal EBTANAS SMA IPS 1997

Nilai $\lim\limits_{x \to 3} \dfrac{x-3}{x^{2}+x-12} =\cdots$

$\begin{align} (A)\ & 4 \\ (B)\ & 3 \\ (C)\ & \dfrac{3}{7} \\ (D)\ & \dfrac{1}{7} \\ (E)\ & 0 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{x-3}{x^{2}+x-12} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)}{ \left( x+4 \right)\left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( 1 \right)}{\left( x+4 \right)} \\ & = \dfrac{ 1 }{3+4}=\dfrac{ 1 }{7} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ \dfrac{1}{7}$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{x-3}{x^{2}+x-12} \\ & = \lim\limits_{x \to 3} \dfrac{1}{2x+1} \\ & = \dfrac{1}{2(3)+1} = \dfrac{1}{7} \end{align} $

8. Soal EBTANAS SMA IPS 1998

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}-x-2} =\cdots$

$\begin{align} (A)\ & 3 \\ (B)\ & 2 \\ (C)\ & 0 \\ (D)\ & -2 \\ (E)\ & -3 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}-x-2} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x+4 \right)\left( x-2 \right)}{ \left( x-2 \right)\left( x+1 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x+4 \right)}{\left( x+1 \right)} \\ & = \dfrac{ 2+4 }{2+1}=\dfrac{ 6 }{3}=2 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ 2$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}-x-2} \\ & = \lim\limits_{x \to 2} \dfrac{2x+2}{2x-1} \\ & = \dfrac{2(2)+2}{2(2)-1} = \dfrac{6}{3}=2 \end{align} $

9. Soal EBTANAS SMA IPS 1999

Nilai $\lim\limits_{x \to 3} \dfrac{\left( x-2 \right)^{2}-1}{x-3} =\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 1 \\ (C)\ & 2 \\ (D)\ & 4 \\ (E)\ & 6 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{\left( x-2 \right)^{2}-1}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}-4x+4-1}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}-4x+3}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-1 \right)\left( x-3 \right)}{ \left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x-1 \right)}{\left( 1 \right)} \\ & = \dfrac{ 3-1 }{1}=2 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 2$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{\left( x-2 \right)^{2}-1}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{2\left( x-2 \right)(1)}{1} \\ & = \dfrac{2\left( 3-2 \right)(1)}{1}=2 \end{align} $

10. Soal EBTANAS SMA IPS 2000

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}+4x-12} =\cdots$

$\begin{align} (A)\ & \infty \\ (B)\ & 1 \\ (C)\ & \dfrac{3}{4} \\ (D)\ & \dfrac{1}{2} \\ (E)\ & 0 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}+4x-12} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x+4 \right)\left( x-2 \right)}{ \left( x+6 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x+4 \right)}{\left( x+6 \right)} \\ & = \dfrac{ 2+4 }{2+6}=\dfrac{ 6 }{8}=\dfrac{ 3 }{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ \dfrac{3}{4}$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}+4x-12} \\ & = \lim\limits_{x \to 2} \dfrac{2x+2}{2x+4} \\ & = \dfrac{2(2)+2}{2(2)+4} = \dfrac{6}{8}=\dfrac{3}{4} \end{align} $

11. Soal EBTANAS Matematika SMA IPA 2002

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}-5x+6}{x^{2}-4} =\cdots$

$\begin{align} (A)\ & -\dfrac{1}{4} \\ (B)\ & -\dfrac{1}{8} \\ (C)\ & \dfrac{1}{8} \\ (D)\ & 1 \\ (E)\ & \dfrac{5}{4} \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-5x+6}{x^{2}-4} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x-2 \right)\left( x-3 \right)}{ \left( x+2 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x-3 \right)}{\left( x+2 \right)} \\ & = \dfrac{ 2-3 }{2+2}=\dfrac{ -1 }{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -\dfrac{1}{4}$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-5x+6}{x^{2}-4} \\ & = \lim\limits_{x \to 2} \dfrac{2x-5}{2x} \\ & = \dfrac{2(2)-5}{2(2)} = \dfrac{-1}{4} \end{align} $

12. Soal UN Matematika SMA IPS 2017

Nilai $\lim\limits_{x \to 3} \dfrac{2x^{2}-4x-6}{x^{2}-2x-3} =\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & 0 \\ (C)\ & 2 \\ (D)\ & 6 \\ (E)\ & 8 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{2x^{2}-4x-6}{x^{2}-2x-3} \\ & = \lim\limits_{x \to 3} \dfrac{2 \left( x+1 \right)\left( x-3 \right)}{ \left( x+1 \right)\left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{2 \left( x+1 \right)}{ \left( x+1 \right)} \\ & = \dfrac{ 2(3+1) }{3+1}=2 \end{align} $

$ \therefore $ Jawaban yang sesuai adalah $(C)\ 2$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 3} \dfrac{2x^{2}-4x-6}{x^{2}-2x-3} \\ & = \lim\limits_{x \to 3} \dfrac{4x-4}{2x-2} \\ & = \dfrac{4(3)-4}{2(3)-2}=\dfrac{8}{4}=2 \end{align} $

13. Soal UN Matematika SMA IPS 2015

Nilai $\lim\limits_{x \to 4} \dfrac{x^{2}-16}{x-4} =\cdots$

$\begin{align} (A)\ & 16 \\ (B)\ & 8 \\ (C)\ & 4 \\ (D)\ & -4 \\ (E)\ & -8 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 4} \dfrac{x^{2}-16}{x-4} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x+4 \right)\left( x-4 \right)}{ \left( x-4 \right)} \\ & = \lim\limits_{x \to 4} \dfrac{ \left( x+4 \right)}{\left( 1 \right)} \\ & = \dfrac{ 4+4 }{1}=8 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ 8$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 4} \dfrac{x^{2}-16}{x-4} \\ & = \lim\limits_{x \to 4} \dfrac{2x}{1} \\ & = \dfrac{2(4)}{1} = 8 \end{align} $

14. Soal UN Matematika SMA IPS 2014

Nilai $\lim\limits_{x \to -4} \dfrac{x^{2}+7x+12}{2x+8} =\cdots$

$\begin{align} (A)\ & -1 \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & \dfrac{7}{8} \\ (D)\ & \dfrac{3}{2} \\ (E)\ & \dfrac{7}{2} \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to -4} \dfrac{x^{2}+7x+12}{2x+8} \\ & = \lim\limits_{x \to -4} \dfrac{\left( x+3 \right)\left( x+4 \right)}{ \left( 2 \right)\left( x+4 \right)} \\ & = \lim\limits_{x \to -4} \dfrac{ \left( x+3 \right)}{\left( 2 \right)} \\ & = \dfrac{ -4+3 }{2}=\dfrac{ -1 }{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ -\dfrac{1}{2}$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to -4} \dfrac{x^{2}+7x+12}{2x+8} \\ & = \lim\limits_{x \to -4} \dfrac{2x+7}{2} \\ & = \dfrac{2(-4)+7}{2} = \dfrac{-1}{2} \end{align} $

15. Soal UN Matematika SMA IPA 2008

Nilai $\lim\limits_{x \to 2} \dfrac{x^{3}-4x}{x-2} =\cdots$

$\begin{align} (A)\ & 32 \\ (B)\ & 16 \\ (C)\ & 8 \\ (D)\ & 4 \\ (E)\ & 2 \end{align}$

Pembahasan:

Kita Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-4x}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{ x\ \left( x-2 \right)\left( x+2 \right)}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{ x\ \left( x+2 \right)}{1} \\ & = \dfrac{2\left( 2+2 \right)}{1}=8 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 8$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-4x}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{3x^{2}-4}{1} \\ & = \dfrac{3(2)^{2}-4}{1} = 8 \end{align} $

16. Soal UN Matematika SMA IPA 2010

Nilai $\lim\limits_{x \to 2} \left( \dfrac{2}{x-2}-\dfrac{8}{x^{2}-4} \right)=\cdots$

$\begin{align} (A)\ & \dfrac{1}{4} \\ (B)\ & \dfrac{1}{2} \\ (C)\ & 2 \\ (D)\ & 4 \\ (E)\ & \infty \end{align}$

Pembahasan:

Kita Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \left( \dfrac{2}{x-2}-\dfrac{8}{x^{2}-4} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{\left( 2x^{2}-8 \right)-\left( 8x-16 \right)}{\left( x-2 \right)\left( x^{2}-4 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2x^{2}-8x+8}{\left( x-2 \right)\left( x-2 \right)\left( x+2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2\left( x-2 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x-2 \right)\left( x+2 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2}{\left( x+2 \right)} \right) \\ & = \dfrac{2}{\left( 2+2 \right)}=\dfrac{2}{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ \dfrac{1}{2}$

17. Soal UN Matematika SMA IPA 2007

Nilai $\lim\limits_{x \to 1} \dfrac{x^{2}-5x+4}{x^{3}-1} =\cdots$

$\begin{align} (A)\ & 3 \\ (B)\ & 2\frac{1}{2} \\ (C)\ & 2 \\ (D)\ & 1 \\ (E)\ & -1 \end{align}$

Pembahasan:

Kita Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2}-5x+4}{x^{3}-1} \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x-1 \right)\left( x-4 \right)}{\left( x-1 \right)\left( x^{2}+x+1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x-4 \right)}{ \left( x^{2}+x+1 \right)} \\ & = \dfrac{ \left( 1-4 \right)}{ \left( (1)^{2}+(1)+1 \right)} \\ & = \dfrac{-3}{3} =-1 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ -1$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2}-5x+4}{x^{3}-1} \\ & = \lim\limits_{x \to 1} \dfrac{2x-5}{3x^{2}} \\ & = \dfrac{2(1)-5}{3(1)^{2}} = \dfrac{-3}{3}=-1 \end{align} $

18. Soal UAN Matematika SMA IPA 2004

Nilai $\lim\limits_{x \to 2} \left( \dfrac{2}{x^{2}-4}-\dfrac{3}{x^{2}+2x-8} \right)=\cdots$

$\begin{align} (A)\ & -\dfrac{7}{12} \\ (B)\ & -\dfrac{1}{4} \\ (C)\ & -\dfrac{1}{12} \\ (D)\ & -\dfrac{1}{24} \\ (E)\ & 0 \end{align}$

Pembahasan:

Kita Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \left( \dfrac{2}{x^{2}-4}-\dfrac{3}{x^{2}+2x-8} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{\left( 2x^{2}+4x-16 \right)-\left( 3x^{2}-12 \right)}{\left( x^{2}+2x-8 \right)\left( x^{2}-4 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-x^{2}+4x-4}{\left( x+2 \right)\left( x-2 \right)\left( x+4 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-\left( x-2 \right)\left( x-2 \right)}{\left( x+2 \right)\left( x-2 \right)\left( x+4 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-1}{\left( x+2 \right) \left( x+4 \right)} \right) \\ & = \dfrac{-1}{\left( 2+2 \right) \left( 2+4 \right)}=\dfrac{-1}{24} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -\dfrac{1}{24}$

19. Soal SPMB 2004

$\lim\limits_{x \to 2} \dfrac{x^{3}-8}{x^{2}+x-6} =\cdots$

$\begin{align} (A)\ & \dfrac{3}{4} \\ (B)\ & \dfrac{2}{15} \\ (C)\ & 1 \dfrac{1}{3} \\ (D)\ & 2 \dfrac{2}{5} \\ (E)\ & 6 \end{align}$

Pembahasan:

Kerjakan dengan cara memfaktorkan terlebih dahulu,
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-8}{x^{2}+x-6} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x^{2}+2x+4 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x^{2}+2x+4 \right)}{\left( x+3 \right)} \\ & = \dfrac{ (2)^{2}+2(2)+4}{2+3} = \dfrac{ 12}{5} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 2 \dfrac{2}{5}$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-8}{x^{2}+x-6} \\ & = \lim\limits_{x \to 2} \dfrac{3x^{2}-0}{2x+1-0} \\ & = \dfrac{3(2)^{2}-0}{2(2)+1-0} = \dfrac{ 12}{5} \end{align} $

20. Soal UM UGM 2005 Kode 621

$\lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{x-x^{3}} \right)=\cdot$

$\begin{align} (A)\ & -\dfrac{3}{2} \\ (B)\ & -\dfrac{2}{3} \\ (C)\ & \dfrac{2}{3} \\ (D)\ & 1 \\ (E)\ & \dfrac{3}{2} \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{x-x^{3}} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{\left( x \right)\left( 1+x \right)-2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{x^{2}+x-2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{ \left( x+2 \right)\left( x-1 \right)}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{ -\left( x+2 \right)\left( 1-x \right)}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{ -\left( x+2 \right)}{\left( x \right)\left( 1+x \right)} \right) \\ & = \dfrac{-\left( (1)+2 \right)}{\left( 1 \right)\left( 1+(1) \right)} \\ & = \dfrac{-3}{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -\dfrac{3}{2}$

21. Soal SIMAK UI 2018 Kode 416

Jika $\lim\limits_{x \to -3} \dfrac{\frac{1}{ax}+\frac{1}{3}}{bx^{3}+27}=-\dfrac{1}{3^{5}}$, nilai $a+b$ untuk $a$ dan $b$ bulat positif adalah…

$\begin{align} (A)\ & -4 \\ (B)\ & -2 \\ (C)\ & 0 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

Untuk menyelesaikan soal limit di atas sehingga hasilnya seperti yang diharapkan, kita coba dengan menggunakan turunan. Kita anggap pada turunan pertama nilai limit untuk $x \to -3$ hasilnya adalah $-\dfrac{1}{3^{5}}$.

$\begin{align} \lim\limits_{x \to -3} \dfrac{\frac{1}{ax}+\frac{1}{3}}{bx^{3}+27} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{\frac{1}{a}x^{-1}+\frac{1}{3}}{bx^{3}+27} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-\frac{1}{ax^{2}}}{3bx^{2}} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-1}{3bx^{2} \cdot ax^{2}} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-1}{3abx^{4}} & = -\dfrac{1}{3^{5}} \\ -\dfrac{1}{3ab(-3)^{4}} & = -\dfrac{1}{3^{5}} \\ -\dfrac{1}{ ab3^{5}} & = -\dfrac{1}{3^{5}} \\ ab & = 1 \end{align}$

Untuk $a$ dan $b$ bilangan bulat positif yang memenuhi $ab=1$ adalah $a=1$ dan $b=1$, maka $a+b=2$

$\therefore$ Jawaban yang benar $(D)\ 2$

22. Soal UNBK Matematika SMA IPA 2019

Diketahui $f(x)=5x^{2}+3$. Hasil dari $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ adalah…

$\begin{align} (A)\ & 0 \\ (B)\ & 5 \\ (C)\ & 10 \\ (D)\ & 10x \\ (E)\ & 5x^{2} \end{align}$

Pembahasan:

Dari informasi pada soal, yang ditanyakan adalah $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ dari sebuah fungsi $f(x)$. Jika kita teliti dalam membaca soal bahwa yang ditanyakan pada soal adalah turunan fungsi $f(x)$. Definisi turunan fungsi $f(x)$ adalah $f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$. $ \begin{align} f(x) &=5x^{2}+3 \\ f'(x) &=10x \end{align}$

Tetapi jika ingin mengerjakannya dengan proses limit fungsi berikut pembahasannya:
$\begin{align} & \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5(x+h)^{2}+3 -\left( 5x^{2}+3 \right)}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5 \left(x^{2}+2hx+h^{2} \right) +3 – 5x^{2}-3}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5 x^{2}+10hx+5h^{2} +3 – 5x^{2}-3}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ +10hx+5h^{2}}{h} \\ & = \lim\limits_{h \to 0} \left( +10 x+5h \right) \\ & = 10 x+5(0) \\ & = 10x \end{align}$

$\therefore$ Jawaban yang benar adalah $(D)\ 10x$

23. Soal UMB PTN 2008 Kode 270

$\lim\limits_{t \to 2} \dfrac{4t^{4}+4t-72}{\left( t-2 \right)\left( t^{2}+3t+2 \right)}=\cdot$

$\begin{align} (A)\ & \dfrac{11}{4} \\ (B)\ & \dfrac{11}{3} \\ (C)\ & 11 \\ (D)\ & 22 \\ (E)\ & 33 \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{t \to 2} \dfrac{4t^{4}+4t-72}{\left( t-2 \right)\left( t^{2}+3t+2 \right)} \\ & = \lim\limits_{t \to 2} \dfrac{4\left( t-2 \right)\left( t^{3}+2t^{2}+4t+9 \right)}{\left( t-2 \right)\left( t^{2}+3t+2 \right)} \\ & = \lim\limits_{t \to 2} \dfrac{4 \left( t^{3}+2t^{2}+4t+9 \right)}{ \left( t^{2}+3t+2 \right)} \\ & = \dfrac{4 \left( (2)^{3}+2(2)^{2}+4(2)+9 \right)}{ (2)^{2}+3(2)+2} \\ & = \dfrac{4 \left( 33 \right)}{ 12} = 11 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 11$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{t \to 2} \dfrac{4t^{4}+4t-72}{\left( t-2 \right)\left( t^{2}+3t+2 \right)} \\ & = \lim\limits_{t \to 2} \dfrac{16t^{3}+4}{\left( 1 \right)\left( t^{2}+3t+2 \right)+\left( t-2 \right)\left( 2t+3 \right)} \\ & = \dfrac{16(2)^{3}+4}{\left( 1 \right)\left( (2)^{2}+3(2)+2 \right)+\left( 2-2 \right)\left( 2(2)+3 \right)} \\ & = \dfrac{16 \cdot 8 +4}{12} = \dfrac{4 (4 \cdot 8) +1}{12} = 11 \end{align} $

24. Soal UM UGM 2010 Kode 461

Nilai $\lim\limits_{x \to 2} \left( \dfrac{6}{x^{2}-x-2}-\dfrac{2}{x-2} \right)$ sama dengan…

$\begin{align} (A)\ & -1 \\ (B)\ & -\dfrac{2}{3} \\ (C)\ & -\dfrac{1}{3} \\ (D)\ & \dfrac{1}{3} \\ (E)\ & \dfrac{2}{3} \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{x \to 2} \left( \dfrac{6}{x^{2}-x-2}-\dfrac{2}{x-2} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6}{\left( x-2 \right)\left( x+1 \right)}-\dfrac{2}{x-2} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6}{\left( x-2 \right)\left( x+1 \right)}-\dfrac{2\left( x+1 \right)}{\left( x-2 \right)\left( x+1 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6-2\left( x+1 \right)}{\left( x-2 \right)\left( x+1 \right)} \right)\\ & = \lim\limits_{x \to 2} \left( \dfrac{4-2x}{\left( x-2 \right)\left( x+1 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-2\left( x-2 \right)}{\left( x-2 \right)\left( x+1 \right)} \right)\\ & = \lim\limits_{x \to 2} \left( \dfrac{-2 }{ \left( x+1 \right)} \right)\\ & = \dfrac{-2 }{ \left( 2+1 \right)} = \dfrac{-2}{3} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ -\dfrac{2}{3}$

25. Soal SIMAK UI 2011 Kode 213

Jika $\lim\limits_{x \to a} \left( f(x)-3g(x) \right)=2$ dan $\lim\limits_{x \to a} \left( 3f(x)+g(x) \right)=1$ maka $\lim\limits_{x \to a} \left( f(x) \cdot g(x) \right)=\cdots$

$\begin{align} (A)\ & -\dfrac{1}{2} \\ (B)\ & -\dfrac{1}{4} \\ (C)\ & \dfrac{1}{4} \\ (D)\ & \dfrac{1}{2} \\ (E)\ & 1 \end{align}$

Pembahasan:

Dari persamaan $\lim\limits_{x \to a} \left( f(x)-3g(x) \right)=2$ dan $\lim\limits_{x \to a} \left( 3f(x)+g(x) \right)=1$ dengan teorema limit dapat kita ubah betuknya menjadi:
$ \begin{align} \lim\limits_{x \to a} \left( f(x)-3g(x) \right) &=2 \\ \lim\limits_{x \to a} f(x)-3\ \lim\limits_{x \to a} g(x) &=2 \\ \hline \lim\limits_{x \to a} \left( 3f(x)+g(x) \right) &=1 \\ 3\ \lim\limits_{x \to a} f(x)+ \lim\limits_{x \to a} g(x) &=1 \end{align} $

Jika kita misalkan $\lim\limits_{x \to a} f(x)=m$ dan $\lim\limits_{x \to a} f(x)=n$, maka kita peroleh:
$\begin{array}{c|c|cc} m-3n=2 & (\times 3) \\ 3m+ n =1 & (\times 1) \\ \hline 3m – 9n =6 & \\ 3m + n =1 \ \ \ (-)& \\ \hline -10n = 5 & \\ n = -\dfrac{1}{2} & \lim\limits_{x \to a} g(x)=-\dfrac{1}{2} \\ m = \dfrac{1}{2} & \lim\limits_{x \to a} f(x) = \dfrac{1}{2} \\ \end{array} $

$ \begin{align} \lim\limits_{x \to a} \left( f(x) \cdot g(x) \right) & = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a} g(x) \\ & = \left( -\dfrac{1}{2} \right) \cdot \left( \dfrac{1}{2} \right) \\ & = -\dfrac{1}{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ -\dfrac{1}{4}$

26. Soal SBMPTN 2016 Kode 355

Jika $a$ dan $b$ adalah dua bilangan real dengan $\lim\limits_{x \to 2} \dfrac{x^{2}+2ax+b}{x-2}=-3$, maka $ab=\cdots$

$\begin{align} (A)\ & -35 \\ (B)\ & -30 \\ (C)\ & -15 \\ (D)\ & -3 \\ (E)\ & -1 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2ax+b}{x-2}=-3$. Jika kita substitusi langsung nilai $x=2$ maka nilai $x^{2}+2ax+b$ harus $0$, karena jika $x^{2}+2ax+b$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $x^{2}+2ax+b$ untuk $x=2$ adalah $0$ maka $x-2$ adalah salah satu faktornya sehingga berlaku;

$\begin{align} x^{2}+2ax+b & \equiv (x-2)(mx+n) \\ x^{2}+2ax+b & \equiv mx^{2}+nx-2mx-2n \\ x^{2}+2ax+b & \equiv mx^{2}+(n-2m)x-2n \\ \hline \therefore & 1 = m \\ \therefore & 2a = n-2m=n-2 \\ \therefore & b = -2n \end{align}$

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2ax+b}{x-2}=-3$, maka:
$\begin{align} \lim\limits_{x \to 2} \dfrac{(x-2)(x+n)}{x-2} & =-3 \\ \lim\limits_{x \to 2} (x+n) & =-3 \\ 2+n & =-3 \\ n &= -3-2 \\ n &= -5 \end{align}$

Untuk $n=-5$ maka $b=10$ dan $a=-\dfrac{7}{2}$

Nilai $ab=-\dfrac{7}{2} \cdot 10 = -35$

$\therefore$ Jawaban yang benar $(A)\ -35$

27. Soal SBMPTN 2016 Kode 333

Jika $f(x)=x^{2}+ax+b$ dengan $f(1)=0$ dan $\lim\limits_{x \to 1} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$, maka $b=\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 2 \end{align}$

Pembahasan:

Diketahui $f(1)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f(1)\ & = 1^{2}+a(1)+b \\ 0\ & = 1+a +b \\ -1\ & = a +b \end{align}$

Nilai $\lim\limits_{x \to 1} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$ sehingga jika kita substitusi langsung nilai $x=1$ maka nilai $f \left( x+1 \right)-f \left( x \right)$ harus $0$, karena jika $f \left( x+1 \right)-f \left( x \right)$ tidak nol maka nilai limit adalah $\infty$.

