Exercise Zone : Limit [2]

Berikut ini adalah kumpulan soal dan pembahasan mengenai limit tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 11

Nilai $\displaystyle\lim_{x\to0}\dfrac{2x}{x+2}=$ ...

1. $0$
2. $1$
   
3. $3$
   

4. $9$
5. $\infty$
   

$\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{2x}{x+2}&=\dfrac{2(0)}{0+2}\\[8pt] &=\dfrac02\\ &=\boxed{\boxed{0}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to0}\dfrac{2x}{x+2}=0$.

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No. 12

Jika $f(x)=\dfrac{3-\sqrt{2x+9}}x$, maka $\displaystyle\lim_{x\to0}f(x)=$

1. $-\dfrac13$
2. $6$
3. $\dfrac23$

4. $12$
5. $\dfrac12$

CARA 1 : KALI SEKAWAN

$\begin{aligned} \displaystyle\lim_{x\to0}f(x)&=\displaystyle\lim_{x\to0}\dfrac{3-\sqrt{2x+9}}x\cdot\dfrac{3+\sqrt{2x+9}}{3+\sqrt{2x+9}}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{9-(2x+9)}{x(3+\sqrt{2x+9})}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{9-2x-9}{x(3+\sqrt{2x+9})}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{-2x}{x(3+\sqrt{2x+9})}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{-2}{3+\sqrt{2x+9}}\\[8pt] &=\dfrac{-2}{3+\sqrt{2(0)+9}}\\[8pt] &=\dfrac{-2}{3+\sqrt{0+9}}\\[8pt] &=\dfrac{-2}{3+\sqrt9}\\[8pt] &=\dfrac{-2}{3+3}\\[8pt] &=\dfrac{-2}6\\ &=\boxed{\boxed{-\dfrac13}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} \displaystyle\lim_{x\to0}f(x)&=\displaystyle\lim_{x\to0}\dfrac{3-\sqrt{2x+9}}x\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{0-\dfrac2{2\sqrt{2x+9}}}1\\[8pt] &=\displaystyle\lim_{x\to0}-\dfrac1{\sqrt{2x+9}}\\[8pt] &=-\dfrac1{\sqrt{2(0)+9}}\\[8pt] &=-\dfrac1{\sqrt{0+9}}\\[8pt] &=-\dfrac1{\sqrt9}\\ &=\boxed{\boxed{-\dfrac13}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to0}f(x)=-\dfrac13$.
JAWAB: A

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No. 13

Nilai dari $\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}$ adalah

1. $1$
2. $2$
   
3. $3$
   

4. $4$
5. $5$
   

CARA 1

$\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}&=\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}\color{red}{\cdot\dfrac{\sqrt{2x-1}+\sqrt{x}}{\sqrt{2x-1}+\sqrt{x}}}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{(2x-2)\left(\sqrt{2x-1}+\sqrt{x}\right)}{2x-1-x}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{2\cancel{(x-1)}\left(\sqrt{2x-1}+\sqrt{x}\right)}{\cancel{x-1}}\\[8pt] &=\displaystyle\lim_{x\to1}2\left(\sqrt{2x-1}+\sqrt{x}\right)\\ &=2\left(\sqrt{2(1)-1}+\sqrt{1}\right)\\ &=2\left(\sqrt{2-1}+1\right)\\ &=2\left(\sqrt1+1\right)\\ &=2\left(1+1\right)\\ &=2\left(2\right)\\ &=\boxed{\boxed{4}} \end{aligned}$

CARA 2

$\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}&=\displaystyle\lim_{x\to1}\dfrac2{\dfrac2{2\sqrt{2x-1}}-\dfrac1{2\sqrt{x}}}\\[21pt] &=\dfrac2{\dfrac2{2\sqrt{2(1)-1}}-\dfrac1{2\sqrt1}}\\[21pt] &=\dfrac2{\dfrac2{2\sqrt{2-1}}-\dfrac1{2(1)}}\\[21pt] &=\dfrac2{\dfrac2{2\sqrt1}-\dfrac12}\\[21pt] &=\dfrac2{\dfrac2{2(1)}-\dfrac12}\\[21pt] &=\dfrac2{\dfrac22-\dfrac12}\\[21pt] &=\dfrac2{\dfrac12}\\ &=\boxed{\boxed{4}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}=4$.
JAWAB: D

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No. 14

Diektahui ${f(x)=\sqrt{1+x}}$. Nilai $\displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}$ adalah

1. $0$
2. $\dfrac67$
3. $\dfrac98$

4. $\dfrac23$
5. $\dfrac54$

CARA 1

$\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\sqrt{1+3+2h^2}-\sqrt{1+3-2h^2}}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\sqrt{4+2h^2}-\sqrt{4-2h^2}}{h^2}\color{red}{\cdot\dfrac{\sqrt{4+2h^2}+\sqrt{4-2h^2}}{\sqrt{4+2h^2}+\sqrt{4-2h^2}}}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\left(4+2h^2\right)-\left(4-2h^2\right)}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{4+2h^2-4+2h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{4h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac4{\sqrt{4+2h^2}+\sqrt{4-2h^2}}\\[8pt] &=\dfrac4{\sqrt{4+2(0)^2}+\sqrt{4-2(0)^2}}\\[8pt] &=\dfrac4{\sqrt{4+0}+\sqrt{4-0}}\\[8pt] &=\dfrac4{\sqrt4+\sqrt4}\\[8pt] &=\dfrac4{2+2}\\[8pt] &=\dfrac44\\ &=\boxed{\boxed{1}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} f(x)&=\sqrt{1+x}\\ f'(x)&=\dfrac1{2\sqrt{1+x}} \end{aligned}$