Nilai $f \left( x+1 \right)-f \left( x \right)$ untuk $x=1$ adalah $0$ sehingga dapat kita tuliskan:
$\begin{align} f \left( x+1 \right)-f \left( x \right) & =0 \\ f \left( 1+1 \right)-f \left( 1 \right) & =0 \\ f \left( 2 \right)- 0 & =0 \\ 2^{2}+a(2)+b & =0 \\ 4+2a+b & =0 \\ \hline 2a+b & = -4 \\ a +b & = -1\ (-) \\ \hline a & = -3 \\ b & = 2 \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 2$

28. Soal SBMPTN 2016 Kode 353

Jika $f(x)=x^{2}+ax+b$ dengan $f(2)=0$ dan $\lim\limits_{x \to 2} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-2}=2$, maka $b=\cdots$

$\begin{align} (A)\ & -6 \\ (B)\ & -5 \\ (C)\ & 0 \\ (D)\ & 5 \\ (E)\ & 6 \end{align}$

Pembahasan:

Diketahui $f(2)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f(2)\ & = 2^{2}+a(2)+b \\ 0\ & = 4+2a +b \\ -4\ & = 2a +b \end{align}$

Nilai $\lim\limits_{x \to 2} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$ sehingga jika kita substitusi langsung nilai $x=2$ maka nilai $f \left( x+1 \right)-f \left( x \right)$ harus $0$. Karena jika $f \left( x+1 \right)-f \left( x \right)$ tidak nol maka nilai limit adalah $\infty$.

Karena nilai $f \left( x+1 \right)-f \left( x \right)$ untuk $x=2$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} f \left( x+1 \right)-f \left( x \right) & =0 \\ f \left( 2+1 \right)-f \left( 2 \right) & =0 \\ f \left( 3 \right)- 0 & =0 \\ 3^{2}+a(3)+b & =0 \\ 9+3a+b & =0 \\ \hline 3a+b & = -9 \\ 2a +b & = -4\ (-)\\ \hline a & = -5 \\ b & = 6 \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 6$

29. Soal SBMPTN 2016 Kode 321

Diketahui $f(x)=x^{2}+ax+b$ dengan $f \left( b+1 \right)=0$ dan $\lim\limits_{x \to 0} \dfrac{ f \left( x+b \right)}{x}=-1$, maka $a+2b=\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 2 \end{align}$

Pembahasan:

Diketahui $f(b+1)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f \left( b+1 \right)\ & = \left( b+1 \right)^{2}+a\left( b+1 \right)+b \\ 0\ & = b^{2}+2b+1+ab+a+b \\ 0\ & = b^{2}+3b+ab+a+1 \end{align}$

Nilai $\lim\limits_{x \to 0} \dfrac{ f \left( x+b \right)}{x}=-1$ sehingga jika kita substitusi langsung nilai $x=0$ maka nilai $f \left( 0+b \right)$ harus $0$. Karena jika $f \left( 0+b \right)$ tidak nol maka nilai limit adalah $\infty$.

Karena nilai $f \left(b \right)$ untuk $x=0$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} f(x) & = x^{2}+ax+b \\ f(b) & = (b)^{2}+a(b)+b \\ 0 & = b^{2}+ab+b \\ \hline 0\ & = b^{2}+2b+ab+a+b+1 \\ 0\ & = b^{2}+ab+b+2b+a+1 \\ 0\ & = 0+2b+a+1 \\ -1\ & = 2b+a \end{align}$

$\therefore$ Jawaban yang benar $(B)\ -1$

30. Soal SBMPTN 2016 Kode 355

Jika $a$ dan $b$ adalah dua bilangan real dengan $\lim\limits_{x \to 2} \dfrac{ x^{2}+2ax+b}{x-2}=-3$, maka $ab=\cdots$

$\begin{align} (A)\ & -35 \\ (B)\ & -30 \\ (C)\ & -15 \\ (D)\ & -3 \\ (E)\ & -1 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 2} \dfrac{ x^{2}+2ax+b}{x-2}=-3$ sehingga jika kita substitusi langsung nilai $x=2$ maka nilai $x^{2}+2ax+b$ harus $0$, karena jika $x^{2}+2ax+b$ tidak nol maka nilai limit adalah $\infty$.

Karena nilai $x^{2}+2ax+b$ untuk $x=2$ adalh $0$ maka $x-2$ adalah salah satu faktornya sehingga $x^{2}+2ax+b \equiv (x-2)( x+n)$, dan dapat kita tuliskan;
$\begin{align} \lim\limits_{x \to 2} \dfrac{ x^{2}+2ax+b}{x-2} &=-3 \\ \lim\limits_{x \to 2} \dfrac{ (x-2)( x+n)}{x-2} &=-3 \\ \lim\limits_{x \to 2} \left( x+n \right) &=-3 \\ 2+n &=-3 \\ n &=-5 \end{align}$

Untuk $n=-5$, kita peroleh:
$\begin{align} x^{2}+2ax+b &= \equiv (x-2)( x+n) \\ x^{2}+2ax+b &= \equiv (x-2)( x-5) \\ x^{2}+2ax+b &= \equiv x^{2}-7x+10 \\ \hline 2a &=-7 \\ b &=10 \\ ab &= -35 \end{align}$

$\therefore$ Jawaban yang benar $(A)\ -35$

31. Soal SBMPTN 2017 Kode 226|*Soal Lengkap

Jika kurva $f(x)=ax^{2}+bx+c$ memotong sumbu-$y$ di titik $(0,1)$ dan $\lim\limits_{x \to 1} \dfrac{f(x)}{x-1}=-4$ maka $\dfrac{b+c}{a}=\cdots$

$\begin{align} (A)\ & -1 \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & \dfrac{3}{2} \end{align}$

Pembahasan:

Kurva $f(x)=ax^{2}+bx+c$ memotong sumbu-$y$ di titik $(0,1)$ maka nilai $c=1$ sehingga $f(x)=ax^{2}+bx+1$.

Nilai $\lim\limits_{x \to 1} \dfrac{ax^{2}+bx+1}{x-1}=-4$
Jika kita substitusi langsung nilai $x=1$ maka nilai $ax^{2}+bx+1$ harus $0$, karena jika $ax^{2}+bx+1$ tidak nol maka nilai limit adalah $\infty$.

Karena nilai $ax^{2}+bx+1$ untuk $x=1$ adalh $0$ maka $x-1$ adalah salah satu faktornya sehingga berlaku;
$\begin{align} ax^{2}+bx+1 & \equiv (x-1)(mx+n) \\ ax^{2}+bx+1 & \equiv mx^{2}+nx-mx-n \\ ax^{2}+bx+1 & \equiv mx^{2}+(n-m)x-n \\ -1 &= n \\ b &= n-m \\ b &= -1-m \\ a &= m \end{align}$

Nilai $\lim\limits_{x \to 1} \dfrac{ax^{2}+bx+1}{x-1}=-4$, maka:
$\begin{align} \lim\limits_{x \to 1} \dfrac{(x-1)(mx+n)}{x-1} & =-4 \\ \lim\limits_{x \to 1} (mx+n) & =-4 \\ \lim\limits_{x \to 1} (mx-1) & =-4 \\ m-1 & =-4 \\ m &= -4+1 \\ m &=-3 \end{align}$

Untuk $m=-3$ nilai $a=-3$, $b=2$ dan $c=1$, maka $\dfrac{b+c}{a}=\dfrac{2+1}{-3}=-1$

$\therefore$ Jawaban yang benar $(A)\ -1$

32. Soal UMPTN 1998 (Rayon C)

$\lim\limits_{x \to 1} \dfrac{x^{2n}- x}{1-x}=\cdots$

$\begin{align} (A)\ & 2n-1 \\ (B)\ & 1-2n \\ (C)\ & 2n \\ (D)\ & 2n-2 \\ (E)\ & 2n+2 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $a^{n}-b^{n}=\left(a-b \right)\left(a^{n-1}+a^{n-2}b+ \cdots + ab^{n-2}+b^{n-1} \right)$ Untuk $n$ bilangan Asli

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2n}- x}{1-x} \\ & = \lim\limits_{x \to 1} \dfrac{x\left( x^{2n-1}- 1 \right)}{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left( x^{2n-1}- 1^{2n-1} \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x-1 \right)\left(x^{2n-1-1}+x^{2n-1-2}(1)+ \cdots + (x)(1)^{2n-1-2}+(1)^{2n-1-1} \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x-1 \right)\left(x^{2n-2}+x^{2n-3} + \cdots + x+1 \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x^{2n-2}+x^{2n-3} + \cdots + x+1 \right) }{-1} \\ & = \dfrac{(1) \left((1)^{2n-2}+(1)^{2n-3} + \cdots + (1)+1 \right) }{-1} \\ & = -(1) \left( 2n-2 +1 \right) \\ & = -(1) \left( 2n-1 \right) \\ & = -2n+1 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ 1-2n$

Peringatan!
Cara alternatif menggunakan Aturan L’Hospital, Sebaiknya digunakan apabila sudah belajar turunan fungsi.
$ \begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2n}- x}{1-x} \\ & = \lim\limits_{x \to 1} \dfrac{2nx^{2n-1}- 1}{0-1} \\ & = \dfrac{2n(1)^{2n-1}- 1}{-1} \\ & = \dfrac{2n- 1}{-1}=-2n+1 \end{align} $

33. Soal UM UNDIP 2019 Kode 324

Jika $\left| f(x)-2 \right| \leq x+3$, maka nilai $\lim\limits_{x \to -3}f(x)=\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & 0 \\ (C)\ & 1 \\ (D)\ & 2 \\ (E)\ & 3 \end{align}$

Pembahasan:

Berdasarkan sifat pertidaksamaan nilai mutlak Himpunan penyelesaian dari $\left| f(x) \right| \leq a$ adalah $-a \leq f(x) \leq a$. Sehingga jika kita terapkan pada fungsi soal, kita akan peroleh:

\begin{align} \left| f(x)-2 \right| & \leq x+3 \\ -(x+3) \leq f(x) & -2 \leq (x+3) \\ – x-3+2 \leq f(x) & \leq x+3+2 \\ – x-1 \leq f(x) & \leq x+5 \\ \lim\limits_{x \to -3} \left(-x-1 \right) \leq \lim\limits_{x \to -3} & f(x) \leq \lim\limits_{x \to -3}\left( x+5 \right) \\ -(-3)-1 \leq \lim\limits_{x \to -3} & f(x) \leq -3+5 \\ 2 \leq \lim\limits_{x \to -3} & f(x) \leq 2 \end{align}

$\therefore$ Jawaban yang benar adalah $(D)\ 2$

34. Soal UM UGM 2019 Kode 923/924

Jika $p \gt 0$ dan $\lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx}{x-p}=12$, maka nilai $p-q$ adalah…

$\begin{align} (A)\ & 14 \\ (B)\ & 10 \\ (C)\ & 8 \\ (D)\ & 5 \\ (E)\ & 3 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx}{x-p}=12$ sehingga jika kita substitusi langsung nilai $x=p$ maka nilai $x^{3}+px^{2}+qx$ harus $0$, karena jika $x^{3}+px^{2}+qx$ tidak nol maka nilai limit adalah $\infty$.

Karena nilai $x^{3}+px^{2}+qx$ untuk $x=p$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} (p)^{3}+p(p)^{2}+q(p) &= 0 \\ 2p^{3} + pq &= 0 \\ 2p^{2} + q &= 0 \\ q &= 2p^{2} \\ \end{align}$

$x-p$ adalah salah satu faktor $x^{3}+px^{2}+qx$ sehingga dapat kita tuliskan;
$\begin{align} x^{3}+px^{2}+qx &= (x-p)( x^{2}+bx+c) \\ x^{3}+px^{2}+qx &= x^{3}+bx^{2}+cx-px^{2}-bpx-pc \\ x^{3}+px^{2}+qx &= x^{3}+ \left( b-p \right)x^{2}+\left( c-bp \right)x-pc \\ \hline -pc=0 & \rightarrow c=0 \\ b-p=p & \rightarrow b=2p \\ \hline x^{3}+px^{2}+qx &= (x-p)( x^{2}+2px) \\ \end{align}$


Dari hasil di atas kita peroleh:
$\begin{align} \lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx }{x-p} &=12 \\ \lim\limits_{x \to p} \dfrac{ (x-p)( x^{2}+2px) }{x-p} &=12 \\ \lim\limits_{x \to p} \dfrac{ ( x^{2}+2px) }{1} &=12 \\ (p)^{2}+2p(p) &=12 \\ 3p^{2} &=12 \\ p^{2} &= 4 \rightarrow p=2 \\ \hline q &= -2p^{2} \rightarrow q=-8 \end{align}$

Nilai $p-q$ adalah $2-(-8)=10$

Peringatan!
Jika sudah belajar turunan fungsi, maka dapat digunakan Aturan L’Hospital yang mungkin dapat menghemat beberapa langkah.

$\therefore$ Jawaban yang benar $(B)\ 10$

35. Soal UN Matematika SMA IPA 2011

$\lim\limits_{x \to 4} \dfrac{\left( x-4 \right)}{\sqrt{x}-2}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 4 \\ (C)\ & 8 \\ (D)\ & 12 \\ (E)\ & 16 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)}{\sqrt{x}-2} \cdot \dfrac{\sqrt{x}+2}{\sqrt{x}+2} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)\left( \sqrt{x}+2 \right)}{x-4} \\ & = \lim\limits_{x \to 4} \dfrac{\left( \sqrt{x}+2 \right)}{1} \\ & = \dfrac{\left( \sqrt{4}+2 \right)}{1}=4 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ 4$

36. Soal UNBK Matematika SMA IPA 2019

Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}-6x+8}{3-\sqrt{17-2x^{2}}}=\cdots$

$\begin{align} (A)\ & 6 \\ (B)\ & \dfrac{2}{3} \\ (C)\ & 0 \\ (D)\ & -\dfrac{3}{2} \\ (E)\ & -6 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-6x+8}{3-\sqrt{17-2x^{2}}} \cdot \dfrac{3+\sqrt{17-2x^{2}}}{3+\sqrt{17-2x^{2}}} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x-4)\left( 3+\sqrt{17-2x^{2}} \right)}{9-\left( 17-2x^{2} \right)} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x-4)\left( 3+\sqrt{17-2x^{2}} \right)}{ 2x^{2}-8} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x-4)\left( 3+\sqrt{17-2x^{2}} \right)}{ 2(x-2)(x+2)} \\ & = \lim\limits_{x \to 2} \dfrac{ (x-4)\left( 3+\sqrt{17-2x^{2}} \right)}{ 2 (x+2)} \\ & = \dfrac{ (2-4)\left( 3+\sqrt{17-2(2)^{2}} \right)}{ 2 (2+2)} \\ & = \dfrac{ (-2)\left( 3+3 \right)}{8} \\ & = -\dfrac{3}{2} \end{align}$

$\therefore$ Jawaban yang benar $(D)\ -\dfrac{3}{2}$

37. Soal EBATANAS Matematika SMA IPA 1999

Nilai $\lim\limits_{x \to 2} \dfrac{x-2}{\sqrt{x+7}-3}=\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & -\dfrac{2}{3} \\ (C)\ & 0 \\ (D)\ & 6 \\ (E)\ & 12 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{x-2}{\sqrt{x+7}-3} \\ & = \lim\limits_{x \to 2} \dfrac{x-2}{\sqrt{x+7}-3} \cdot \dfrac{\sqrt{x+7}+3}{\sqrt{x+7}+3} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x-2 \right)\left( \sqrt{x+7}+3 \right)}{ x+7 +9} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x-2 \right)\left( \sqrt{x+7}+3 \right)}{ x-2} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( \sqrt{x+7}+3 \right)}{ 1 } \\ & = \dfrac{ \left( \sqrt{2+7}+3 \right)}{ 1 }=6 \end{align}$

$\therefore$ Jawaban yang benar $(D)\ 6$

38. Soal SPMB 2004

$\lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( \sqrt{x}+\sqrt{3} \right)}{\sqrt{x} – \sqrt{3}} =\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 3 \\ (C)\ & 6 \\ (D)\ & 12 \\ (E)\ & 15 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( \sqrt{x}+\sqrt{3} \right)}{\sqrt{x} – \sqrt{3}} \cdot \dfrac{ \left( \sqrt{x}+\sqrt{3} \right)}{ \left( \sqrt{x}+\sqrt{3} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( \sqrt{x}+\sqrt{3} \right)\left( \sqrt{x}+\sqrt{3} \right)}{x – 3} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( \sqrt{x}+\sqrt{3} \right)\left( \sqrt{x}+\sqrt{3} \right)}{1} \\ & = \dfrac{ \left( \sqrt{3}+\sqrt{3} \right)\left( \sqrt{3}+\sqrt{3} \right)}{1} \\ & = \left( 2\sqrt{3} \right)\left( 2\sqrt{3} \right)=12 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 12$

39. Soal SIMAK UI 2018 Kode 421

Jika $\lim\limits_{x \to 4} \dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\cdots$

$\begin{align} (A)\ & \dfrac{1}{64} \\ (B)\ & \dfrac{1}{128} \\ (C)\ & \dfrac{1}{256} \\ (D)\ & \dfrac{1}{512} \\ (E)\ & \dfrac{1}{1024} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

Untuk menyelesaikan soal limit diatas, kita coba dengan memisalkan $\sqrt{x}=p$ sehingga $x^{2}=p^{4}$ dan karena $x \to 4$ maka $p \to 2$, perubahan pada soal menjadi;

$\begin{align} & \lim\limits_{x \to 4} \dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16} \\ & = \lim\limits_{p \to 2} \dfrac{p-\sqrt{3p-2}}{p^{4}-16} \\ & = \lim\limits_{p \to 2} \dfrac{p-\sqrt{3p-2}}{p^{4}-16} \\ & = \lim\limits_{p \to 2} \dfrac{p-\sqrt{3p-2}}{p^{4}-16} \times \dfrac{p+\sqrt{3p-2}}{p+\sqrt{3p-2}} \\ & = \lim\limits_{p \to 2} \dfrac{p^{2}-(3p-2)}{\left (p^{2}-4 \right )\left (p^{2}+4 \right )\left (p+\sqrt{3p-2} \right )} \\ & = \lim\limits_{p \to 2} \dfrac{(p-2)(p-1)}{\left (p-2 \right )\left (p+2 \right )\left (p^{2}+4 \right )\left (p+\sqrt{3p-2} \right )} \\ & = \lim\limits_{p \to 2} \dfrac{ (p-1)}{ \left (p+2 \right )\left (p^{2}+4 \right )\left (p+\sqrt{3p-2} \right )} \\ & = \dfrac{ (2-1)}{ \left (2+2 \right )\left (2^{2}+4 \right )\left (2+\sqrt{3(2)-2} \right )} \\ & = \dfrac{1}{ \left (4 \right )\left (8 \right )\left (4 \right )} = \dfrac{1}{128} \end{align}$

$\therefore$ Jawaban yang benar $(B)\ \dfrac{1}{128}$

40. Soal UN Matematika SMA IPA 2003 – Soal UMB PTN 2014 Kode 573

$ \lim\limits_{x \to 2} \dfrac{4-x^{2}}{3-\sqrt{x^{2}+5}}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 2 \\ (C)\ & 3 \\ (D)\ & 4 \\ (E)\ & 6 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{4-x^{2}}{3-\sqrt{x^{2}+5}} \\ & = \lim\limits_{x \to 2} \dfrac{4-x^{2}}{3-\sqrt{x^{2}+5}} \times \dfrac{3+\sqrt{x^{2}+5}}{3+\sqrt{x^{2}+5}} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( 4-x^{2} \right) \left( 3+\sqrt{x^{2}+5} \right)}{9-\left( x^{2}+5 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( 4-x^{2} \right) \left( 3+\sqrt{x^{2}+5} \right)}{4- x^{2} } \\ & = \lim\limits_{x \to 2} \dfrac{ \left( 3+\sqrt{x^{2}+5} \right)}{1} \\ & = 3+\sqrt{2^{2}+5} \\ & = 3+\sqrt{9}=6 \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 6$