$\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{4h\ f'\left(3+2h^2\right)-(-4h)f'\left(3-2h^2\right)}{2h}\\[8pt] &=\displaystyle\lim_{h\to0}\left(2f'\left(3+2h^2\right)+2f'\left(3-2h^2\right)\right)\\ &=2f'\left(3+2(0)^2\right)+2f'\left(3-2(0)^2\right)\\ &=2f'(3)+2f'(3)\\ &=4f'(3)\\ &=4\left(\dfrac1{2\sqrt{1+3}}\right)\\[8pt] &=\dfrac4{2\sqrt4}\\[8pt] &=\dfrac4{2(2)}\\[8pt] &=\dfrac44\\ &=\boxed{\boxed{1}} \end{aligned}$

Jadi, $\displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}=1$.
JAWAB: -

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No. 15

Nilai dari $\displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x}$ adalah

1. $3$
2. $-3$
   
3. $-2$
   

4. $2$
5. $0$
   

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x}&=\displaystyle\lim_{x\to3}\dfrac{(x+6)\cancel{(x-3)}}{x\cancel{(x-3)}}\\[8pt] &=\displaystyle\lim_{x\to3}\dfrac{x+6}x\\[8pt] &=\dfrac{3+6}3\\[8pt] &=\dfrac93\\ &=\boxed{\boxed{3}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x}=3$.
JAWAB: A

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No. 16

Diketahui ${\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}=\dfrac14}$, nilai $a$ adalah

1. $0$
2. $-4$
3. $4$

4. $-2$
5. $2$

$\begin{aligned} \displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}&=\dfrac14\\[8pt] \displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}\color{red}{\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2}}&=\dfrac14\\[8pt] \displaystyle\lim_{x\to a}\dfrac{\cancel{x-4}}{\cancel{(x-4)}\left(\sqrt{x}+2\right)}&=\dfrac14\\[8pt] \displaystyle\lim_{x\to a}\dfrac1{\sqrt{x}+2}&=\dfrac14\\[8pt] \dfrac1{\sqrt{a}+2}&=\dfrac14\\[8pt] \sqrt{a}+2&=4\\ \sqrt{a}&=2\\ a&=4 \end{aligned}$

Jadi, $a=4$.
JAWAB: C

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No. 17

Jika $\displaystyle\lim_{\theta\to-1}f(\theta)$ ada dan ${\dfrac{\theta^2+\theta-2}{\theta+3}\leq\dfrac{f(\theta)}{\theta+3}\leq\dfrac{\theta^2+2\theta-1}{\theta+3}}$, tentukan $\displaystyle\lim_{\theta\to-1}f(\theta)$!

$\begin{array}{rcccll} \dfrac{\theta^2+\theta-2}{\theta+3}&\leq&\dfrac{f(\theta)}{\theta+3}&\leq&\dfrac{\theta^2+2\theta-1}{\theta+3}\\[8pt] \displaystyle\lim_{\theta\to-1}\dfrac{\theta^2+\theta-2}{\theta+3}&\leq&\displaystyle\lim_{\theta\to-1}\dfrac{f(\theta)}{\theta+3}&\leq&\displaystyle\lim_{\theta\to-1}\dfrac{\theta^2+2\theta-1}{\theta+3}\\[8pt] \dfrac{(-1)^2+(-1)-2}{-1+3}&\leq&\dfrac{\displaystyle\lim_{\theta\to-1}f(\theta)}{-1+3}&\leq&\dfrac{(-1)^2+2(-1)-1}{-1+3}\\[8pt] \dfrac{-2}2&\leq&\dfrac{\displaystyle\lim_{\theta\to-1}f(\theta)}2&\leq&\dfrac{-2}{2}&\quad\color{red}{\times2}\\[8pt] -2&\leq&\displaystyle\lim_{\theta\to-1}f(\theta)&\leq&-2 \end{array}$
$\displaystyle\lim_{\theta\to-1}f(\theta)=-2$

Jadi, $\displaystyle\lim_{\theta\to-1}f(\theta)=-2$.

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No. 18

Jika ${\displaystyle\lim_{x\to2}\dfrac{x^2-3x+a}{x-2}=1}$ maka nilai $a$ adalah ....

1. $-2$
2. $-1$
   
3. $0$
   

4. $1$
5. $2$
   

dari Dyah Mesha

Jika ${f(x)=x^2-3x+a}$, maka ${f(2)=0}$.
$\begin{aligned} 2^2-3(2)+a&=0\\ 4-6+a&=0\\ -2+a&=0\\ a&=\boxed{\boxed{2}} \end{aligned}$

Jadi, $a=2$.

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