41. Soal UN Matematika SMA IPA 2019

Nilai $\lim\limits_{x \to 3} \dfrac{x^{2}-x-6}{\sqrt{3x^{2}-2}-5}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{25}{9} \\ (C)\ & \dfrac{25}{6} \\ (D)\ & \dfrac{25}{3} \\ (E)\ & \infty \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 3} \dfrac{x^{2}-x-6}{\sqrt{3x^{2}-2}-5} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( x+2 \right)}{\sqrt{3x^{2}-2}-5} \cdot \dfrac{\sqrt{3x^{2}-2}+5}{\sqrt{3x^{2}-2}+5}\\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( x+2 \right)\left( \sqrt{3x^{2}-2}+5 \right)}{3x^{2}-2 -25} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( x+2 \right)\left( \sqrt{3x^{2}-2}+5 \right)}{3 \left( x^{2}-9 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)\left( x+2 \right)\left( \sqrt{3x^{2}-2}+5 \right)}{3 \left( x-3 \right)\left( x+3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x+2 \right)\left( \sqrt{3x^{2}-2}+5 \right)}{3 \left( x+3 \right)} \\ & = \dfrac{\left( 3+2 \right)\left( \sqrt{3(3)^{2}-2}+5 \right)}{3 \left( 3+3 \right)} \\ & = \dfrac{\left( 5 \right)\left( \sqrt{27-2}+5 \right)}{18} \\ & = \dfrac{\left( 5 \right)\left( 10 \right)}{18} \\ & = \dfrac{25}{9} \end{align}$

$\therefore$ Jawaban yang benar $(B)\ \dfrac{25}{9}$

42. Soal UN Matematika SMA IPA 2006

Nilai $\lim\limits_{x \to 6} \dfrac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6}=\cdots$

$\begin{align} (A)\ & -\dfrac{1}{4} \\ (B)\ & -\dfrac{1}{8} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{8} \\ (E)\ & \dfrac{1}{4} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 6} \dfrac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6} \\ & = \lim\limits_{x \to 6} \dfrac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6} \cdot \dfrac{\sqrt{3x-2}+\sqrt{2x+4}}{\sqrt{3x-2}+\sqrt{2x+4}} \\ & = \lim\limits_{x \to 6} \dfrac{\left( 3x-2 \right)-\left( 2x+4 \right)}{\left( x-6 \right)\left( \sqrt{3x-2}+\sqrt{2x+4} \right)} \\ & = \lim\limits_{x \to 6} \dfrac{x-6}{\left( x-6 \right)\left( \sqrt{3x-2}+\sqrt{2x+4} \right)} \\ & = \lim\limits_{x \to 6} \dfrac{1}{\left(\sqrt{3x-2}+\sqrt{2x+4} \right)} \\ & = \dfrac{1}{\left(\sqrt{3(6)-2}+\sqrt{2(6)+4} \right)} \\ & = \dfrac{1}{\left(\sqrt{16}+\sqrt{16} \right)} = \dfrac{1}{8} \end{align}$

$\therefore$ Jawaban yang benar $(D)\ \dfrac{1}{8}$

43. Soal EBTANAS Matematika SMA IPA 2000

$ \lim\limits_{x \to 0} \dfrac{x^{2}}{1-\sqrt{1+x^{2}}}=\cdots$

$\begin{align} (A)\ & 2 \\ (B)\ & 0 \\ (C)\ & -1 \\ (D)\ & -2 \\ (E)\ & -3 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 0} \dfrac{x^{2}}{1-\sqrt{1+x^{2}}} \\ & = \lim\limits_{x \to 0} \dfrac{x^{2}}{1-\sqrt{1+x^{2}}} \times \dfrac{1+\sqrt{1+x^{2}}}{1+\sqrt{1+x^{2}}} \\ & = \lim\limits_{x \to 0} \dfrac{x^{2} \left( 1+\sqrt{1+x^{2}} \right)}{1-\left( 1+x^{2} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{x^{2} \left( 1+\sqrt{1+x^{2}} \right)}{ x^{2} } \\ & = \lim\limits_{x \to 0} \dfrac{ \left( 1+\sqrt{1+x^{2}} \right)}{ 1 } \\ & = \dfrac{ \left( 1+\sqrt{1+0} \right)}{ 1 } = 2 \end{align}$

$\therefore$ Jawaban yang benar $(A)\ 2$

44. Soal EBATANAS Matematika SMA IPA 1995

Nilai $\lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-\sqrt{3x-2}}{x-2}=\cdots$

$\begin{align} (A)\ & 2 \\ (B)\ & 1 \\ (C)\ & \dfrac{1}{2} \\ (D)\ & 0 \\ (E)\ & -\dfrac{1}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-\sqrt{3x-2}}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{\sqrt{x+2}-\sqrt{3x-2}}{x-2} \cdot \dfrac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{x+2}+\sqrt{3x-2}} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x+2\right)-\left(3x-2\right)}{\left(x-2 \right)\left( \sqrt{x+2}+\sqrt{3x-2} \right)} \\ & = \lim\limits_{x \to 2} \dfrac{4-2x}{\left(x-2 \right)\left( \sqrt{x+2}+\sqrt{3x-2} \right)} \\ & = \lim\limits_{x \to 2} \dfrac{-2\left(x-2 \right)}{\left(x-2 \right)\left( \sqrt{x+2}+\sqrt{3x-2} \right)} \\ & = \lim\limits_{x \to 2} \dfrac{-2 }{\left( \sqrt{x+2}+\sqrt{3x-2} \right)} \\ & = \dfrac{-2 }{\left( \sqrt{2+2}+\sqrt{3(2)-2} \right)} \\ & = \dfrac{-2}{2+2}= -\dfrac{1}{2} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ -\dfrac{1}{2}$

45. Soal UMB PTN 2014 Kode 672

$ \lim\limits_{x \to 2} \dfrac{3-\sqrt{2x+5}}{x+1-\sqrt{2x+5}}=\cdots$

$\begin{align} (A)\ & -2 \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{2} \\ (E)\ & 2 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{3-\sqrt{2x+5}}{x+1-\sqrt{2x+5}} \\ & = \lim\limits_{x \to 2} \dfrac{3-\sqrt{2x+5}}{x+1-\sqrt{2x+5}} \times \dfrac{x+1+\sqrt{2x+5}}{x+1+\sqrt{2x+5}} \\ & = \lim\limits_{x \to 2} \dfrac{\left (3-\sqrt{2x+5} \right )\left (x+1+\sqrt{2x+5} \right )}{(x+1)^{2}-(2x+5)} \\ & = \lim\limits_{x \to 2} \dfrac{\left (3-\sqrt{2x+5} \right )\left (x+1+\sqrt{2x+5} \right )}{ x^{2}-2x+1- 2x-5 } \\ & = \lim\limits_{x \to 2} \dfrac{\left (3-\sqrt{2x+5} \right )\left (x+1+\sqrt{2x+5} \right )}{ x^{2}-4 } \times \dfrac{\left (3+\sqrt{2x+5} \right ) }{ \left (3+\sqrt{2x+5} \right ) } \\ & = \lim\limits_{x \to 2} \dfrac{\left (9- 2x-5 \right )\left (x+1+\sqrt{2x+5} \right )}{ (x-2)(x+2)\left (3+\sqrt{2x+5} \right ) } \\ & = \lim\limits_{x \to 2} \dfrac{-2\left (x-2 \right )\left (x+1+\sqrt{2x+5} \right )}{ (x-2)(x+2)\left (3+\sqrt{2x+5} \right ) } \\ & = \lim\limits_{x \to 2} \dfrac{-2 \left (x+1+\sqrt{2x+5} \right )}{ (x+2)\left (3+\sqrt{2x+5} \right ) } \\ & = \dfrac{-2 \left( 2+1+\sqrt{2(2)+5 } \right )}{ (2+2)\left (3+\sqrt{2(2)+5} \right ) } \\ & = \dfrac{-2 \left( 3+3 \right)}{(2+2) \left( 3+3 \right)}=-\dfrac{1}{2} \end{align}$

$\therefore$ Jawaban yang benar $(B)\ -\dfrac{1}{2}$

46. Soal UM UNDIP 2009 Kode 192 – Soal SPM UNNES 2009 Kode 9763

Jika $f(x)=\dfrac{x-\sqrt{x}}{x+\sqrt{x}}$, maka $\lim\limits_{x \to 0} f(x)=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & -1 \\ (D)\ & -2 \\ (E)\ & \infty \end{align}$

Pembahasan:

Karena $f(x)=\dfrac{x-\sqrt{x}}{x+\sqrt{x}}$ maka:

$\begin{align} \lim\limits_{x \to 0} f(x) & = \lim\limits_{x \to 0} \dfrac{x-\sqrt{x}}{x+\sqrt{x}} \\ & = \lim\limits_{x \to 0} \dfrac{x-\sqrt{x}}{x+\sqrt{x}} \cdot \dfrac{\dfrac{1}{\sqrt{x}}}{\dfrac{1}{\sqrt{x}}} \\ & = \lim\limits_{x \to 0} \dfrac{\dfrac{x}{\sqrt{x}}-1}{\dfrac{x}{\sqrt{x}}+1} \\ & = \lim\limits_{x \to 0} \dfrac{\sqrt{x}-1}{\sqrt{x}+1} \\ & = \dfrac{\sqrt{0}-1}{\sqrt{0}+1} \\ & = \dfrac{-1}{1}=-1 \end{align}$

$\therefore$ Jawaban yang benar $(C)\ -1$

47. Soal UMB PTN 2013 Kode 172

$\lim\limits_{x \to 2} \dfrac{x^{2}-2x}{\sqrt{x^{2}+5}-3}=\cdots $

$\begin{align} (A)\ & -6 \\ (B)\ & -3 \\ (C)\ & 0 \\ (D)\ & 3 \\ (E)\ & 6 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-2x}{\sqrt{x^{2}+5}-3} \\ & = \lim\limits_{x \to 2} \dfrac{x^{2}-2x}{\sqrt{x^{2}+5}-3} \times \dfrac{\sqrt{x^{2}+5}+3}{\sqrt{x^{2}+5}+3} \\ & = \lim\limits_{x \to 2} \dfrac{x(x-2) \left( \sqrt{x^{2}+5}+3 \right)}{ x^{2}+5-9} \\ & = \lim\limits_{x \to 2} \dfrac{x(x-2) \left( \sqrt{x^{2}+5}+3 \right)}{ (x-2)(x+2)} \\ & = \lim\limits_{x \to 2} \dfrac{x \left( \sqrt{x^{2}+5}+3 \right)}{ (x+2)} \\ & = \dfrac{2 \left( \sqrt{2^{2}+5}+3 \right)}{ (2+2)} \\ & = \dfrac{2 \left( 3+3 \right)}{ 4}= 3 \end{align}$

$\therefore$ Jawaban yang benar $(D)\ 3$

48. Soal UM UGM 2013 Kode 251

Jika $a=\lim\limits_{x \to 2} \dfrac{x^{2}-4}{2-\sqrt{x+2}}$ maka nilai $4-a$ adalah…

$\begin{align} (A)\ & -20 \\ (B)\ & -12 \\ (C)\ & -4 \\ (D)\ & 12 \\ (E)\ & 20 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} a & = \lim\limits_{x \to 2} \dfrac{x^{2}-4}{2-\sqrt{x+2}} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x+2)}{2-\sqrt{x+2}} \times \dfrac{2+\sqrt{x+2}}{2+\sqrt{x+2}} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x+2) \left( 2+\sqrt{x+2} \right)}{4-(x+2)} \\ & = \lim\limits_{x \to 2} \dfrac{(x-2)(x+2) \left( 2+\sqrt{x+2} \right)}{-(x-2)} \\ & = \lim\limits_{x \to 2} \dfrac{ (x+2) \left( 2+\sqrt{x+2} \right)}{-1} \\ & = \dfrac{ (2+2) \left( 2+\sqrt{2+2} \right)}{-1} \\ & = -16 \\ 4-a & = 4-(-16)=20 \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 20$

49. Soal SPMB 2004

$\lim\limits_{x \to 2} \dfrac{x\sqrt{x}-2\sqrt{x}-2\sqrt{2}+x\sqrt{2}}{\sqrt{x} – \sqrt{2}} =\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 2 \\ (C)\ & 4 \\ (D)\ & 8 \\ (E)\ & 10 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 2} \dfrac{x\sqrt{x}-2\sqrt{x}-2\sqrt{2}+x\sqrt{2}}{\sqrt{x} – \sqrt{2}} \\ & = \lim\limits_{x \to 2} \dfrac{\left( \sqrt{x} + \sqrt{2} \right)\left( x-2 \right)}{\sqrt{x} – \sqrt{2}} \\ & = \lim\limits_{x \to 2} \dfrac{\left( \sqrt{x} + \sqrt{2} \right)\left( x-2 \right)}{\sqrt{x} – \sqrt{2}} \cdot \dfrac{\left( \sqrt{x} + \sqrt{2} \right)}{\left( \sqrt{x} + \sqrt{2} \right)}\\ & = \lim\limits_{x \to 2} \dfrac{\left( \sqrt{x} + \sqrt{2} \right)\left( x-2 \right)}{x-2}\\ & = \lim\limits_{x \to 2} \dfrac{\left( \sqrt{x} + \sqrt{2} \right)^{2}}{1} \\ & = \left( 2\sqrt{2} \right)^{2} = 8 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 8$

50. Soal SPMB 2005

$\lim\limits_{x \to 0} \dfrac{5x^{2}+x}{\sqrt{4+x} – 2} =\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 1 \\ (C)\ & 2 \\ (D)\ & 4 \\ (E)\ & 6 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 0} \dfrac{5x^{2}+x}{\sqrt{4+x} – 2} \cdot \dfrac{\sqrt{4+x} – 2}{\sqrt{4+x} + 2} \\ & = \lim\limits_{x \to 0} \dfrac{x \left(5x +1 \right) \left( \sqrt{4+x} + 2 \right)}{ 4+x – 4} \\ & = \lim\limits_{x \to 0} \dfrac{x \left(5x +1 \right) \left( \sqrt{4+x} + 2 \right)}{ x } \\ & = \lim\limits_{x \to 0} \dfrac{ \left(5x +1 \right) \left( \sqrt{4+x} + 2 \right)}{ 1 } \\ & = \dfrac{ \left(5(0) +1 \right) \left( \sqrt{4+0} + 2 \right)}{ 1 } \\ & = \left( 1 \right)\left( 2 \right) = 2 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 2$

51. Soal SPMB 2005 Kode 470

$\lim\limits_{x \to 2} \dfrac{4-x^{2}}{3-\sqrt{x^{2}+5}} =\cdots$

$\begin{align} (A)\ & -1 \\ (B)\ & 0 \\ (C)\ & 2 \\ (D)\ & 6 \\ (E)\ & 8 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{4-x^{2}}{3-\sqrt{x^{2}+5}} \cdot \dfrac{3+\sqrt{x^{2}+5}}{3+\sqrt{x^{2}+5}} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 4-x^{2} \right) \left( 3+\sqrt{x^{2}+5} \right)}{ 9- \left( x^{2}+5 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 4-x^{2} \right) \left( 3+\sqrt{x^{2}+5} \right)}{ 4- x^{2} } \\ & = \lim\limits_{x \to 2} \dfrac{ \left( 3+\sqrt{x^{2}+5} \right)}{1} \\ & = 3+\sqrt{(2)^{2}+5} = 3+\sqrt{9} = 6 \end{align}$

$ \therefore $ Jawaban yang benar adalah $(C)\ 6$

52. Soal SIMAK UI 2010 Kode 506

Untuk $t \gt 0$ maka $\lim\limits_{t \to 0} \left( \dfrac{1}{t}+\dfrac{1}{\sqrt{t}} \right)\left( \sqrt{t+1}-1 \right)=\cdots$

$\begin{align} (A)\ & -\infty \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{2} \\ (E)\ & \infty \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{t \to 0} \left( \dfrac{1}{t}+\dfrac{1}{\sqrt{t}} \right)\left( \sqrt{t+1}-1 \right) \\ & = \lim\limits_{t \to 0} \left( \dfrac{1}{t}+\dfrac{\sqrt{t}}{t} \right)\left( \sqrt{t+1}-1 \right) \times \dfrac{\sqrt{t+1}+1}{\sqrt{t+1}+1} \\ & = \lim\limits_{t \to 0} \left( \dfrac{\sqrt{t}+1}{t} \right)\left( \dfrac{t+1-1}{\sqrt{t+1}+1} \right) \\ & = \lim\limits_{t \to 0} \left( \dfrac{\sqrt{t}+1}{t} \right)\left( \dfrac{t}{\sqrt{t+1}+1} \right) \\ & = \lim\limits_{t \to 0} \left( \dfrac{\sqrt{t}+1}{\sqrt{t+1}+1} \right) \\ & = \dfrac{\sqrt{0}+1}{\sqrt{0+1}+1} \\ & = \dfrac{1}{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ \dfrac{1}{2}$

53. Soal UM UGM 2004

$\lim\limits_{x \to 3} \dfrac{9-x^{2}}{4-\sqrt{x^{2}+7}} =\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 1 \\ (C)\ & 6 \\ (D)\ & 8 \\ (E)\ & \infty \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 3} \dfrac{9-x^{2}}{4-\sqrt{x^{2}+7}} \cdot \dfrac{4+\sqrt{x^{2}+7}}{4+\sqrt{x^{2}+7}} \\ & = \lim\limits_{x \to 3} \dfrac{\left( 9-x^{2} \right) \left( 4+\sqrt{x^{2}+7} \right) }{ 16- \left( x^{2}+7 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{\left( 9-x^{2} \right) \left( 4+\sqrt{x^{2}+7} \right)}{ 9- x^{2} } \\ & = \lim\limits_{x \to 3} \dfrac{ \left( 4+\sqrt{x^{2}+7} \right)}{1} \\ & = 4+\sqrt{(3)^{2}+7} = 4+\sqrt{16} = 8 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 8$

54. Soal SPMB 2005 Kode 370

$\lim\limits_{x \to q} \dfrac{x\sqrt{x}-q\sqrt{q}}{\sqrt{x}-\sqrt{q}} =\cdots$

$\begin{align} (A)\ & 3\sqrt{q} \\ (B)\ & \sqrt{q} \\ (C)\ & q \\ (D)\ & q\sqrt{q} \\ (E)\ & 3q \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to q} \dfrac{x\sqrt{x}-q\sqrt{q}}{\sqrt{x}-\sqrt{q}} \cdot \dfrac{\sqrt{x}+ \sqrt{q}}{\sqrt{x}+\sqrt{q}} \\ & = \lim\limits_{x \to q} \dfrac{x^{2}+x\sqrt{qx}-q\sqrt{qx}-q^{2}}{x-q} \\ &= \lim\limits_{x \to q} \dfrac{\left ( x-q \right )\left ( x+q \right )+\sqrt{qx}\left ( x-q \right )}{x-q} \\ &= \lim\limits_{x \to q} \dfrac{ \left ( x+q \right )+\sqrt{qx} }{1} \\ &= \dfrac{ \left ( q+q \right )+\sqrt{q(q)} }{1} \\ &= 2q+ q =3q \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ 3q$

55. Soal UM UGM 2006 Kode 372

Jika $f(x)= \dfrac{1-x}{2-\sqrt{x^{2}+3}}$, maka $\lim\limits_{x \to 1}\ f(x)=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{2} \\ (C)\ & 1 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} \lim\limits_{x \to 1}\ f(x) & = \lim\limits_{x \to 1} \dfrac{1-x}{2-\sqrt{x^{2}+3}} \cdot \dfrac{2+\sqrt{x^{2}+3}}{2+\sqrt{x^{2}+3}} \\ & = \lim\limits_{x \to 1} \dfrac{\left( 1-x \right) \left( 2+\sqrt{x^{2}+3} \right)}{ 4- \left( x^{2}+3 \right) } \\ & = \lim\limits_{x \to 1} \dfrac{\left( 1-x \right) \left( 2+\sqrt{x^{2}+3} \right)}{ 1-x^{2} } \\ & = \lim\limits_{x \to 1} \dfrac{\left( 1-x \right) \left( 2+\sqrt{x^{2}+3} \right)}{ \left( 1-x \right)\left( 1+x \right) } \\ & = \lim\limits_{x \to 1} \dfrac{ 2+\sqrt{x^{2}+3} }{ 1+x } \\ & = \dfrac{ 2+\sqrt{(1)^{2}+3} }{ 1+(1) } \\ & = \dfrac{ 2+2}{2}=2 \\ \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 2$

56. Soal SPMB 2006 Kode 411

$\lim\limits_{x \to 7} \dfrac{\sqrt{x} \left(x-7 \right)}{\sqrt{x}-\sqrt{7}} =\cdots$

$\begin{align} (A)\ & 14 \\ (B)\ & 7 \\ (C)\ & 2\sqrt{q} \\ (D)\ & \sqrt{7} \\ (E)\ & \dfrac{1}{2} \sqrt{7} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 7} \dfrac{\sqrt{x} \left(x-7 \right)}{\sqrt{x}-\sqrt{7}} \cdot \dfrac{\sqrt{x}+\sqrt{7}}{\sqrt{x}+\sqrt{7}} \\ & = \lim\limits_{x \to 7} \dfrac{\sqrt{x} \left(x-7 \right) \left( \sqrt{x}+\sqrt{7} \right)}{x-7} \\ & = \lim\limits_{x \to 7} \dfrac{\sqrt{x} \left( \sqrt{x}+\sqrt{7} \right)}{1} \\ & = \dfrac{\sqrt{7} \left( \sqrt{7}+\sqrt{7} \right)}{1} \\ &= \sqrt{7} \left( 2\sqrt{7} \right) = 14 \end{align} $

$\therefore$ Jawaban yang benar adalah $(A)\ 14$

57. Soal SPMB 2007 Kode 341

$\lim\limits_{x \to 4} \dfrac{x^{2}-16}{5-\sqrt{x^{2}+9}}=\cdot$

$\begin{align} (A)\ & -20 \\ (B)\ & -10 \\ (C)\ & 0 \\ (D)\ & 8 \\ (E)\ & 20 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 4} \dfrac{x^{2}-16}{5-\sqrt{x^{2}+9}} \\ & = \lim\limits_{x \to 4} \dfrac{x^{2}-16}{5-\sqrt{x^{2}+9}} \cdot \dfrac{5+\sqrt{x^{2}+9}}{5+\sqrt{x^{2}+9}} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x^{2}-16 \right) \left( 5+\sqrt{x^{2}+9} \right)}{ 25- \left( x^{2} +9 \right) } \\ & = \lim\limits_{x \to 4} \dfrac{\left( x^{2}-16 \right) \left( 5+\sqrt{x^{2}+9} \right)}{ 16- x^{2}} \\ & = \lim\limits_{x \to 4} \dfrac{ – \left( 5+\sqrt{x^{2}+9} \right)}{ 1 } \\ & = \dfrac{ – \left( 5+\sqrt{(4)^{2}+9} \right)}{ 1 } \\ & = – \left( 5+\sqrt{25} \right) = -10 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ -10$

58. Soal SPMB 2007 Kode 141

$\lim\limits_{x \to 5} \dfrac{x^{2}-25}{\sqrt{x^{2}+24}-7}=\cdot$

$\begin{align} (A)\ & 0 \\ (B)\ & 5 \\ (C)\ & 7 \\ (D)\ & 14 \\ (E)\ & 18 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 5} \dfrac{x^{2}-25}{\sqrt{x^{2}+24}-7} \\ & = \lim\limits_{x \to 5} \dfrac{x^{2}-25}{\sqrt{x^{2}+24}-7} \cdot \dfrac{\sqrt{x^{2}+24}+7}{\sqrt{x^{2}+24}+7} \\ & = \lim\limits_{x \to 5} \dfrac{\left( x^{2}-25 \right) \left( \sqrt{x^{2}+24}+7 \right)}{ x^{2}+24 -49 } \\ & = \lim\limits_{x \to 5} \dfrac{\left( x^{2}-25 \right) \left( \sqrt{x^{2}+24}+7 \right)}{ x^{2} – 25 } \\ & = \lim\limits_{x \to 5} \dfrac{ \sqrt{x^{2}+24}+7 }{ 1 } \\ & = \sqrt{(5)^{2}+24}+7 = \sqrt{25+24}+7 = 14 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 14$

59. Soal SPMB 2007 Kode 541

$\lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x}+1 \right)}{\sqrt{x}+1}=\cdot$

$\begin{align} (A)\ & 0 \\ (B)\ & 1 \\ (C)\ & 2 \\ (D)\ & 4 \\ (E)\ & 8 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x}+1 \right)}{\sqrt{x}-1} \\ & = \lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x}+1 \right)}{\sqrt{x}-1} \cdot \dfrac{\sqrt{x}+1}{\sqrt{x}+1} \\ & = \lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x}+1 \right)^{2}}{ x-1 } \\ & = \lim\limits_{x \to 1} \dfrac{\left( \sqrt{x}+1 \right)^{2}}{ 1 } \\ & = \left(\sqrt{1}+1 \right)^{2} = 2^{2} = 4 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 4$

60. Soal SPMB 2007 Kode 441

$\lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{x+3}-2}=\cdot$

$\begin{align} (A)\ & \dfrac{1}{4} \\ (B)\ & \dfrac{1}{2} \\ (C)\ & 1 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{x+3}-2} \\ & = \lim\limits_{x \to 1} \dfrac{x-1}{\sqrt{x+3}-2} \cdot \dfrac{\sqrt{x+3}+2}{\sqrt{x+3}+2} \\ & = \lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x+3}+2 \right) }{ x+3-4 } \\ & = \lim\limits_{x \to 1} \dfrac{\left( x-1 \right) \left( \sqrt{x+3}+2 \right) }{ x-1 } \\ & = \lim\limits_{x \to 1} \dfrac{\sqrt{x+3}+2}{ 1 } \\ & = \sqrt{1+3}+2 = 2+2 = 4 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ 4$

61. Soal SPMB 2007 Kode 741

$\lim\limits_{x \to 1} \dfrac{\sqrt{2-x}-x}{x^{2}-x}=\cdot$

$\begin{align} (A)\ & -1\dfrac{1}{2} \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 1\dfrac{1}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{\sqrt{2-x}-x}{x^{2}-x} \\ & = \lim\limits_{x \to 1} \dfrac{\sqrt{2-x}-x}{x^{2}-x} \cdot \dfrac{\sqrt{2-x}+x}{\sqrt{2-x}+x} \\ & = \lim\limits_{x \to 1} \dfrac{2-x+x^{2} }{ \left( x^{2}-x \right) \left( \sqrt{2-x}+x \right) } \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x+2 \right) \left( x-1 \right) }{x \left( x -1 \right) \left( \sqrt{2-x}+x \right) } \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x+2 \right) }{x \left( \sqrt{2-x}+x \right) } \\ & = \dfrac{ \left( 1+2 \right) }{1 \left( \sqrt{2-1}+1 \right) } = \dfrac{ 3 }{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ 1\dfrac{1}{2}$

62. Soal SPMB 2007 Kode 641

$\lim\limits_{x \to 1} \dfrac{1-\sqrt{3x-2}}{1-x}=\cdot$

$\begin{align} (A)\ & -1\dfrac{1}{2} \\ (B)\ & -1 \\ (C)\ & 0 \\ (D)\ & 1 \\ (E)\ & 1\dfrac{1}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{1-\sqrt{3x-2}}{1-x} \\ & = \lim\limits_{x \to 1} \dfrac{1-\sqrt{3x-2}}{1-x} \cdot \dfrac{1+\sqrt{3x-2}}{1+\sqrt{3x-2}} \\ & = \lim\limits_{x \to 1} \dfrac{1-\left( 3x-2 \right) }{ \left( 1-x \right)\left( 1+\sqrt{3x-2} \right) } \\ & = \lim\limits_{x \to 1} \dfrac{1-3x+2 }{ \left( 1-x \right)\left( 1+\sqrt{3x-2} \right) } \\ & = \lim\limits_{x \to 1} \dfrac{3-3x }{ \left( 1-x \right)\left( 1+\sqrt{3x-2} \right) } \\ & = \lim\limits_{x \to 1} \dfrac{3\left( 1-x \right) }{ \left( 1-x \right)\left( 1+\sqrt{3x-2} \right) } \\ & = \lim\limits_{x \to 1} \dfrac{3 }{ \left( 1+\sqrt{3x-2} \right) } \\ & = \dfrac{3 }{ \left( 1+\sqrt{3(1)-2} \right) } \\ & = \dfrac{3 }{ \left( 1+\sqrt{1} \right) } = \dfrac{3}{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ 1\dfrac{1}{2}$

63. Soal SNMPTN 2008 Kode 111

$\lim\limits_{x \to 2} \dfrac{\sqrt{3x-2}-2}{2x-4}=\cdot$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{3}{8} \\ (C)\ & \dfrac{3}{4} \\ (D)\ & 1 \\ (E)\ & 1\dfrac{1}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 2} \dfrac{\sqrt{3x-2}-2}{2x-4} \\ & = \lim\limits_{x \to 2} \dfrac{\sqrt{3x-2}-2}{2x-4} \cdot \dfrac{\sqrt{3x-2}+2}{\sqrt{3x-2}+2} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( 3x-2 \right)-4 }{ \left( 2x-4 \right)\left( \sqrt{3x-2}+2 \right) } \\ & = \lim\limits_{x \to 2} \dfrac{ 3x-6 }{ \left( 2x-4 \right)\left( \sqrt{3x-2}+2 \right) } \\ & = \lim\limits_{x \to 2} \dfrac{ 3(x-2) }{ 2\left( x-2 \right)\left( \sqrt{3x-2}+2 \right) } \\ & = \lim\limits_{x \to 2} \dfrac{ 3 }{ 2 \left( \sqrt{3x-2}+2 \right) } \\ & = \dfrac{ 3 }{ 2 \left( \sqrt{3(2)-2}+2 \right) } \\ & = \dfrac{ 3 }{ 2 \left( \sqrt{4}+2 \right)} = \dfrac{3}{8} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ \dfrac{3}{8}$

64. Soal SNMPTN 2008 Kode 201

$\lim\limits_{x \to 1} \dfrac{3x+x\sqrt{x}-4}{\sqrt{x}-1}=\cdot$

$\begin{align} (A)\ & 6 \\ (B)\ & 7 \\ (C)\ & 8 \\ (D)\ & 9 \\ (E)\ & 10 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{3x+x\sqrt{x}-4}{\sqrt{x}-1} \\ & = \lim\limits_{x \to 1} \dfrac{3x+x\sqrt{x}-4}{\sqrt{x}-1} \cdot \dfrac{\sqrt{x}+1}{\sqrt{x}+1} \\ & = \lim\limits_{x \to 1} \dfrac{ 3x\sqrt{x}+3x+x^{2}+x\sqrt{x}-4\sqrt{x}-4 }{x-1} \\ & = \lim\limits_{x \to 1} \dfrac{ 4x \sqrt{x} – 4\sqrt{x}+x^{2}+3x-4 }{x-1} \\ & = \lim\limits_{x \to 1} \dfrac{ 4\sqrt{x} \left( x-1 \right)+\left( x+4 \right)\left( x-1 \right) }{x-1} \\ & = \lim\limits_{x \to 1} \dfrac{ 4\sqrt{x} +\left( x+4 \right) }{1} \\ & = \dfrac{ 4\sqrt{1} +\left( 1+4 \right) }{1} = \dfrac{ 4 +\left( 5 \right) }{1} = 9 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 9$

65. Soal UM UGM 2008 Kode 482

$\lim\limits_{x \to p} \dfrac{x\sqrt{x}-p\sqrt{p}}{\sqrt{x}-\sqrt{p}} =\cdots$

$\begin{align} (A)\ & 3\sqrt{p} \\ (B)\ & \sqrt{p} \\ (C)\ & p \\ (D)\ & p\sqrt{p} \\ (E)\ & 3p \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to p} \dfrac{x\sqrt{x}-p \sqrt{p}}{\sqrt{x}-\sqrt{p}} \cdot \dfrac{\sqrt{x}+ \sqrt{p}}{\sqrt{x}+\sqrt{p}} \\ & = \lim\limits_{x \to p} \dfrac{x^{2}+x\sqrt{px}-p\sqrt{px}-p^{2}}{x-p} \\ &= \lim\limits_{x \to p} \dfrac{\left ( x-p \right )\left ( x+p \right )+\sqrt{px}\left ( x-p \right )}{x-p} \\ &= \lim\limits_{x \to p} \dfrac{ \left ( x+p \right )+\sqrt{px} }{1} \\ &= \dfrac{ \left ( p+p \right )+\sqrt{p(p)} }{1} \\ &= 2p+ p =3p \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ 3p$

66. Soal UM UGM 2007 Kode 741

$\lim\limits_{x \to 2} \dfrac{\sqrt{x^{2}+5}-3}{x^{2}-2x}=\cdot$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{3} \\ (C)\ & \dfrac{1}{2} \\ (D)\ & \dfrac{3}{4} \\ (E)\ & \infty \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 2} \dfrac{\sqrt{x^{2}+5}-3}{x^{2}-2x} \\ & = \lim\limits_{x \to 2} \dfrac{\sqrt{x^{2}+5}-3}{x\left(x-2 \right)} \cdot \dfrac{\sqrt{x^{2}+5}+3}{\sqrt{x^{2}+5}+3} \\ & = \lim\limits_{x \to 2} \dfrac{x^{2}+5-9}{x\left(x-2 \right)\left( \sqrt{x^{2}+5}+3 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{x^{2}-4}{x\left(x-2 \right)\left( \sqrt{x^{2}+5}+3 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x-2 \right)\left(x + 2 \right)}{x\left(x-2 \right)\left( \sqrt{x^{2}+5}+3 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x+2 \right)}{x \left( \sqrt{x^{2}+5}+3 \right)} \\ & = \dfrac{ \left( 2+2 \right)}{2 \left( \sqrt{(2)^{2}+5}+3 \right)} \\ & = \dfrac{ 4 }{2 \left( \sqrt{9}+3 \right)} = \dfrac{1}{3} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ \dfrac{1}{3}$

67. Soal UMB PTN 2012 Kode 270

$ \lim\limits_{x \to 3} \dfrac{x-3}{3-\sqrt{x+6}}=\cdots$

$\begin{align} (A)\ & -6 \\ (B)\ & -3 \\ (C)\ & 0 \\ (D)\ & 3 \\ (E)\ & 6 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 3} \dfrac{x-3}{3-\sqrt{x+6}} \\ & = \lim\limits_{x \to 3} \dfrac{x-3}{3-\sqrt{x+6}} \cdot \dfrac{3+\sqrt{x+6}}{3+\sqrt{x+6}} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x-3 \right) \left( 3+\sqrt{x+6} \right)}{9- \left(x+6 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x-3 \right) \left( 3+\sqrt{x+6} \right)}{3-x } \\ & = \lim\limits_{x \to 3} \dfrac{- \left( 3-x \right) \left( 3+\sqrt{x+6} \right)}{3-x } \\ & = \lim\limits_{x \to 3} \dfrac{- \left( 3+\sqrt{x+6} \right)}{1 } \\ & = \dfrac{- \left( 3+\sqrt{3+6} \right)}{1 } = -6 \end{align}$

$\therefore$ Jawaban yang benar $(A)\ -6$

68. Soal UM UNDIP 2011 Kode 111

Nilai $\lim\limits_{x \to 0} \dfrac{3-\sqrt{3x^{2}+2x+9}}{2x}=\cdots$

$\begin{align} (A)\ & -\dfrac{1}{6} \\ (B)\ & -\dfrac{1}{3} \\ (C)\ & \dfrac{1}{6} \\ (D)\ & \dfrac{1}{3} \\ (E)\ & \dfrac{2}{3} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 0} \dfrac{3-\sqrt{3x^{2}+2x+9}}{2x} \\ & = \lim\limits_{x \to 0} \dfrac{3-\sqrt{3x^{2}+2x+9}}{2x} \cdot \dfrac{3+\sqrt{3x^{2}+2x+9}}{3+\sqrt{3x^{2}+2x+9}} \\ & = \lim\limits_{x \to 0} \dfrac{\left( 3-\sqrt{3x^{2}+2x+9} \right) \left( 3+\sqrt{3x^{2}+2x+9} \right)}{ \left(2x \right) \left( 3+\sqrt{3x^{2}+2x+9} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{9- \left( 3x^{2}+2x+9 \right)}{\left( 2x \right) \left( 3+\sqrt{3x^{2}+2x+9} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{-\left(3x^{2}+2x \right)}{\left(2x \right) \left( 3+\sqrt{3x^{2}+2x+9} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{-\left(2x \right) \left(\dfrac{3}{2}x + 1 \right)}{\left(2x \right) \left( 3+\sqrt{3x^{2}+2x+9} \right)} \\ & = \lim\limits_{x \to 0} \dfrac{-\left(\dfrac{3}{2}x + 1 \right)}{\left( 3+\sqrt{3x^{2}+2x+9} \right)} \\ & = \dfrac{-\left(\dfrac{3}{2}(0) + 1 \right)}{\left( 3+\sqrt{3(0)+2(0)+9} \right)} \\ & = \dfrac{-1}{3+\sqrt{9}} = – \dfrac{1}{6} \end{align}$

$\therefore$ Jawaban yang benar $(A)\ -\dfrac{1}{6}$

69. Soal UM UPI 2009

Nilai dari $ \lim\limits_{x \to 2}\ \dfrac{\sqrt{8x}-4}{\sqrt{x+2}-2}$ adalah$ \cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{4} \\ (C)\ & 1 \\ (D)\ & 4 \\ (E)\ & 4 \dfrac{1}{4} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$\begin{align} & \lim\limits_{x \to 2} \dfrac{\sqrt{8x}-4}{\sqrt{x+2}-2} \\ & = \lim\limits_{x \to 2} \dfrac{\sqrt{8x}-4}{\sqrt{x+2}-2} \cdot \dfrac{ \left(\sqrt{8x}+4 \right)+\left(\sqrt{x+2}+2 \right)}{\left(\sqrt{8x}+4 \right)\left(\sqrt{x+2}+2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left(8x-16 \right)\left(\sqrt{x+2}+2 \right)}{\left(\sqrt{8x}+4 \right)\left(x+2-4 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ 8 \left(x-2 \right)\left(\sqrt{x+2}+2 \right)}{\left(\sqrt{8x}+4 \right)\left(x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ 8 \left(\sqrt{x+2}+2 \right)}{\left(\sqrt{8x}+4 \right)} \\ & = \dfrac{ 8 \left(\sqrt{2+2}+2 \right)}{\left(\sqrt{8(2)}+4 \right)} \\ & = \dfrac{ 8 \left(2+2 \right)}{\left(4+4 \right)}=4 \end{align}$

$\therefore$ Jawaban yang benar $(D)\ 4$

70. Soal SBMPTN 2015 Kode 507

Nilai $\lim\limits_{x \to 1} \dfrac{\left( \sqrt{5-x}-2 \right) \left( \sqrt{2-x}+1 \right)}{1-x}$ adalah$\cdots$

$\begin{align} (A)\ & -\dfrac{1}{2} \\ (B)\ & -\dfrac{1}{4} \\ (C)\ & -\dfrac{1}{8} \\ (D)\ & \dfrac{1}{4} \\ (E)\ & \dfrac{1}{2} \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 1} \dfrac{\left( \sqrt{5-x}-2 \right) \left( \sqrt{2-x}+1 \right)}{1-x}$ adalah

$\begin{align} & = \lim\limits_{x \to 1} \dfrac{\left( \sqrt{5-x}-2 \right) \left( \sqrt{2-x}+1 \right)}{1-x} \cdot \dfrac{\left( \sqrt{5-x}+2 \right)}{\left( \sqrt{5-x}+2 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{\left( 5-x-4 \right) \left( \sqrt{2+x}+1 \right)}{\left( \sqrt{5-x}+2 \right) \left(1-x \right)} \\ & = \lim\limits_{x \to 1} \dfrac{ \left( \sqrt{2-x}+1 \right)}{\left( \sqrt{5-x}+2 \right)} \\ & = \dfrac{ \sqrt{2-(1)}+1 }{ \sqrt{5-(1)}+2 } \\ & = \dfrac{2}{ \sqrt{4}+2 } \\ & = \dfrac{2}{4} =\dfrac{1}{2} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ \dfrac{1}{2}$

71. Soal Latihan Limit Fungsi Matematika SMA

Jika $\lim\limits_{x \to 4} \dfrac{\sqrt{ax-3}-\sqrt{bx+5}}{x-4}=\dfrac{1}{3}$, maka nilai $a+b=\cdots$

$\begin{align} (A)\ & 1 \\ (B)\ & 2 \\ (C)\ & 3 \\ (D)\ & 4 \\ (E)\ & 5 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 4} \dfrac{\sqrt{ax-3}-\sqrt{bx+5}}{x-4}=\dfrac{1}{3}$. Jika kita substitusi langsung nilai $x=4$ maka nilai $\sqrt{ax-3}-\sqrt{bx+5}$ harus $0$, karena jika $\sqrt{ax-3}-\sqrt{bx+5}$ tidak nol maka nilai limit adalah $\infty$. Untuk $x=4$

$\begin{align} \sqrt{ax-3}-\sqrt{bx+5} & = 0 \\ \sqrt{4a-3}-\sqrt{4b+5} & = 0\\ \sqrt{4a-3} & = \sqrt{4b+5} \\ 4a-3 & = 4b+5 \\ 4a-4b & = 8 \\ a- b & = 2 \end{align}$ Jika soal limit di atas kita kalikan dengan akar sekawan menjadi seperti berikut ini:

$\begin{align} \lim\limits_{x \to 4} \dfrac{\sqrt{ax-3}-\sqrt{bx+5}}{x-4} &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{\sqrt{ax-3}-\sqrt{bx+5}}{x-4} \cdot \dfrac{\sqrt{ax-3}+\sqrt{bx+5}}{\sqrt{ax-3}+\sqrt{bx+5}} &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{\left(ax-3 \right)-\left(bx+5 \right)}{\left( x-4 \right)\left( \sqrt{ax-3}+\sqrt{bx+5} \right) } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ ax-3 – bx-5}{\left( x-4 \right)\left( \sqrt{ax-3}+\sqrt{bx+5} \right) } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ \left( a-b \right)x – 8}{\left( x-4 \right)\left( \sqrt{ax-3}+\sqrt{bx+5} \right) } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ 2x – 8}{\left( x-4 \right)\left( \sqrt{ax-3}+\sqrt{bx+5} \right) } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ 2 \left( x – 4 \right) }{\left( x-4 \right)\left( \sqrt{ax-3}+\sqrt{bx+5} \right) } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ 2 }{ \sqrt{ax-3}+\sqrt{bx+5} } &= \dfrac{1}{3} \\ \lim\limits_{x \to 4} \dfrac{ 2 }{ \sqrt{4a-3}+\sqrt{4b+5} } &= \dfrac{1}{3} \\ \sqrt{4a-3}+\sqrt{4b+5} &= 6 \\ \sqrt{4a-3}+\sqrt{4(a-2) +5} &= 6 \\ \sqrt{4a-3}+\sqrt{4a-8+5} &= 6 \\ \sqrt{4a-3}+\sqrt{4a-3} &= 6 \\ 2\sqrt{4a-3} &= 6 \\ \sqrt{4a-3} &= 3 \\ 4a-3 &= 9 \\ a &= 3 \end{align}$

Untuk $a=3$ kita peroleh $b=a-2=1$ sehingga nilai $a+b=4$

$\therefore$ Jawaban yang benar $(D)\ 4$

72. Soal UM UNPAD 2006

$ \lim\limits_{x \to -6} \dfrac{\sqrt{x+a}-2}{x+6}=b $ maka $2a+4b= \cdots $

$\begin{align} (A)\ & -\dfrac{1}{4} \\ (B)\ & \dfrac{1}{4} \\ (C)\ & 10 \\ (D)\ & 20 \\ (E)\ & 21 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to -6} \dfrac{\sqrt{x+a}-2}{x+6}=b $. Jika kita substitusi langsung nilai $x=-6$ maka nilai $\sqrt{-6+a}-2$ harus $0$, karena jika $\sqrt{-6+a}-2$ tidak nol maka nilai limit adalah $\infty$.

$\begin{align} \sqrt{-6+a}-2 &= 0 \\ \sqrt{-6+a} &= 2 \\ -6+a &= 4 \\ a &= 6+4=10 \end{align}$ Untuk $a=10$ maka $\lim\limits_{x \to -6} \dfrac{\sqrt{x+10}-2}{x+6}=b $

$\begin{align} \lim\limits_{x \to -6} \dfrac{\sqrt{x+10}-2}{x+6} &= b \\ \lim\limits_{x \to -6} \dfrac{\sqrt{x+10}-2}{x+6} \cdot \dfrac{ \left(\sqrt{x+10}+2 \right)}{\left(\sqrt{x+10}+2 \right)} &= b \\ \lim\limits_{x \to -6} \dfrac{x+10-4}{(x+6)\left(\sqrt{x+10}+2 \right)} &= b \\ \lim\limits_{x \to -6} \dfrac{x+6}{(x+6)\left(\sqrt{x+10}+2 \right)} &= b \\ \lim\limits_{x \to -6} \dfrac{1}{\left(\sqrt{x+10}+2 \right)} &= b \\ \dfrac{1}{\left(\sqrt{-6+10}+2 \right)} &= b \\ \dfrac{1}{4} &= b \\ 2a+4b &= 2(10)+4(\dfrac{1}{4}) \\ &= 21 \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 21$

73. Soal UMPTN 1993 (Rayon A,B,C) – Soal SPM UNNES 2008 Kode 140208

Jika $\lim\limits_{x \to 4} \dfrac{ax+b -\sqrt{x}}{x-4}=\dfrac{3}{4}$, maka $a+b$ sama dengan…

$\begin{align} (A)\ & 3 \\ (B)\ & 2 \\ (C)\ & 1 \\ (D)\ & -1 \\ (E)\ & -2 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 4} \dfrac{ax+b -\sqrt{x}}{x-4}=\dfrac{3}{4}$. Jika kita substitusi langsung nilai $x=4$ maka nilai $ax+b -\sqrt{x}$ harus $0$, karena jika $ax+b -\sqrt{x}$ tidak nol maka nilai limit adalah $\infty$. sehingga untuk $x=4$ berlaku:

$\begin{align} ax+b -\sqrt{x} & = 0 \\ 4a+b -\sqrt{4} & = 0 \\ 4a+b & = 2 \end{align}$ Lalu dengan mengalikan akar sekawan maka akan kita peroleh:

$\begin{align} \lim\limits_{x \to 4} \dfrac{ax+b -\sqrt{x}}{x-4} &= \dfrac{3}{4} \\ \lim\limits_{x \to 4} \dfrac{ax+b -\sqrt{x}}{x-4} \times \dfrac{ax+b +\sqrt{x}}{ax+b +\sqrt{x}} &= \dfrac{3}{4} \\ \lim\limits_{x \to 4} \dfrac{a^{2}x^{2}+2abx+b^{2}-x}{\left(x-4 \right)\left( ax+b +\sqrt{x} \right)} &= \dfrac{3}{4} \\ \lim\limits_{x \to 4} \dfrac{\left(x-4 \right)\left(a^{2}x+2ab+4a^{2}-1 \right)+\left(4a+b\right)^{2}-4}{\left(x-4 \right)\left( ax+b +\sqrt{x} \right)} &= \dfrac{3}{4} \\ \lim\limits_{x \to 4} \dfrac{\left(x-4 \right)\left(a^{2}x+2ab+4a^{2}-1 \right)+\left( 2 \right)^{2}-4}{ \left(x-4 \right)\left( ax+b +\sqrt{x} \right)} &= \dfrac{3}{4} \\ \lim\limits_{x \to 4} \dfrac{\left(a^{2}x+2ab+4a^{2}-1 \right)}{ \left( ax+b +\sqrt{x} \right)} &= \dfrac{3}{4} \\ \dfrac{\left(4a^{2} +2ab+4a^{2}-1 \right)}{ \left( 4a+b +\sqrt{4} \right)} &= \dfrac{3}{4} \\ \dfrac{\left(8a^{2} +2ab-1 \right)}{ \left( 2 + 2 \right)} &= \dfrac{3}{4} \\ \dfrac{\left( 8a^{2} +2ab -1 \right)}{4} &= \dfrac{3}{4} \\ \hline 8a^{2} +2ab -1 &= 3 \\ 8a^{2} +2a\left( 2-4a\right) -1 &= 3 \\ 8a^{2} +4a-8a^{2} &= 4 \\ 4a &= 4 \\ a &= 1 \end{align}$ Untuk $a=1$ dan $4a+b=2$ maka $b=-2$. Nilai $a+b=-1$

Peringatan!
Jika sudah belajar turunan fungsi, maka dapat digunakan Aturan L’Hospital yang mungkin dapat menghemat beberapa langkah.

$\therefore$ Jawaban yang benar $(D)\ -1$

74. Soal SBMPTN 2014 Kode 542

Jika $\lim\limits_{x \to 0} \dfrac{\sqrt{Ax+B}-2}{x}=1$, maka$\cdots$

$\begin{align} (A)\ & B=A^{2} \\ (B)\ & 4B^{2}=A \\ (C)\ & 4B=A^{2} \\ (D)\ & 4B=A \\ (E)\ & A+B=0 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 0} \dfrac{\sqrt{Ax+B}-2}{x}=1$. Jika kita substitusi langsung nilai $x=0$ maka nilai $ \sqrt{Ax+B}-2$ harus $0$, karena jika $\sqrt{Ax+B}-2$ tidak nol maka nilai limit adalah $\infty$. Untuk $x=0$

$\begin{align} \sqrt{Ax+B}-2 & = 0 \\ \sqrt{A(0)+B}-2 & = 0\\ \sqrt{B} & = 2\\ B & = 4 \end{align}$

Untuk $B=4$ Nilai $\lim\limits_{x \to 0} \dfrac{\sqrt{Ax+B}-2}{x}=1$, maka

$\begin{align} \lim\limits_{x \to 0} \dfrac{\sqrt{Ax+4}-2}{x} & = 1 \\ \lim\limits_{x \to 0} \dfrac{\sqrt{Ax+4}-2}{x} \cdot \dfrac{\sqrt{Ax+4}+2}{\sqrt{Ax+4}+2} & = 1 \\ \lim\limits_{x \to 0} \dfrac{\left( \sqrt{Ax+4}-2 \right) \left( \sqrt{Ax+4}+2 \right)}{x\left( \sqrt{Ax+4}+2 \right)} & = 1 \\ \lim\limits_{x \to 0} \dfrac{\left( Ax+4 -4 \right)}{x\left( \sqrt{Ax+4}+2 \right)} & = 1 \\ \lim\limits_{x \to 0} \dfrac{Ax}{x\left( \sqrt{Ax+4}+2 \right)} & = 1 \\ \lim\limits_{x \to 0} \dfrac{A}{ \sqrt{Ax+4}+2 } & = 1 \\ \dfrac{A}{ \sqrt{A(0)+4}+2 } & = 1 \\ \dfrac{A}{ \sqrt{4}+2 } & = 1 \\ \dfrac{A}{4} & = 1 \\ A & = 4 \end{align}$

$\therefore$ Jawaban yang benar $(C)\ 4B=A^{2}$

75. Soal SPMB 2006 Kode 320

Agar $\lim\limits_{x \to 1} \dfrac{\sqrt{p(x-1)+q}-3}{x-1}=-\dfrac{3}{2}$, maka nilai $p+2q=\cdots$

$\begin{align} (A)\ & -27 \\ (B)\ & -9 \\ (C)\ & 9 \\ (D)\ & 18\dfrac{1}{9} \\ (E)\ & 27 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 1} \dfrac{\sqrt{p(x-1)+q}-3}{x-1}=-\dfrac{3}{2}$. Jika kita substitusi langsung nilai $x=1$ maka nilai $\sqrt{p(x-1)+q}-3$ harus $0$, karena jika $\sqrt{p(x-1)+q}-3$ tidak nol maka nilai limit adalah $\infty$. sehingga untuk $x=1$ berlaku:

$\begin{align} \sqrt{p(x-1)+q}-3 & = 0 \\ \sqrt{p(1-1)+q}-3 & = 0 \\ \sqrt{q}-3 & = 0 \\ q & = 9 \end{align}$ untuk $q=9$ bentuk soal kita sekarang adalah $\lim\limits_{x \to 1} \dfrac{\sqrt{p(x-1)+9}-3}{x-1}=-\dfrac{3}{2}$ dengan mengalikan akar sekawan, maka akan kita peroleh:

$\begin{align} \lim\limits_{x \to 1} \dfrac{\sqrt{p(x-1)+9}-3}{x-1} \cdot \dfrac{\sqrt{p(x-1)+9}+3}{\sqrt{p(x-1)+9}+3} &= -\dfrac{3}{2} \\ \lim\limits_{x \to 1} \dfrac{p(x-1)+9-9}{(x-1) \left( \sqrt{p(x-1)+9}+3 \right)} &= -\dfrac{3}{2} \\ \lim\limits_{x \to 1} \dfrac{p(x-1)}{(x-1) \left( \sqrt{p(x-1)+9}+3 \right)} &= -\dfrac{3}{2} \\ \lim\limits_{x \to 1} \dfrac{p }{ \left( \sqrt{p(x-1)+9}+3 \right)} &= -\dfrac{3}{2} \\ \dfrac{p }{ \sqrt{p((1)-1)+9}+3 } &= -\dfrac{3}{2} \\ \dfrac{p }{ \sqrt{9}+3 } &= -\dfrac{3}{2} \\ \dfrac{p }{ 6 } &= -\dfrac{3}{2} \\ 2p &= – 18 \\ p &=-9 \end{align}$

Nilai $p+2q=-9+18=9$

$\therefore$ Jawaban yang benar $(C)\ 9$

76. Soal SIMAK UI 2013 Kode 333

$\lim\limits_{x \to 5} \dfrac{\sqrt{x+2 \sqrt{ x+1}}}{\sqrt{x-2 \sqrt{ x+1}}}=\cdots $

$\begin{align} (A)\ & \sqrt{3}+\sqrt{2} \\ (B)\ & 5-2\sqrt{6} \\ (C)\ & 2\sqrt{6} \\ (D)\ & 5 \\ (E)\ & 5+2\sqrt{6} \end{align}$

Pembahasan:

$\begin{align} & \lim\limits_{x \to 5} \dfrac{\sqrt{x+2 \sqrt{ x+1}}}{\sqrt{x-2 \sqrt{ x+1}}} \\ & = \dfrac{\sqrt{5+2 \sqrt{ 5+1}}}{\sqrt{5-2 \sqrt{ 5+1}}} \\ & = \dfrac{\sqrt{5+2 \sqrt{6}}}{\sqrt{5-2 \sqrt{6}}} \\ & = \dfrac{\sqrt{(3+2)+2 \sqrt{3 \cdot 2}}}{\sqrt{(3+2)-2 \sqrt{3 \cdot 2}}} \\ & = \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}- \sqrt{2}} \times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+ \sqrt{2}}\\ & = \dfrac{3+2+2\sqrt{6}}{3- 2} \\ & = 5+2\sqrt{6} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ 5+2\sqrt{6}$

77. Soal SIMAK UI 2012 Kode 224

$\lim\limits_{x \to 0} \dfrac{\sqrt{5+2\sqrt{x}}-\sqrt{5-2\sqrt{x}}}{\sqrt{x}}=\cdots $

$\begin{align} (A)\ & \dfrac{2}{2\sqrt{5}} \\ (B)\ & 2\sqrt{5} \\ (C)\ & \dfrac{2}{5}\sqrt{5} \\ (D)\ & \dfrac{4}{5\sqrt{5}} \\ (E)\ & \dfrac{4}{5} \sqrt{5} \end{align}$

Pembahasan:

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sqrt{5+2\sqrt{x}}-\sqrt{5-2\sqrt{x}}}{\sqrt{x}} \\ & = \lim\limits_{x \to 0} \dfrac{\sqrt{5+2\sqrt{x}}-\sqrt{5-2\sqrt{x}}}{\sqrt{x}} \times \dfrac{\sqrt{5+2\sqrt{x}}+\sqrt{5-2\sqrt{x}}}{\sqrt{5+2\sqrt{x}}+\sqrt{5-2\sqrt{x}}} \\ & = \lim\limits_{x \to 0} \dfrac{\left (5+2\sqrt{x} \right )-\left (5-2\sqrt{x} \right )}{\sqrt{x} \left (\sqrt{5+2\sqrt{x}}+\sqrt{5-2\sqrt{x}} \right )} \\ & = \lim\limits_{x \to 0} \dfrac{4\sqrt{x}}{\sqrt{x} \left (\sqrt{5+2\sqrt{x}}+\sqrt{5-2\sqrt{x}} \right )} \\ & = \lim\limits_{x \to 0} \dfrac{4 }{ \left (\sqrt{5+2\sqrt{x}}+\sqrt{5-2\sqrt{x}} \right )} \\ & = \dfrac{4 }{ \left (\sqrt{5+2\sqrt{0}}+\sqrt{5-2\sqrt{0}} \right )} \\ & = \dfrac{4 }{ \sqrt{5} +\sqrt{5}} \\ & = \dfrac{4 }{ 2\sqrt{5} } = \dfrac{2}{\sqrt{5} } \end{align}$

$\therefore$ Jawaban yang benar $(C)\ \dfrac{2}{5}\sqrt{5} $

78. Soal SPMB 2005 Kode 580

Jika $a \neq 0$, maka $\lim\limits_{x \to a} \dfrac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a} =\cdots$

$\begin{align} (A)\ & 3a \sqrt[3]{a} \\ (B)\ & 2a \sqrt[3]{a} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{2a}\sqrt[3]{a} \\ (E)\ & \dfrac{1}{3a}\sqrt[3]{a} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

$ \begin{align} & \lim\limits_{x \to a} \dfrac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a} \cdot \dfrac{\sqrt[3]{x^{2}}+\sqrt[3]{ax}+\sqrt[3]{a^{2}}}{\sqrt[3]{x^{2}}+\sqrt[3]{ax}+\sqrt[3]{a^{2}}} \\ & = \lim\limits_{x \to a} \dfrac{\left( x-a \right)}{\left(x-a \right) \left( \sqrt[3]{x^{2}}+\sqrt[3]{ax}+\sqrt[3]{a^{2}} \right)} \\ & = \dfrac{1}{\left( \sqrt[3]{a^{2}}+\sqrt[3]{a(a)}+\sqrt[3]{a^{2}} \right)} \\ & = \dfrac{1}{3\sqrt[3]{a^{2}}} \cdot \dfrac{\sqrt[3]{a}}{\sqrt[3]{a}} \\ & = \dfrac{1}{3a}\sqrt[3]{a^{2}} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(E)\ \dfrac{1}{3a}\sqrt[3]{a}$

79. Soal USM STIS 2017

$\lim\limits_{x \to 0} \dfrac{\sqrt{1-x}-1}{1-\sqrt[3]{1-x}}$ adalah$\cdots$

$\begin{align} (A)\ & \dfrac{3}{2} \\ (B)\ & \dfrac{2}{3} \\ (C)\ & 0 \\ (D)\ & -\dfrac{2}{3} \\ (E)\ & -\dfrac{3}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

Untuk menghemat penulisan kita coba dengan memisalkan $m=\sqrt{1-x}$, karena $x \to 0$ maka $m \to 1$. Soal limit $\lim\limits_{x \to 0} \dfrac{\sqrt{1-x}-1}{1-\sqrt[3]{1-x}}$ sudah bisa kita tuliskan menjadi $\lim\limits_{m \to 1} \dfrac{m-1}{1-\sqrt[3]{m^{2}}}$ atau $\lim\limits_{m \to 1} \dfrac{m-1}{-\left(\sqrt[3]{m^{2}}-1 \right)}$.

Sedikit catatan kita tentang perkalian akar sekawan yaitu $\left( \sqrt[3]{a}-1 \right)\left( \sqrt[3]{a^{2}}+\sqrt[3]{a}+1 \right)=a-1$

$\begin{align} & \lim\limits_{m \to 1} \dfrac{m-1}{-\left(\sqrt[3]{m^{2}}-1 \right)} \\ & = \lim\limits_{m \to 1} \dfrac{m-1}{-\left(\sqrt[3]{m^{2}}-1 \right)} \cdot \dfrac{\left( \sqrt[3]{m^{2}}+\sqrt[3]{m}+1 \right)}{\left(\sqrt[3]{m^{2}}+\sqrt[3]{m}+1 \right)} \\ & = \lim\limits_{m \to 1} \dfrac{\left(m-1 \right)\left( \sqrt[3]{m^{2}}+\sqrt[3]{m}+1 \right)}{-\left(m^{2} -1 \right)} \\ & = \lim\limits_{m \to 1} \dfrac{\left(m-1 \right)\left( \sqrt[3]{m^{2}}+\sqrt[3]{m}+1 \right)}{-\left(m -1 \right)\left(m +1 \right)} \\ & = \lim\limits_{m \to 1} \dfrac{\left( \sqrt[3]{m^{2}}+\sqrt[3]{m}+1 \right)}{- \left(m +1 \right)} \\ & = \dfrac{\left( \sqrt[3]{(1)^{2}}+\sqrt[3]{(1)}+1 \right)}{- \left(1 +1 \right)} \\ & = \dfrac{\left( 1+1+1 \right)}{- \left( 2 \right)} \\ & = -\dfrac{3}{2} \end{align}$

Peringatan!
Jika soal ini kita kerjakan dengan dengan menggunakan Aturan L’Hospital, penyelesaian seperti berikut ini;

$\begin{align} \lim\limits_{m \to 1} \dfrac{m-1}{1-m^{\frac{2}{3}}} & = \lim\limits_{m \to 1} \dfrac{1}{\frac{2}{3}m^{-\frac{1}{3}}} \\ & = \dfrac{1}{-\frac{2}{3}(1)^{-\frac{1}{3}}} \\ & = \dfrac{1}{-\dfrac{2}{3}} = – \dfrac{3}{2} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ – \dfrac{3}{2}$

80. Soal SIMAK UI 2009 Kode 951

$\lim\limits_{t \to 8} \dfrac{\sqrt{2+\sqrt[3]{x}}-2}{x-8}=\cdots$

$\begin{align} (A)\ & \dfrac{1}{64} \\ (B)\ & \dfrac{1}{48} \\ (C)\ & \dfrac{1}{24} \\ (D)\ & \dfrac{1}{16} \\ (E)\ & \infty \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

$ \begin{align} & \lim\limits_{t \to 8} \dfrac{\sqrt{2+\sqrt[3]{x}}-2}{x-8} \\ & = \lim\limits_{t \to 8} \dfrac{\sqrt{2+\sqrt[3]{x}}-2}{x-8} \cdot \dfrac{\sqrt{2+\sqrt[3]{x}}+2}{\sqrt{2+\sqrt[3]{x}}+2} \\ & = \lim\limits_{t \to 8} \dfrac{2+\sqrt[3]{x} -4}{\left( x-8 \right) \left( \sqrt{2+\sqrt[3]{x}}+2 \right)} \\ & = \lim\limits_{t \to 8} \dfrac{\sqrt[3]{x} -2}{\left( x-8 \right) \left( \sqrt{2+\sqrt[3]{x}}+2 \right)} \\ & = \lim\limits_{t \to 8} \dfrac{\sqrt[3]{x} -\sqrt[3]{8}}{\left( x-8 \right) \left( \sqrt{2+\sqrt[3]{x}}+2 \right)} \\ & = \lim\limits_{t \to 8} \dfrac{\sqrt[3]{x} -\sqrt[3]{8}}{\left (\sqrt[3]{x}-\sqrt[3]{8} \right )\left ( \sqrt[3]{x^{2}}+\sqrt[3]{8x}+\sqrt[3]{8^{2}} \right ) \left( \sqrt{2+\sqrt[3]{x}}+2 \right)} \\ & = \lim\limits_{t \to 8} \dfrac{1}{\left ( 1 \right )\left ( \sqrt[3]{x^{2}}+\sqrt[3]{8x}+\sqrt[3]{8^{2}} \right ) \left( \sqrt{2+\sqrt[3]{x}}+2 \right)} \\ & = \dfrac{1}{\left ( \sqrt[3]{8^{2}}+\sqrt[3]{8 (8)}+\sqrt[3]{8^{2}} \right ) \left( \sqrt{2+\sqrt[3]{(8)}}+2 \right)} \\ & = \dfrac{1}{ \left ( 4+4+4 \right ) \left( \sqrt{2+2}+2 \right)} \\ & = \dfrac{1}{48} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ \dfrac{1}{48}$

81. Soal SIMAK UI 2011 Kode 214

$\lim\limits_{x \to 3} \dfrac{x-3}{\sqrt{x+3-2\sqrt{3x}}}$ sama dengan…

$\begin{align} (A)\ & -2\sqrt{3} \\ (B)\ & -\sqrt{3} \\ (C)\ & \sqrt{3} \\ (D)\ & 2\sqrt{3} \\ (E)\ & 3\sqrt{3} \end{align}$

Pembahasan:

Sebagai catatan, kita ingatkan sedikit tentang bentuk akar yaitu: $\sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}$ dengan $a,\ b \geq 0$ dan $a \geq b$, atau $\sqrt{(a+b)-2\sqrt{ab}}=\left| \sqrt{a}-\sqrt{b} \right|$ $\sqrt{x+3-2\sqrt{3x}}=\left| \sqrt{x}-\sqrt{3} \right|$

Dari kesamaan bentuk akar di atas dan soal limit, maka kita peroleh: $ \begin{align} & \lim\limits_{x \to 3} \dfrac{x-3}{\sqrt{x+3-2\sqrt{3x}}} \\ & = \lim\limits_{x \to 3} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right|} \end{align} $

Untuk $x \gt 3$ maka $\sqrt{x}-\sqrt{3} \gt 0$, sehingga kita peroleh limit kanan; $ \begin{align} & \lim\limits_{x \to 33^{+}} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right| } \\ & = \lim\limits_{x \to 3^{+}} \dfrac{\left( \sqrt{x} – \sqrt{3} \right) \left( \sqrt{x} + \sqrt{3} \right)}{\sqrt{x}-\sqrt{3}} \\ & = \lim\limits_{x \to 3^{+}} \dfrac{ \left( \sqrt{x} + \sqrt{3} \right)}{1} \\ & = \dfrac{ \left( \sqrt{3} + \sqrt{3} \right)}{1}=2\sqrt{3} \\ \end{align} $

Untuk $0 \lt x \lt 3$ maka $\sqrt{x}-\sqrt{3} \lt 0$, sehingga kita peroleh limit kiri; $ \begin{align} &\lim\limits_{x \to 3^{-}} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right| } \\ & = \lim\limits_{x \to 3^{-}} \dfrac{\left( \sqrt{x} – \sqrt{3} \right) \left( \sqrt{x} + \sqrt{3} \right)}{-\left( \sqrt{x} – \sqrt{3} \right)} \\ & = \lim\limits_{x \to 3^{-}} \dfrac{ \left( \sqrt{x} + \sqrt{3} \right)}{-1} \\ & = \dfrac{ \left( \sqrt{3} + \sqrt{3} \right)}{-1}=-2\sqrt{3} \\ \end{align} $

Karena nilai $\lim\limits_{x \to 3^{+}} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right| } \neq \lim\limits_{x \to 3^{+}} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right| }$ atau Limit Kiri $\neq$ Limit Kanan maka nilai $\lim\limits_{x \to 3} \dfrac{x-3}{\left| \sqrt{x}-\sqrt{3} \right| }$ tidak ada.

$ \therefore $ Jawaban yang benar adalah $-$

82. Soal UM UGM 2014 Kode 531

Diketahui $f(x)=\sqrt{1+x}$. Nilai $\lim\limits_{x \to 0} \dfrac{f(3+2h^{2})-f(3-3h^{2})}{h^{2}}$ adalah$\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{2}{3} \\ (C)\ & \dfrac{6}{7} \\ (D)\ & \dfrac{9}{8} \\ (E)\ & \dfrac{5}{4} \end{align}$

Pembahasan:

Fungsi $f(x)=\sqrt{1+x}$, maka

$\begin{align} f(3+2h^{2}) & = \sqrt{1+3+2h^{2}} \\ & = \sqrt{4+2h^{2}} \\ f(3-3h^{2}) & = \sqrt{1+3-3h^{2}} \\ & = \sqrt{4-3h^{2}} \end{align}$ Nilai $\lim\limits_{x \to 0} \dfrac{f(3+2h^{2})-f(3-3h^{2})}{h^{2}}$ adalah

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sqrt{4+2h^{2}}-\sqrt{4-3h^{2}}}{h^{2}} \\ & = \lim\limits_{x \to 0} \dfrac{\sqrt{4+2h^{2}}-\sqrt{4-2h^{2}}}{h^{2}} \cdot \dfrac{\sqrt{4+2h^{2}}+\sqrt{4-3h^{2}}}{\sqrt{4+2h^{2}}+\sqrt{4-3h^{2}}} \\ & = \lim\limits_{x \to 0} \dfrac{\left(4+2h^{2}\right) -\left(4-3h^{2}\right)}{h^{2} \left(\sqrt{4+2h^{2}}+\sqrt{4-3h^{2}} \right)}\\ & = \lim\limits_{x \to 0} \dfrac{4+2h^{2}-4+3h^{2}}{h^{2} \left(\sqrt{4+2h^{2}}+\sqrt{4-3h^{2}} \right)}\\ & = \lim\limits_{x \to 0} \dfrac{5h^{2}}{h^{2} \left(\sqrt{4+2h^{2}}+\sqrt{4-3h^{2}} \right)}\\ & = \lim\limits_{x \to 0} \dfrac{5}{ \sqrt{4+2h^{2}}+\sqrt{4-3h^{2}} }\\ & = \dfrac{5}{\sqrt{4+0}+\sqrt{4-0} } \\ & = \dfrac{5}{4} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ \dfrac{5}{4}$

83. Soal SBMPTN 2018 Kode 526

Diketahui $O(0,0)$, $A(1,0)$, $B(2,0)$, $C(2,y)$, dan $D(0,y)$. Nilai $\lim\limits_{y \to 1} \dfrac{keliling\ \square ABCD}{keliling\ \bigtriangleup ACD}$ adalah…

$ \begin{align} (A)\ & \dfrac{1}{2}(2\sqrt{3}+3) \\ (B)\ & \dfrac{1}{4}(3\sqrt{2}+2) \\ (C)\ & \dfrac{1}{2}( \sqrt{3}+1) \\ (D)\ & \dfrac{1}{2}(3\sqrt{2}-2) \\ (E)\ & \dfrac{1}{4}(3\sqrt{2}-2) \end{align} $

Pembahasan:

Sebelum kita hitung nilai $\lim\limits_{y \to 1} \dfrac{keliling\ \square ABCD}{keliling\ \bigtriangleup ACD}$, coba kita hitung keliling $\square ABCD$ dan keliling $\bigtriangleup ACD$. Jarak dua titik dapat kita hitung dengan $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$,

  • Jarak titik $A(1,0)$ ke $B(2,0)$ adalah $AB=\sqrt{(2-1)^{2}+(0-0)^{2}}$$=\sqrt{1}=1$
  • Jarak titik $B(2,0)$ ke $C(2,y)$ adalah $BC=\sqrt{(2-2)^{2}+(y-0)^{2}}$$=\sqrt{y^{2}}=y$
  • Jarak titik $C(2,y)$ ke $D(0,y)$ adalah $CD=\sqrt{(0-2)^{2}+(y-y)^{2}}$$=\sqrt{4}=2$
  • Jarak titik $A(1,0)$ ke $D(0,y)$ adalah $AD=\sqrt{(1-0)^{2}+(0-y)^{2}}$$=\sqrt{1+y^{2}}$
  • Jarak titik $A(1,0)$ ke $C(2,y)$ adalah $AC=\sqrt{(2-1)^{2}+(y-0)^{2}}$$=\sqrt{1+y^{2}}$
$ \begin{align} & \text{keliling}\ \square ABCD \\ & = AB+BC+CD+DA \\ & = 1+y+2+\sqrt{1+y^{2}} \\ & = 3+y+\sqrt{1+y^{2}} \end{align} $ $ \begin{align} & \text{keliling}\ \bigtriangleup ACD \\ & = AC+CD+DA \\ & = \sqrt{1+y^{2}} +2+\sqrt{1+y^{2}} \\ & = 2+2\sqrt{1+y^{2}} \end{align} $

$ \begin{align} & \lim\limits_{y \to 1} \dfrac{keliling\ \square ABCD}{keliling\ \bigtriangleup ACD} \\ & = \lim\limits_{y \to 1} \dfrac{3+y+\sqrt{1+y^{2}}}{2+2\sqrt{1+y^{2}}} \\ & = \dfrac{3+1+\sqrt{1+1^{2}}}{2+2\sqrt{1+1^{2}}} \\ & = \dfrac{4+\sqrt{2}}{2+2\sqrt{2}} \times \dfrac{2-2\sqrt{2}}{2-2\sqrt{2}} \\ & = \dfrac{8-8\sqrt{2}+2\sqrt{2}-4}{4-8} \\ & = \dfrac{4-6\sqrt{2}}{-4} \\ & = \dfrac{2-3\sqrt{2}}{-2} \\ & =\dfrac{1}{2}\left(3\sqrt{2}-2 \right) \end{align} $

$\therefore$ Jawaban yang benar adalah $(D)\ \dfrac{1}{2}(3\sqrt{2}-2)$

84. Soal SBMPTN 2018 Kode 527

Diketahui $O(0,0)$, $A(2,0)$, $B(2,y)$, $C(0,y)$, dan $D(0,\frac{1}{2}y)$. Nilai $\lim\limits_{y \to 2} \dfrac{keliling\ \bigtriangleup BCD}{keliling\ \square OABD}$ adalah…

$ \begin{align} (A)\ & \dfrac{5+2\sqrt{5}}{5} \\ (B)\ & \dfrac{5+\sqrt{5}}{10} \\ (C)\ & \dfrac{1}{2}\sqrt{5} \\ (D)\ & \dfrac{5-2\sqrt{5}}{5} \\ (E)\ & \dfrac{5-\sqrt{5}}{5} \end{align} $

Pembahasan:

Sebelum kita hitung nilai $\lim\limits_{y \to 2} \dfrac{keliling\ \bigtriangleup BCD}{keliling\ \square OABD}$, coba kita hitung keliling $\square OABD$ dan keliling $\bigtriangleup BCD$. Jarak dua titik dapat kita hitung dengan $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$,

  • Jarak titik $O(0,0)$ ke $A(2,0)$ adalah $OA=\sqrt{(2-0)^{2}+(0-0)^{2}}$$=\sqrt{4}=2$
  • Jarak titik $A(2,0)$ ke $B(2,y)$ adalah $AB=\sqrt{(2-2)^{2}+(y-0)^{2}}$$=\sqrt{y^{2}}=y$
  • Jarak titik $B(2,y)$ ke $D(0,\frac{1}{2}y)$ adalah $BD=\sqrt{(0-2)^{2}+(\frac{1}{2}y-y)^{2}}$$=\sqrt{4+\frac{1}{4}y^{2}}$
  • Jarak titik $O(0,0)$ ke $D(0,\frac{1}{2}y)$ adalah $OD=\sqrt{(0-0)^{2}+(0-\frac{1}{2}y)^{2}}$$=\sqrt{\frac{1}{4}y^{2}}=\frac{1}{2}y$
  • Jarak titik $B(2,y)$ ke $C(0,y)$ adalah $BC=\sqrt{(2-0)^{2}+(y-y)^{2}}$$=\sqrt{4}=2$
  • Jarak titik $C(0,y)$ ke $D(0,\frac{1}{2}y)$ adalah $CD=\sqrt{(0-0)^{2}+(\frac{1}{2}y-y)^{2}}$$=\sqrt{\frac{1}{4}y^{2}}=\frac{1}{2}y$
$ \begin{align} & \text{keliling}\ \bigtriangleup BCD \\ & = BC+CD+DB \\ & = 2+\frac{1}{2}y +\sqrt{4+\frac{1}{4}y^{2}} \end{align} $ $ \begin{align} & \text{keliling}\ \square OABD \\ & = OA+AB+BD+DO \\ & = 2+y+\sqrt{4+\frac{1}{4}y^{2}}+\frac{1}{2}y \\ & = 2+\frac{3}{2}y+\sqrt{4+\frac{1}{4}y^{2}} \end{align} $

$ \begin{align} & \lim\limits_{y \to 2} \dfrac{keliling\ \bigtriangleup BCD}{keliling\ \square OABD} \\ & = \lim\limits_{y \to 2} \dfrac{2+\frac{1}{2}y +\sqrt{4+\frac{1}{4}y^{2}}}{2+\frac{3}{2}y+\sqrt{4+\frac{1}{4}y^{2}}} \\ & = \dfrac{2+\frac{1}{2}(2) +\sqrt{4+\frac{1}{4}(2)^{2}}}{2+\frac{3}{2}(2)+\sqrt{4+\frac{1}{4}(2)^{2}}} \\ & = \dfrac{2+1+\sqrt{4+1}}{2+3+\sqrt{4+1}} \\ & = \dfrac{3+\sqrt{5}}{5+\sqrt{5}} \times \dfrac{5-\sqrt{5}}{5-\sqrt{5}} \\ & = \dfrac{15-3\sqrt{5}+5\sqrt{5}-5}{25-5} \\ & = \dfrac{10+2\sqrt{5}}{20} \\ & = \dfrac{5+\sqrt{5}}{10} \end{align} $

$\therefore$ Jawaban yang benar adalah $(B)\ \dfrac{5+\sqrt{5}}{10}$

85. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}}-2 \right )=A$, maka nilai $\lim\limits_{t \to 2} \left (\sqrt[3]{\dfrac{a}{8}+\dfrac{b}{8t^{3}}}-t+1 \right )=\cdots$

$\begin{align} (A)\ & \dfrac{A}{2} \\ (B)\ & \dfrac{A}{3} \\ (C)\ & 0 \\ (D)\ & \dfrac{A+2}{2} \\ (E)\ & \dfrac{A+3}{3} \\ \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap
$ \begin{align} \lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}}-2 \right ) &= A \\ \lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}} \right )-\lim\limits_{t \to 2} \left (2 \right ) &= A \\ \lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}} \right )- 2 &= A \\ \lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}} \right ) &= A+2 \end{align} $

$ \begin{align} & \lim\limits_{t \to 2} \left (\sqrt[3]{\dfrac{a}{8}+\dfrac{b}{t^{3}}}-t+1 \right ) \\ &= \dfrac{1}{2} \cdot \lim\limits_{t \to 2} \left (\sqrt[3]{a+\dfrac{b}{t^{3}}} \right )-\lim\limits_{t \to 2} \left (t \right )+\lim\limits_{t \to 2} \left (1 \right ) \\ &= \dfrac{1}{2} \cdot \left (A+2 \right )-2+1 \\ &= \dfrac{A}{2}+1-1 = \dfrac{A}{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ \dfrac{A}{2}$

86. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2 }{x-1} \right )=A$, maka nilai $\lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2x }{x^{2}+2x-3} \right )=\cdots$

$\begin{align} (A)\ & \dfrac{2-A}{2} \\ (B)\ & -\dfrac{A}{2} \\ (C)\ & \dfrac{A-2}{4} \\ (D)\ & \dfrac{A}{4} \\ (E)\ & \dfrac{A+2}{4} \\ \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap

$ \begin{align} & \lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2x }{x^{2}+2x-3} \right) \\ & = \lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2x }{(x-1)(x+3)} \right) \\ & = \lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2+2-2x }{(x-1)(x+3)} \right) \\ & = \lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2 }{(x-1)(x+3)}+\dfrac{2-2x }{(x-1)(x+3)} \right) \\ & = \lim\limits_{x \to 1} \left (\dfrac{\sqrt{ax^{4}+b}-2 }{(x-1)} \cdot \dfrac{1}{(x+3)}+\dfrac{2-2x }{(x-1)} \cdot \dfrac{1}{(x+3)} \right) \\ & = \lim\limits_{x \to 1} \dfrac{\sqrt{ax^{4}+b}-2}{(x-1)} \cdot \lim\limits_{x \to 1} \dfrac{1} {(x+3)}+\lim\limits_{x \to 1} \dfrac{2-2x }{(x-1)} \cdot \lim\limits_{x \to 1} \dfrac{1}{(x+3)} \\ & = A \cdot \dfrac{1} {4}+ \lim\limits_{x \to 1} \dfrac{-2(x-1) }{(x-1)} \cdot \dfrac{1}{4} \\ & = A \cdot \dfrac{1} {4}+ (-2) \cdot \dfrac{1}{4} \\ & = \dfrac{A}{4}- \dfrac{2}{4} = \dfrac{A-2}{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ \dfrac{A-2}{4}$

87. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{ax+b}}{x+1} \right )=2$, maka nilai $\lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{\dfrac{ax}{8}+\dfrac{b}{8}}-2x+1}{x^{2}+4x+3} \right )=\cdots$

$\begin{align} (A)\ & -\dfrac{2}{15} \\ (B)\ & -\dfrac{1}{15} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{15} \\ (E)\ & \dfrac{2}{15} \\ \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap
$ \begin{align} \lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{ax+b}}{x+1} \right ) &= 2 \\ \dfrac{\sqrt[3]{2a+b}}{2+1} &= 2 \\ \sqrt[3]{2a+b} &= 6 \end{align} $

$ \begin{align} & \lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{\dfrac{ax}{8}+\dfrac{b}{8}}-2x+1}{x^{2}+4x+3} \right ) \\ &= \lim\limits_{x \to 2} \left (\dfrac{\dfrac{1}{2} \cdot \sqrt[3]{ ax + b }-2x+1}{(x+1)(x+3)} \right ) \\ &= \dfrac{\dfrac{1}{2} \cdot \sqrt[3]{ 2a + b }-2(2) +1}{(2+1)(2+3)} \\ &= \dfrac{\dfrac{1}{2} \cdot 6-3}{15} \\ &= \dfrac{0}{15}=0 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 0$

88. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{ax^{3}+b}}{x-1} \right )=A$, maka nilai $\lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{\dfrac{ax^{3}}{8}+\dfrac{b}{8}}-2x}{x^{2}+2x-2} \right )=\cdots $

$\begin{align} (A)\ & \dfrac{1}{12}A \\ (B)\ & \dfrac{1}{12}(A-2) \\ (C)\ & \dfrac{1}{12}(A-1) \\ (D)\ & \dfrac{1}{12}(A-6) \\ (E)\ & \dfrac{1}{12}(A-8) \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap
$ \begin{align} \lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{ax^{3}+b}}{x-1} \right ) &= A \\ \dfrac{\sqrt[3]{a(2)^{3}+b}}{2-1} &= A \\ \sqrt[3]{8a +b} &= A \end{align} $

$ \begin{align} & \lim\limits_{x \to 2} \left (\dfrac{\sqrt[3]{\dfrac{ax^{3}}{8}+\dfrac{b}{8}}-2x}{x^{2}+2x-2} \right ) \\ &= \lim\limits_{x \to 2} \left (\dfrac{\frac{1}{2} \cdot \sqrt[3]{ax^{3}+b}-2x}{x^{2}+2x-2} \right ) \\ &= \dfrac{\frac{1}{2} \cdot \sqrt[3]{8a +b}-2(2)}{(2)^{2}+2(2)-2} \\ &= \dfrac{\frac{1}{2} \cdot A-4}{6} \\ &= \dfrac{A-8}{12} \end{align} $

$ \therefore $ Jawaban yang benar adalah $\dfrac{1}{12}(A-8)$

89. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{ax^{3}+b}-2}{x-\frac{1}{2}} \right )=A$, maka nilai $\lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{\dfrac{ax^{3}}{8}+\dfrac{b}{8}}-2x}{4x^{2}-1} \right )=\cdots $

$\begin{align} (A)\ & \dfrac{1}{8}A-2 \\ (B)\ & \dfrac{1}{8}A-1 \\ (C)\ & \dfrac{1}{8}A-\dfrac{1}{2} \\ (D)\ & \dfrac{1}{8}A-\dfrac{1}{4} \\ (E)\ & \dfrac{1}{8}A-\dfrac{1}{8} \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$
  • $\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{ \lim\limits_{x \to c} f(x)}$ dimana $\lim\limits_{x \to c} f(x) \gt 0$ bilamana $n$ genap

$ \begin{align} \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{ax^{3}+b}-2}{x-\frac{1}{2}} \right ) &= A \\ \dfrac{1}{2} \cdot \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{ax^{3}+b}-2}{x-\frac{1}{2}} \right ) &= \dfrac{1}{2} A \\ \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{ax^{3}+b}-2}{2 \left( x-\frac{1}{2} \right)} \right ) &= \dfrac{A}{2} \\ \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{ax^{3}+b}-2}{ \left( 2x- 1 \right)} \right ) &= \dfrac{A}{2} \\ \end{align} $

Dengan beberapa manipulasi aljabar, penjabaran soal menjadi seperti berikut ini:
$ \begin{align} & \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\sqrt[3]{\dfrac{ax^{3}}{8}+\dfrac{b}{8}}-2x}{4x^{2}-1} \right ) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\frac{1}{2} \sqrt[3]{ ax^{3} + b }-2x}{ \left( 2x-1 \right)\left( 2x+1 \right)} \right ) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{\frac{1}{2} \left( \sqrt[3]{ ax^{3} + b }- 4x \right)}{ \left( 2x-1 \right)\left( 2x+1 \right)} \right ) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 4x }{2 \left( 2x-1 \right) \left( 2x+1 \right)} \right ) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 4x }{ \left( 2x-1 \right)} \right ) \cdot \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ 1 }{2 \left( 2x+1 \right)} \right )\\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 2 + 2 – 4x }{ \left( 2x-1 \right)} \right ) \cdot \left( \dfrac{ 1 }{2 \left( 2 \cdot \frac{1}{2} +1 \right)} \right) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 2 – 2 \left(2x -1 \right) }{ \left( 2x-1 \right)} \right ) \cdot \left( \dfrac{ 1 }{4} \right) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 2}{ \left( 2x-1 \right)} – \dfrac{ 2 \left(2x -1 \right) }{ \left( 2x-1 \right)} \right ) \cdot \left( \dfrac{ 1 }{4} \right) \\ &= \lim\limits_{x \to \frac{1}{2}} \left (\dfrac{ \sqrt[3]{ ax^{3} + b }- 2}{ \left( 2x-1 \right)} – 2 \right ) \cdot \left( \dfrac{ 1 }{4} \right) \\ &= \left (\lim\limits_{x \to \frac{1}{2}} \dfrac{ \sqrt[3]{ ax^{3} + b }- 2}{ \left( 2x-1 \right)} – \lim\limits_{x \to \frac{1}{2}} 2 \right ) \cdot \left( \dfrac{ 1 }{4} \right) \\ &= \left ( \dfrac{A}{2} – 2 \right ) \cdot \left( \dfrac{ 1 }{4} \right) \\ &= \dfrac{A}{8} – \dfrac{ 1 }{2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ \dfrac{1}{8}A-\dfrac{1}{2}$

90. Soal UTBK-SBMPTN 2019

Jika $\lim\limits_{t \to a} \left (\dfrac{\left (\left | t \right |-1 \right )^{2}-\left (\left | a \right |-1 \right )^{2}}{t^{2}-a^{2}} \right )=K$, maka nilai $\lim\limits_{t \to a} \left (\dfrac{\left (\left | t \right |-1 \right )^{4}-\left (\left | a \right |-1 \right )^{4}}{t-a} \right )=\cdots$

$\begin{align} (A)\ & 2K \left( \left | a \right |-1 \right )^{2} \\ (B)\ & K \left( \left | a \right |-1 \right )^{2} \\ (C)\ & 4aK \left( \left | a \right |-1 \right )^{2} \\ (D)\ & aK \left( \left | a \right |-1 \right )^{2} \\ (E)\ & K^{2} \left( \left | a+K \right |-1 \right )^{2} \end{align}$

Pembahasan:

Catatan calon guru tentang limit fungsi yang mungkin kita butuhkan adalah:

  • $\lim\limits_{x \to c} k=k$
  • $\lim\limits_{x \to c} \left( f(x)\pm g(x) \right) = \lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$

Dengan beberapa manipulasi aljabar, penjabaran soal menjadi seperti berikut ini:
$ \begin{align} & \lim\limits_{t \to a} \left (\dfrac{\left (\left | t \right |-1 \right )^{4}-\left (\left | a \right |-1 \right )^{4}}{t-a} \right ) \\ & = \lim\limits_{t \to a} \left (\dfrac{\left [\left (\left | t \right |-1 \right )^{2}-\left (\left | a \right |-1 \right )^{2} \right ]\left [\left (\left | t \right |-1 \right )^{2}+\left (\left | a \right |-1 \right )^{2} \right ]}{t-a} \right ) \\ & = \lim\limits_{t \to a} \left (\dfrac{\left [\left (\left | t \right |-1 \right )^{2}-\left (\left | a \right |-1 \right )^{2} \right ]\left [\left (\left | t \right |-1 \right )^{2}+\left (\left | a \right |-1 \right )^{2} \right ]}{t-a} \cdot \dfrac{\left (t+a \right )}{\left (t+a \right )} \right ) \\ &= \lim\limits_{t \to a} \left (\dfrac{\left [\left (\left | t \right |-1 \right )^{2}-\left (\left | a \right |-1 \right )^{2} \right ]\left [\left (\left | t \right |-1 \right )^{2}+\left (\left | a \right |-1 \right )^{2} \right ] \cdot \left (t+a \right ) }{t^{2}-a^{2}} \right ) \\ &= \lim\limits_{t \to a} \left (\dfrac{\left [\left (\left | t \right |-1 \right )^{2}-\left (\left | a \right |-1 \right )^{2} \right ]}{t^{2}-a^{2}} \right ) \cdot \lim\limits_{t \to a} \left (\dfrac{\left [\left (\left | t \right |-1 \right )^{2}+\left (\left | a \right |-1 \right )^{2} \right ] \cdot \left (t+a \right ) }{1} \right ) \\ &= K \cdot \left ( \left [\left (\left | a \right |-1 \right )^{2}+\left (\left | a \right |-1 \right )^{2} \right ] \cdot \left (a+a \right ) \right ) \\ &= K \cdot \left ( 2 \left ( \left | a \right |-1 \right )^{2} \cdot \left (2a \right ) \right ) \\ &= 4aK \cdot \left ( \left | a \right |-1 \right )^{2} \\ \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ 4aK \left( \left | a \right |-1 \right )^{2}$

91. Soal Latihan Limit Fungsi Matematika SMA

Jika $\lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2}{x-1}=A$, maka nilai dari $\lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2x}{x^{2}+2x-3}=\cdots$

$\begin{align} (A)\ & \dfrac{ A-2 }{2} \\ (B)\ & \dfrac{ A-2 }{4} \\ (C)\ & A-2 \\ (D)\ & 2A-4 \\ (E)\ & 2A \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2}{x-1}=A$. Jika kita substitusi langsung nilai $x=1$ maka nilai $\sqrt{ax^{2}+b}-2$ harus $0$, karena jika $\sqrt{ax^{2}+b}-2$ tidak nol maka nilai limit adalah $\infty$. sehingga untuk $x=1$ berlaku:

$\begin{align} \sqrt{ax^{2}+b}-2 & = 0 \\ \sqrt{a(1)^{2}+b}-2 & = 0 \\ \sqrt{a+b} & = 2 \\ \end{align}$

$\begin{align} \lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2}{x-1} \times \dfrac{\sqrt{ax^{2}+b}+2}{\sqrt{ax^{2}+b}+2} &= A \\ \lim\limits_{x \to 1} \dfrac{ax^{2}+b -4}{\left( x-1 \right)\left( \sqrt{ax^{2}+b}+2 \right)} &= A \\ \lim\limits_{x \to 1} \dfrac{a\left( x+1 \right)\left( x-1 \right)+\left( a+b-4 \right)}{\left( x-1 \right)\left( \sqrt{ax^{2}+b}+2 \right)} &= A \\ \lim\limits_{x \to 1} \dfrac{a\left( x+1 \right)\left( x-1 \right)+\left( 4-4 \right)}{\left( x-1 \right)\left( \sqrt{ax^{2}+b}+2 \right)} &= A \\ \lim\limits_{x \to 1} \dfrac{a\left( x+1 \right)\left( x-1 \right) }{\left( x-1 \right)\left( \sqrt{ax^{2}+b}+2 \right)} &= A \\ \lim\limits_{x \to 1} \dfrac{a\left( x+1 \right) }{ \left( \sqrt{ax^{2}+b}+2 \right)} &= A \\ \dfrac{a\left( 1+1 \right) }{ \left( \sqrt{a(1)^{2}+b}+2 \right)} &= A \\ \dfrac{2a}{ \left( \sqrt{a+b}+2 \right)} &= A \\ \dfrac{2a}{ \left( 2+2 \right)} &= A \\ \dfrac{a}{ 2} &= A \\ a &= 2A \\ \end{align}$

$\begin{align} & \lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2x}{x^{2}+2x-3} \\ &= \lim\limits_{x \to 1} \dfrac{\sqrt{ax^{2}+b}-2x}{x^{2}+2x-3} \times \dfrac{\sqrt{ax^{2}+b}+2x}{\sqrt{ax^{2}+b}+2x} \\
&= \lim\limits_{x \to 1} \dfrac{ ax^{2}+b-4x^{2}}{\left( x-1 \right)\left( x+3 \right)\left( \sqrt{ax^{2}+b}+2x \right)} \\
&= \lim\limits_{x \to 1} \dfrac{ \left( a-4 \right)\left( x-1 \right)\left( x+1 \right)+a+b-4}{\left( x-1 \right)\left( x+3 \right)\left( \sqrt{ax^{2}+b}+2x \right)} \\
&= \lim\limits_{x \to 1} \dfrac{ \left( a-4 \right)\left( x-1 \right)\left( x+1 \right)+4-4}{\left( x-1 \right)\left( x+3 \right)\left( \sqrt{ax^{2}+b}+2x \right)} \\
&= \lim\limits_{x \to 1} \dfrac{ \left( a-4 \right) \left( x+1 \right)}{ \left( x+3 \right)\left( \sqrt{ax^{2}+b}+2x \right)} \\
&= \dfrac{ \left( a-4 \right) \left( 1+1 \right)}{ \left( 1+3 \right)\left( \sqrt{a(1)^{2}+b}+2(1) \right)} \\
&= \dfrac{ 2A-4 }{ \left( 2 \right)\left( 2+2 \right)} \\
&= \dfrac{ A-2 }{4} \\ \end{align}$

Peringatan!
Jika sudah belajar turunan fungsi, maka dapat digunakan Aturan L’Hospital yang mungkin dapat menghemat beberapa langkah.

$\therefore$ Jawaban yang benar $(B)\ \dfrac{ A-2 }{4}$

92. Soal UM UNDIP 2019 Kode 431

Nilai $\lim\limits_{x \to 4} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & 1 \\ (C)\ & 2 \\ (D)\ & 3 \\ (E)\ & 4 \end{align}$

Pembahasan:

Untuk menyelesaikan soal di atas, kita coba pinjam catatan bentuk akar yaitu $\sqrt{a}- \sqrt{b}=\sqrt{a+b-2\sqrt{ab}}$ untuk $a \gt b$.

$\begin{align} & \lim\limits_{x \to 4} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\sqrt{x+4- 2\sqrt{4x}}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\sqrt{x}-\sqrt{4}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\sqrt{x}-2}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} 1 = 1 \\ \hline & \lim\limits_{x \to 4} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\sqrt{x+4- 2\sqrt{4x}}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{\sqrt{4}-\sqrt{x}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} \dfrac{2-\sqrt{x}}{\sqrt{x}-2} \\ & = \lim\limits_{x \to 4} -1 = -1 \end{align}$

Dari hasil di atas kita peroleh limit kiri $\lim\limits_{x \to 4^{-}} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2}=-1$ dan limit kanan $\lim\limits_{x \to 4^{+}} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2}=1$ sehingga nilai $\lim\limits_{x \to 4} \dfrac{\sqrt{x+4-4\sqrt{x}}}{\sqrt{x}-2}$ tidak ada.

Tetapi jika pada saat situasi ujian, pilihlah jawaban yang memenuhi salah satu nilai limitnya. Karena pada saat ujian dengan soal pilihan ganda, soalnya dirancang untuk mempunyai satu jawaban yang benar.

$\therefore$ Jawaban yang benar $(B)\ 1$

93. Soal UM UGM 2019 Kode 634

$\lim\limits_{x \to 1} \dfrac{\sqrt[3]{9x-9}}{\left( \sqrt[3]{x}-1 \right)^{\frac{1}{3}}}=\cdots$

$\begin{align} (A)\ & 27 \\ (B)\ & 9 \\ (C)\ & 5 \\ (D)\ & 3 \\ (E)\ & 1 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

$\begin{align} & \lim\limits_{x \to 1} \dfrac{\sqrt[3]{9x-9}}{\left( \sqrt[3]{x}-1 \right)^{\frac{1}{3}}} \\ & = \lim\limits_{x \to 1} \dfrac{\left(9x-9\right)^{\frac{1}{3}}}{\left( \sqrt[3]{x}-1 \right)^{\frac{1}{3}}} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ \left( 9x-9\right)}{\left( \sqrt[3]{x}-1 \right)} \right)^{\frac{1}{3}} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ 9\left( x-1\right)}{\left( \sqrt[3]{x}-1 \right)} \cdot \dfrac{ \left( \sqrt[3]{x^{2}}+\sqrt[3]{x}+1 \right) }{\left( \sqrt[3]{x^{2}}+\sqrt[3]{x}+1 \right)} \right)^{\frac{1}{3}} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ 9\left( x-1\right) \left( \sqrt[3]{x^{2}}+\sqrt[3]{x}+1 \right)}{\left( x-1 \right)} \right)^{\frac{1}{3}} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ 9 \left( \sqrt[3]{x^{2}}+\sqrt[3]{x}+1 \right)}{1} \right)^{\frac{1}{3}} \\ & = \left( 9 \left( \sqrt[3]{(1)^{2}}+\sqrt[3]{1}+1\right) \right)^{\frac{1}{3}} \\ & = \left( 9 \left( 1+1+1 \right) \right)^{\frac{1}{3}} \\ & = \left( 27 \right)^{\frac{1}{3}} =3 \end{align}$

$\therefore$ Jawaban yang benar $(D)\ 3$

94. Soal UN Matematika SMA IPS 2017

$\lim\limits_{x \to 4} \dfrac{x^{2}-16}{1-\sqrt{x-3}}=\cdots$

$\begin{align} (A)\ & -16 \\ (B)\ & -4 \\ (C)\ & 4 \\ (D)\ & 16 \\ (E)\ & 32 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 4} \dfrac{x^{2}-16}{1-\sqrt{x-3}} \\ & = \lim\limits_{x \to 4} \dfrac{x^{2}-16}{1-\sqrt{x-3}} \cdot \dfrac{1+\sqrt{x-3}}{1+\sqrt{x-3}} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)\left( x+4 \right)\left( 1+\sqrt{x-3} \right)}{1-\left( x-3 \right)} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)\left( x+4 \right)\left( 1+\sqrt{x-3} \right)}{4-x} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x-4 \right)\left( x+4 \right)\left( 1+\sqrt{x-3} \right)}{-\left( x-4 \right)} \\ & = \dfrac{ \left( 4+4 \right)\left( 1+\sqrt{4-3} \right)}{-1} \\ & = \dfrac{ \left( 8 \right)\left( 2 \right)}{-1} =-16 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -16$

95. Soal UN Matematika SMA IPA 2012

$\lim\limits_{x \to 3} \dfrac{2-\sqrt{x+1}}{x-3}=\cdot$

$\begin{align} (A)\ & -\dfrac{1}{4} \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & 1 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 3} \dfrac{2-\sqrt{x+1}}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{2-\sqrt{x+1}}{x-3} \cdot \dfrac{2+\sqrt{x+1}}{2+\sqrt{x+1}} \\ & = \lim\limits_{x \to 3} \dfrac{4-(x+1)}{\left( x-3 \right)\left( 2+\sqrt{x+1} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{3-x}{\left( x-3 \right)\left( 2+\sqrt{x+1} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{-\left( x-3 \right)}{\left( x-3 \right)\left( 2+\sqrt{x+1} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{-1}{\left( 2+\sqrt{x+1} \right)} \\ & = \dfrac{-1}{\left( 2+\sqrt{3+1} \right)}=\dfrac{-1}{4} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ -\dfrac{1}{4}$

96. Soal UM UGM 2019 Kode 934

Nilai $\lim\limits_{x \to 1} \dfrac{x-1+\sqrt[3]{1-x}}{\sqrt[3]{1-x^{2}}}$ adalah…

$\begin{align} (A)\ & -\sqrt[3]{2} \\ (B)\ & 0 \\ (C)\ & \dfrac{1}{\sqrt[3]{2}} \\ (D)\ & \dfrac{1}{\sqrt[3]{3}} \\ (E)\ & \sqrt[3]{2} \end{align}$

Pembahasan:

$\begin{align} & \lim\limits_{x \to 1} \dfrac{x-1+\sqrt[3]{1-x}}{\sqrt[3]{1-x^{2}}} \\ & = \lim\limits_{x \to 1} \dfrac{ -\left ( 1-x \right )+\sqrt[3]{1-x}}{\sqrt[3]{\left (1+x \right )\left (1-x \right )}} \\ & = \lim\limits_{x \to 1} \dfrac{ \sqrt[3]{\left (1-x \right )} \left (-\left ( 1-x \right )^{2}+1 \right )}{\sqrt[3]{\left (1-x \right )} \cdot \sqrt[3]{\left (1+x \right )}} \\ & = \lim\limits_{x \to 1} \dfrac{ \left (-\left ( 1-x \right )^{2}+1 \right )}{ \sqrt[3]{\left (1+x \right )}} \\ & = \dfrac{ \left (-\left ( 1-1 \right )^{2}+1 \right )}{ \sqrt[3]{\left (1+1 \right )}} \\ & = \dfrac{ 1}{ \sqrt[3]{2}} \end{align}$

$\therefore$ Jawaban yang benar $(C)\ \dfrac{1}{\sqrt[3]{2}}$

97. Soal UMPTN 1992 (Rayon C)

$\lim\limits_{x \to 1} \dfrac{\sqrt{x}- \sqrt{2x-1}}{x-1} =\cdots$

$\begin{align} (A)\ & -1 \\ (B)\ & -\dfrac{1}{2} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{2} \\ (E)\ & 1 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 1} \dfrac{\sqrt{x}- \sqrt{2x-1}}{x-1} \\ & = \lim\limits_{x \to 1} \dfrac{\sqrt{x}- \sqrt{2x-1}}{x-1} \times \dfrac{\sqrt{x} + \sqrt{2x-1}}{\sqrt{x} + \sqrt{2x-1}} \\ & = \lim\limits_{x \to 1} \dfrac{x- \left(2x-1 \right)}{\left( x-1 \right)\left( \sqrt{x} + \sqrt{2x-1} \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x- 2x+1 }{\left( x-1 \right)\left( \sqrt{x} + \sqrt{2x-1} \right)} \\ & = \lim\limits_{x \to 1} \dfrac{-x+1 }{\left( x-1 \right)\left( \sqrt{x} + \sqrt{2x-1} \right)} \\ & = \lim\limits_{x \to 1} \dfrac{- \left( x-1 \right) }{\left( x-1 \right)\left( \sqrt{x} + \sqrt{2x-1} \right)} \\ & = \lim\limits_{x \to 1} \dfrac{- 1 }{ \left( \sqrt{x} + \sqrt{2x-1} \right)} \\ & = \dfrac{- 1 }{ \left( \sqrt{1} + \sqrt{2(1)-1} \right)} \\ & = \dfrac{- 1 }{ \left( 1 + 1 \right)}=\dfrac{- 1 }{ 2} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ -\dfrac{1}{2}$

98. Soal UMPTN 1997 (Rayon B)

$\lim\limits_{x \to 3} \dfrac{x- \sqrt{2x+3}}{x^{2}-9} =\cdots$

$\begin{align} (A)\ & \dfrac{1}{3} \\ (B)\ & \dfrac{1}{9} \\ (C)\ & 0 \\ (D)\ & \dfrac{1}{2} \\ (E)\ & \dfrac{2}{2} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 3} \dfrac{x- \sqrt{2x+3}}{x^{2}-9} \\ & = \lim\limits_{x \to 3} \dfrac{x- \sqrt{2x+3}}{x^{2}-9} \times \dfrac{x + \sqrt{2x+3}}{x + \sqrt{2x+3}} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}- \left(2x+3\right)}{\left( x^{2}-9 \right)\left( x + \sqrt{2x+3} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}-2x-3 }{\left( x-3 \right)\left( x+3 \right)\left( x + \sqrt{2x+3} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x-3 \right)\left( x+1 \right) }{\left( x-3 \right)\left( x+3 \right)\left( x + \sqrt{2x+3} \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x+1 \right) }{ \left( x+3 \right)\left( x + \sqrt{2x+3} \right)} \\ & = \dfrac{ \left( 3+1 \right) }{ \left( 3+3 \right)\left( 3 + \sqrt{2(3)+3} \right)} \\ & = \dfrac{ \left( 4 \right) }{ \left( 6 \right)\left( 3 + 3 \right)} = \dfrac{ 4 }{ 36}= \dfrac{ 1 }{ 9} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(B)\ \dfrac{1}{9}$

99. Soal UMPTN 1998 (Rayon B)

Nilai $\lim\limits_{x \to 0} \dfrac{2x^{2}- 5x}{3-\sqrt{9+x}}$ adalah…

$\begin{align} (A)\ & 30 \\ (B)\ & 1 \\ (C)\ & 0 \\ (D)\ & -1 \\ (E)\ & -30 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to 0} \dfrac{2x^{2}- 5x}{3-\sqrt{9+x}} \\ & = \lim\limits_{x \to 0} \dfrac{2x^{2}- 5x}{3-\sqrt{9+x}} \times \dfrac{3+\sqrt{9+x}}{3+\sqrt{9+x}} \\ & = \lim\limits_{x \to 0} \dfrac{\left( 2x^{2}- 5x \right)\left( 3+\sqrt{9+x} \right)}{9-\left(9+x \right) } \\ & = \lim\limits_{x \to 0} \dfrac{x\left( 2x- 5 \right)\left( 3+\sqrt{9+x} \right)}{-x} \\ & = \lim\limits_{x \to 0} (-1)\left( 2x- 5 \right)\left( 3+\sqrt{9+x} \right) \\ & = (-1)\left( 2(0)- 5 \right)\left( 3+\sqrt{9+0} \right) \\ & = (-1)\left( – 5 \right)\left( 3+ 3 \right) = 30 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ 30$

100. Soal UN Matematika SMA IPA 2009

Nilai $\lim\limits_{x \to -2} \dfrac{x+2}{\sqrt{5x+14}-2}=\cdots$

$\begin{align} (A)\ & 4 \\ (B)\ & 2 \\ (C)\ & 1,2 \\ (D)\ & 0,8 \\ (E)\ & 0,4 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$

$ \begin{align} & \lim\limits_{x \to -2} \dfrac{x+2}{\sqrt{5x+14}-2} \\ & = \lim\limits_{x \to -2} \dfrac{x+2}{\sqrt{5x+14}-2} \cdot \dfrac{\sqrt{5x+14}+2}{\sqrt{5x+14}+2}\\ & = \lim\limits_{x \to -2} \dfrac{\left( x+2 \right)\left( \sqrt{5x+14}+2 \right)}{5x+14-4} \\ & = \lim\limits_{x \to -2} \dfrac{\left( x+2 \right)\left( \sqrt{5x+14}+2 \right)}{5x+10} \\ & = \lim\limits_{x \to -2} \dfrac{\left( x+2 \right)\left( \sqrt{5x+14}+2 \right)}{5 \left( x+2 \right)} \\ & = \lim\limits_{x \to -2} \dfrac{\left( \sqrt{5x+14}+2 \right)}{5} \\ & = \dfrac{\left( \sqrt{5(-2)+14}+2 \right)}{5} \\ & = \dfrac{\left( 2+2 \right)}{5}= \dfrac{4}{5}=0,8 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(D)\ 0,8$

101. Soal UMPTN 1997 (Rayon A)

$\lim\limits_{x \to 0} \dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{3} \\ (C)\ & \dfrac{2}{3} \\ (D)\ & \dfrac{3}{2} \\ (E)\ & 2 \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

$\begin{align} & \lim\limits_{x \to 0} \dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1} \\ & = \lim\limits_{x \to 0} \dfrac{ \sqrt{1+x}-1 }{\sqrt[3]{1+x}-\sqrt[3]{1}} \times \dfrac{ \sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}} }{\sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}}} \\ & = \lim\limits_{x \to 0} \dfrac{ \left( \sqrt{1+x}-1 \right) \left(\sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}} \right)}{1+x-1} \\ & = \lim\limits_{x \to 0} \dfrac{ \left( \sqrt{1+x}-1 \right) \left(\sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}} \right)}{x} \times \dfrac{ \sqrt{1+x}+1 }{ \sqrt{1+x}+1 } \\ & = \lim\limits_{x \to 0} \dfrac{ \left( 1+x -1 \right) \left(\sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}} \right) }{(x) \left(\sqrt{1+x}+1 \right) } \\ & = \lim\limits_{x \to 0} \dfrac{ \left(\sqrt[3]{1^{2}}+\sqrt[3]{1+x}+\sqrt[3]{(1+x)^{2}} \right)}{ \left(\sqrt{1+x}+1 \right) } \\ & = \dfrac{ \sqrt[3]{1^{2}}+\sqrt[3]{1+0}+\sqrt[3]{(1+0)^{2}}}{ \left(\sqrt{1+0}+1 \right) } \\ & = \dfrac{1+1+1}{1+1}= \dfrac{3}{2} \end{align}$

$\therefore$ Jawaban yang benar $(D)\ \dfrac{3}{2}$

102. Soal UMPTN 1998 (Rayon A)

$\lim\limits_{x \to 1} \dfrac{\sqrt[3]{x^{2}}-2\sqrt[3]{x}+1}{\left(x-1\right)^{2}}=\cdots$

$\begin{align} (A)\ & 0 \\ (B)\ & \dfrac{1}{3} \\ (C)\ & \dfrac{1}{5} \\ (D)\ & \dfrac{1}{7} \\ (E)\ & \dfrac{1}{9} \end{align}$

Pembahasan:

Berikut beberapa sifat aljabar yang akan membantu dalam menyelesaikan soal limit fungsi di atas:

  • $\left(\sqrt{a} + \sqrt{b} \right) \left(\sqrt{a} – \sqrt{b} \right)=a-b$
  • $\left(\sqrt{a} + b \right) \left(\sqrt{a} – b \right)=a-b^{2}$
  • $\left(a + \sqrt{b} \right) \left( a – \sqrt{b} \right)=a^{2}-b$
  • $\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \left( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right)=a-b$

$\begin{align} & \lim\limits_{x \to 1} \dfrac{\sqrt[3]{x^{2}}-2\sqrt[3]{x}+1}{\left(x-1\right)^{2}} \\ & = \lim\limits_{x \to 1} \dfrac{\left( \sqrt[3]{x}-1 \right)^{2}}{\left(x-1\right)^{2}} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ \sqrt[3]{x}-1 }{ x-1 } \right)^{2} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ \sqrt[3]{x}-1 }{ x-1 } \times \dfrac{\sqrt[3]{x^{2}}+\sqrt[3]{x}+\sqrt[3]{1}}{\sqrt[3]{x^{2}}+\sqrt[3]{x}+\sqrt[3]{1}} \right)^{2} \\ & = \lim\limits_{x \to 1} \left( \dfrac{ x-1 }{ x-1 } \times \dfrac{1}{\sqrt[3]{x^{2}}+\sqrt[3]{x}+\sqrt[3]{1}} \right)^{2} \\ & = \lim\limits_{x \to 1} \left( \dfrac{1}{\sqrt[3]{x^{2}}+\sqrt[3]{x}+\sqrt[3]{1}} \right)^{2} \\ & = \left( \dfrac{1}{\sqrt[3]{(1)^{2}}+\sqrt[3]{1}+\sqrt[3]{1}} \right)^{2} \\ & = \left( \dfrac{1}{1+1+1} \right)^{2} =\dfrac{1}{9} \end{align}$

$\therefore$ Jawaban yang benar $(E)\ \dfrac{1}{9}$

103. Soal SNMPTN 2010 Kode 528

Diketahui fungsi $g$ kontinu di $x=3$ dan $\lim\limits_{x \to 3} g(x)=2$. Nilai $\lim\limits_{x \to 3} \left( g(x) \cdot \dfrac{x-3}{\sqrt{x}-\sqrt{3}} \right)$ adalah…

$\begin{align} (A)\ & 4\sqrt{3} \\ (B)\ & 2\sqrt{3} \\ (C)\ & \sqrt{3} \\ (D)\ & 4 \\ (E)\ & 2 \end{align}$

Pembahasan:

Sebuah fungsi $f$ dikatakan kontinu dititik $a$ jika memenuhi ketiga syarat berikut:

  • $f(a)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x) = f(a)$

$ \begin{align} & \lim\limits_{x \to 3} \left( g(x) \cdot \dfrac{x-3}{\sqrt{x}-\sqrt{3}} \right) \\ & = \lim\limits_{x \to 3} g(x) \cdot \lim\limits_{x \to 3} \left( \dfrac{x-3}{\sqrt{x}-\sqrt{3}} \right) \\ & = \left( 2 \right) \cdot \lim\limits_{x \to 3} \left( \dfrac{x-3}{\sqrt{x}-\sqrt{3}} \cdot \dfrac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}} \right) \\ & = \left( 2 \right) \cdot \lim\limits_{x \to 3} \left( \dfrac{\left( x-3 \right) \left( \sqrt{x}+\sqrt{3} \right)}{x-3} \right) \\ & = \left( 2 \right) \cdot \lim\limits_{x \to 3} \dfrac{\left( \sqrt{x}+\sqrt{3} \right)}{1} \\ & = \left( 2 \right) \cdot \left( \sqrt{3}+\sqrt{3} \right) = 4\sqrt{3} \end{align} $

$ \therefore $ Jawaban yang benar adalah $(A)\ 4\sqrt{3}$

104. Soal SNMPTN 2011 Kode 678

Diketahui suku banyak $g(x)=ax^{2}+(a-b)x+a$ habis dibagi $x-1$. Jika $\lim\limits_{x \to 1} \dfrac{g(x)}{x^{2}-2x+1}=\dfrac{1}{3}$, maka nilai $a+b$ adalah…

$\begin{align} (A)\ & \dfrac{4}{3} \\ (B)\ & \dfrac{3}{2} \\ (C)\ & 0 \\ (D)\ & \dfrac{2}{3} \\ (E)\ & \dfrac{3}{4} \end{align}$

Pembahasan:

Sedikit catatan dari suku banyak (polinomial), jika suku banyak banyak $g(x)$ habis dibagi $x-a$ maka $g(a)=0$.

Sehingga untuk $g(x)=ax^{2}+(a-b)x+a$ habis dibagi $x-1$ kita peroleh:
$ \begin{align} g(x) &= ax^{2}+(a-b)x+a \\ g(1) &= a(1)^{2}+(a-b)(1)+a \\ 0 &= a + a-b +a \\ 0 &= 3a -b \\ b &= 3a \end{align} $

$ \begin{align} \lim\limits_{x \to 1} \dfrac{g(x)}{x^{2}-2x+1} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{ax^{2}+(a-b)x+a}{\left(x-1 \right)\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{\left(x-1 \right)\left(ax+2a-b \right)+\left( 3a-b \right)}{\left(x-1 \right)\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{\left(x-1 \right)\left( ax+2a-b \right)}{\left(x-1 \right)\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{\left( ax+2a-b \right)}{\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{\left( ax+2a-3a \right)}{\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{\left( ax-a \right)}{\left(x-1 \right)} &= \dfrac{1}{3} \\ \lim\limits_{x \to 1} \dfrac{a\left( x-1 \right)}{\left(x-1 \right)} &= \dfrac{1}{3} \\ a &= \dfrac{1}{3} \end{align} $

Untuk $a = \dfrac{1}{3}$ kita peroleh $b = 1$ sehingga nilai $a+b= \dfrac{1}{3}+1=\dfrac{4}{3}$

$ \therefore $ Jawaban yang benar adalah $(A)\ \dfrac{4}{3}$

105. Soal SNMPTN 2011 Kode 678

Jika $\lim\limits_{x \to 0} \dfrac{g(x)}{x}=\dfrac{1}{2}$, maka nilai $\lim\limits_{x \to 0} \dfrac{g(x)}{\sqrt{1-x}-1}$ adalah…

$\begin{align} (A)\ & -4 \\ (B)\ & -2 \\ (C)\ & -1 \\ (D)\ & 2 \\ (E)\ & 4 \end{align}$

Pembahasan:

$ \begin{align} & \lim\limits_{x \to 0} \dfrac{g(x)}{\sqrt{1-x}-1} \\ &= \lim\limits_{x \to 0} \dfrac{g(x)}{\sqrt{1-x}-1} \times \dfrac{\sqrt{1-x}+1}{\sqrt{1-x}+1} \\ &= \lim\limits_{x \to 0} \dfrac{g(x) \left( \sqrt{1-x}+1 \right)}{1-x-1} \\ &= \lim\limits_{x \to 0} \dfrac{g(x) \left( \sqrt{1-x}+1 \right)}{-x} \\ &= \lim\limits_{x \to 0} \dfrac{g(x)}{x} \cdot \dfrac{\left( \sqrt{1-x}+1 \right)}{-1} \\ &= \dfrac{1}{2} \cdot \dfrac{\left( \sqrt{1-0}+1 \right)}{-1}=-1 \end{align} $

$ \therefore $ Jawaban yang benar adalah $(C)\ -1$

106. Soal Simulasi UTBK SBMPTN

Jika $\lim\limits_{x \to 2} \dfrac{\sqrt{ax+b}-2}{x^{2}-4}=\dfrac{3}{16}$, maka nilai $a+b=\cdots$

$\begin{align} (A)\ & 4 \\ (B)\ & 3 \\ (C)\ & 2 \\ (D)\ & 1 \\ (E)\ & 0 \end{align}$

Pembahasan:

Nilai $\lim\limits_{x \to 2} \dfrac{\sqrt{ax+b}-2}{x^{2}-4}=\dfrac{3}{16}$. Jika kita substitusi langsung nilai $x=2$ maka nilai $\sqrt{ax+b}-2$ harus $0$, karena jika $\sqrt{ax+b}-2$ tidak nol maka nilai limit adalah $\infty$. sehingga untuk $x=2$ berlaku:

$\begin{align} \sqrt{ax+b}-2 & = 0 \\ \sqrt{2a+b}-2 & = 0 \\ \sqrt{2a+b} & = 2 \\ 2a+b & = 4 \\ \end{align}$ Lalu dengan mengalikan akar sekawan maka akan kita peroleh:

$\begin{align} \lim\limits_{x \to 2} \dfrac{\sqrt{ax+b}-2}{x^{2}-4} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{\sqrt{ax+b}-2}{x^{2}-4} \times \dfrac{\sqrt{ax+b}+2}{\sqrt{ax+b}+2} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{ ax+b -4}{\left( x-2 \right)\left( x+2 \right)\left( \sqrt{ax+b}+2 \right)} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{ a\left( x-2 \right)+ 2a+b-4}{\left( x-2 \right)\left( x+2 \right)\left( \sqrt{ax+b}+2 \right)} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{ a\left( x-2 \right)+ 4-4}{\left( x-2 \right)\left( x+2 \right)\left( \sqrt{ax+b}+2 \right)} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{ a\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)\left( \sqrt{ax+b}+2 \right)} &= \dfrac{3}{16} \\ \lim\limits_{x \to 2} \dfrac{ a }{ \left( x+2 \right)\left( \sqrt{ax+b}+2 \right)} &= \dfrac{3}{16} \\ \dfrac{ a }{ \left( 2+2 \right)\left( \sqrt{2a+b}+2 \right)} &= \dfrac{3}{16} \\ \dfrac{ a }{ \left( 4 \right)\left( 4 \right)} &= \dfrac{3}{16} \\ a &= 3 \\ \end{align}$ Untuk $a=3$ dan $2a+b=4$ maka $b=-2$. Nilai $a+b=1$

Peringatan!
Jika sudah belajar turunan fungsi, maka dapat digunakan Aturan L’Hospital yang mungkin dapat menghemat beberapa langkah.

$\therefore$ Jawaban yang benar $(D)\ 1$

107. Soal Simulasi UTBK SBMPTN

Diketahui $f(x)=\begin{cases}\dfrac{x}{x-2},\ x \neq 2 \\ 2,\ x = 2 \end{cases}$, Semua pernyataan berikut adalah benar, kecuali…

$\begin{align} (A)\ & \lim\limits_{x \to 2} f(x)= 1 \\ (B)\ & \lim\limits_{x \to 2} f(x) \neq f(2) \\ (C)\ & f\ \text{tidak mempunyai turunan di}\ x=2 \\ (D)\ & f\ \text{tidak kontinu di}\ x=2 \\ (E)\ & f\ \text{kontinu di}\ x=0 \end{align}$

Pembahasan:

Sebuah fungsi $f$ dikatakan kontinu dititik $a$ jika memenuhi ketiga syarat berikut:

  • $f(a)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x) = f(a)$

Untuk menentukan jawaban, kita coba tentukan kebenaran dari setiap pilihan:

  • $(A)\ \lim\limits_{x \to 2} f(x)= 1$ salah, karena $\lim\limits_{x \to 2} f(x)$ tidak ada, $\lim\limits_{x \to 2^{+}} f(x)= \infty $ dan $\lim\limits_{x \to 2^{-}} f(x)= – \infty $
  • $(B)\ \lim\limits_{x \to 2} f(x) \neq f(2)$ benar, karena $f(2)=2$
  • $(C)\ f\ \text{tidak mempunyai turunan di}\ x=2$ benar, karena $f$ tidak kontinu pada $x=2$ Jika suatu fungsi tidak kontinu pada $x = c$, maka fungsi tersebut tidak memiliki turunan di $x = c$
  • $(D)\ f\ \text{tidak kontinu di}\ x=2$ benar, karena karena $\lim\limits_{x \to 2} f(x)$ tidak ada
  • $(E)\ f\ \text{kontinu di}\ x=0$ benar, karena $\lim\limits_{x \to 0} f(x) = f(0)=0$

$\therefore$ Jawaban yang benar $(A)\ \lim\limits_{x \to 2} f(x)= 1$

108. Soal SNMPTN 2010 Kode 528

Diketahui fungsi $f$ dengan $f(x)=\begin{cases}\dfrac{x^{2}-1}{x-1},\ x \neq 1 \\ 3,\ x = 1 \end{cases}$, Semua pernyataan berikut benar, kecuali…

$\begin{align} (A)\ & \lim\limits_{x \to 1} f(x)= 2 \\ (B)\ & \lim\limits_{x \to 1} f(x) \neq f(1) \\ (C)\ & f\ \text{mempunyai turunan di}\ x=1 \\ (D)\ & f\ \text{tidak kontinu di}\ x=1 \\ (E)\ & f\ \text{kontinu di}\ x=0 \end{align}$

Pembahasan:

Sebuah fungsi $f$ dikatakan kontinu dititik $a$ jika memenuhi ketiga syarat berikut:

  • $f(a)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x)$ mempunyai nilai (ada)
  • $\lim\limits_{x \to a} f(x) = f(a)$

Untuk menentukan jawaban, kita coba tentukan kebenaran dari setiap pilihan:

  • $(A)\ \lim\limits_{x \to 1} f(x)= 2$ benar,

    $\begin{align} \lim\limits_{x \to 1} f(x) &= \lim\limits_{x \to 1} \dfrac{x^{2}-1}{x-1} \\ &= \lim\limits_{x \to 1} \dfrac{\left( x+1 \right)}{1} \\ &= \dfrac{\left( 1+1 \right)}{1}=2 \end{align}$

  • $(B)\ \lim\limits_{x \to 1} f(x) \neq f(1)$ benar, karena $\lim\limits_{x \to 1} f(x)= 2$ dan $f(1)=3$
  • $(C)\ f\ \text{mempunyai turunan di}\ x=1$ salah, karena $f\ \text{tidak kontinu di}\ x=1$
  • $(D)\ f\ \text{tidak kontinu di}\ x=1$ benar, karena ketiga syarat sebuah fungsi dikatakan kontinu ketiganya tidak dipenuhi.
  • $(E)\ f\ \text{kontinu di}\ x=0$ benar, ketiga syarat sebuah fungsi dikatakan kontinu ketiganya dipenuhi.

$\therefore$ Jawaban yang benar $(C)\ f\ \text{mempunyai turunan di}\ x=1$

109. Soal SBMPTN 2014 Kode 532

Jika $\lim\limits_{x \to a} \left( f(x)+ \dfrac{1}{g(x)} \right)=4$ dan $\lim\limits_{x \to a} \left( f(x)- \dfrac{1}{g(x)} \right)=-3$, maka $\lim\limits_{x \to a} f(x)g(x) =\cdots$

$\begin{align} (A)\ & \dfrac{1}{14} \\ (B)\ & \dfrac{2}{14} \\ (C)\ & \dfrac{3}{14} \\ (D)\ & \dfrac{4}{14} \\ (E)\ & \dfrac{5}{14} \end{align}$

Pembahasan:

$\begin{align} \lim\limits_{x \to a} \left( f(x)+ \dfrac{1}{g(x)} \right) & = 4 \\ \lim\limits_{x \to a} \left( f(x)- \dfrac{1}{g(x)} \right) & = -3 \\ \hline \lim\limits_{x \to a} f(x) + \lim\limits_{x \to a} \dfrac{1}{g(x)} & = 4 \\ \lim\limits_{x \to a} f(x) – \lim\limits_{x \to a} \dfrac{1}{g(x)} & = -3\ \ \ (+) \\ \hline 2\lim\limits_{x \to a} f(x) & = 1 \\ \lim\limits_{x \to a} f(x) & = \dfrac{1}{2} \end{align}$

Untuk $\lim\limits_{x \to a} f(x) = \dfrac{1}{2}$ maka kita peroleh $\lim\limits_{x \to a} \dfrac{1}{g(x)}=\dfrac{7}{2}$ atau $\lim\limits_{x \to a} g(x) =\dfrac{2}{7}$.

Nilai $\lim\limits_{x \to a} f(x)g(x)= \dfrac{1}{2} \cdot \dfrac{2}{7} = \dfrac{2}{14}$

$\therefore$ Jawaban yang benar $(B)\ \dfrac{2}{14}$

110. Soal SBMPTN 2014 Kode 514

Jika $\lim\limits_{x \to a} \left( f(x)+ \dfrac{1}{g(x)} \right)=4$ dan $\lim\limits_{x \to a} \left( f(x)- \dfrac{1}{g(x)} \right)=-3$, maka $\lim\limits_{x \to a} \left( f^{2}(x)+\dfrac{1}{g^{2}(x)} \right) =\cdots$

$\begin{align} (A)\ & \dfrac{24}{3} \\ (B)\ & \dfrac{23}{5} \\ (C)\ & \dfrac{25}{3} \\ (D)\ & \dfrac{25}{2} \\ (E)\ & \dfrac{27}{2} \end{align}$

Pembahasan:

$\begin{align} \lim\limits_{x \to a} \left( f(x)+ \dfrac{1}{g(x)} \right) & = 4 \\ \lim\limits_{x \to a} \left( f(x)- \dfrac{1}{g(x)} \right) & = -3 \\ \hline \lim\limits_{x \to a} f(x) + \lim\limits_{x \to a} \dfrac{1}{g(x)} & = 4 \\ \lim\limits_{x \to a} f(x) – \lim\limits_{x \to a} \dfrac{1}{g(x)} & = -3\ \ \ (+) \\ \hline 2\lim\limits_{x \to a} f(x) & = 1 \\ \lim\limits_{x \to a} f(x) & = \dfrac{1}{2} \end{align}$

Untuk $\lim\limits_{x \to a} f(x) = \dfrac{1}{2}$ maka kita peroleh $\lim\limits_{x \to a} \dfrac{1}{g(x)}=\dfrac{7}{2}$.

$\begin{align} \text{Nilai}\ \lim\limits_{x \to a} \left( f^{2}(x)+\dfrac{1}{g^{2}(x)} \right) &= \left( \dfrac{1}{2} \right)^{2}+\left( \dfrac{7}{2} \right)^{2} \\ & = \dfrac{1}{4}+\dfrac{49}{4} \\ & =\dfrac{50}{4}=\dfrac{25}{2} \end{align}$

$\therefore$ Jawaban yang benar $(D)\ \dfrac{25}{2}$