Exercise Zone : Limit [2]

Berikut ini adalah kumpulan soal dan pembahasan mengenai limit tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

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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
   

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2x}{x+2}}}& ={\displaystyle \frac{2(0)}{0+2}}\\ & ={\displaystyle \frac{0}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 0}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}f(x)}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3-\sqrt{2x+9}}{x}}\cdot {\displaystyle \frac{3+\sqrt{2x+9}}{3+\sqrt{2x+9}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{9-(2x+9)}{x(3+\sqrt{2x+9})}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{9-2x-9}{x(3+\sqrt{2x+9})}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{-2x}{x(3+\sqrt{2x+9})}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{-2}{3+\sqrt{2x+9}}}}\\ & ={\displaystyle \frac{-2}{3+\sqrt{2(0)+9}}}\\ & ={\displaystyle \frac{-2}{3+\sqrt{0+9}}}\\ & ={\displaystyle \frac{-2}{3+\sqrt{9}}}\\ & ={\displaystyle \frac{-2}{3+3}}\\ & ={\displaystyle \frac{-2}{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{3}}}}}}\end{array}$

CARA 2 : L'HOPITAL

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}f(x)}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3-\sqrt{2x+9}}{x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{0-{\displaystyle \frac{2}{2\sqrt{2x+9}}}}{1}}}\\ & ={\displaystyle \underset{x\to 0}{lim}-{\displaystyle \frac{1}{\sqrt{2x+9}}}}\\ & =-{\displaystyle \frac{1}{\sqrt{2(0)+9}}}\\ & =-{\displaystyle \frac{1}{\sqrt{0+9}}}\\ & =-{\displaystyle \frac{1}{\sqrt{9}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{3}}}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{2x-2}{\sqrt{2x-1}-\sqrt{x}}}}& ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{2x-2}{\sqrt{2x-1}-\sqrt{x}}}\cdot {\displaystyle \frac{\sqrt{2x-1}+\sqrt{x}}{\sqrt{2x-1}+\sqrt{x}}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{(2x-2)(\sqrt{2x-1}+\sqrt{x})}{2x-1-x}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{2\cancel{(x-1)}(\sqrt{2x-1}+\sqrt{x})}{\cancel{x-1}}}}\\ & ={\displaystyle \underset{x\to 1}{lim}2(\sqrt{2x-1}+\sqrt{x})}\\ & =2(\sqrt{2(1)-1}+\sqrt{1})\\ & =2(\sqrt{2-1}+1)\\ & =2(\sqrt{1}+1)\\ & =2(1+1)\\ & =2\left(2\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

CARA 2

$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{2x-2}{\sqrt{2x-1}-\sqrt{x}}}}& ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{2}{{\displaystyle \frac{2}{2\sqrt{2x-1}}}-{\displaystyle \frac{1}{2\sqrt{x}}}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{2}{2\sqrt{2(1)-1}}}-{\displaystyle \frac{1}{2\sqrt{1}}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{2}{2\sqrt{2-1}}}-{\displaystyle \frac{1}{2(1)}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{2}{2\sqrt{1}}}-{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{2}{2(1)}}-{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{2}{2}}-{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{2}{{\displaystyle \frac{1}{2}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(3+2h^2)-f(3-2h^2)}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{1+3+2h^2}-\sqrt{1+3-2h^2}}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{4+2h^2}-\sqrt{4-2h^2}}{h^2}}\cdot {\displaystyle \frac{\sqrt{4+2h^2}+\sqrt{4-2h^2}}{\sqrt{4+2h^2}+\sqrt{4-2h^2}}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(4+2h^2)-(4-2h^2)}{h^2(\sqrt{4+2h^2}+\sqrt{4-2h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4+2h^2-4+2h^2}{h^2(\sqrt{4+2h^2}+\sqrt{4-2h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4h^2}{h^2(\sqrt{4+2h^2}+\sqrt{4-2h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4}{\sqrt{4+2h^2}+\sqrt{4-2h^2}}}}\\ & ={\displaystyle \frac{4}{\sqrt{4+2(0{)}^2}+\sqrt{4-2(0{)}^2}}}\\ & ={\displaystyle \frac{4}{\sqrt{4+0}+\sqrt{4-0}}}\\ & ={\displaystyle \frac{4}{\sqrt{4}+\sqrt{4}}}\\ & ={\displaystyle \frac{4}{2+2}}\\ & ={\displaystyle \frac{4}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

CARA 2 : L'HOPITAL

$\begin{array}{rl}f(x)& =\sqrt{1+x}\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2\sqrt{1+x}}}\end{array}$

$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(3+2h^2)-f(3-2h^2)}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4h{f}^{\prime }(3+2h^2)-(-4h)f^{\prime }(3-2h^2)}{2h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}(2f^{\prime }(3+2h^2)+2f^{\prime }(3-2h^2))}\\ & =2f^{\prime }(3+2(0{)}^2)+2f^{\prime }(3-2(0{)}^2)\\ & =2f^{\prime }(3)+2f^{\prime }(3)\\ & =4f^{\prime }(3)\\ & =4\left({\displaystyle \frac{1}{2\sqrt{1+3}}}\right)\\ & ={\displaystyle \frac{4}{2\sqrt{4}}}\\ & ={\displaystyle \frac{4}{2(2)}}\\ & ={\displaystyle \frac{4}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

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
   

$\begin{array}{rl}{\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{x^2+3x-18}{x^2-3x}}}& ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(x+6)\cancel{(x-3)}}{x\cancel{(x-3)}}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{x+6}{x}}}\\ & ={\displaystyle \frac{3+6}{3}}\\ & ={\displaystyle \frac{9}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}{\displaystyle \frac{\sqrt{x}-2}{x-4}}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \underset{x\to a}{lim}{\displaystyle \frac{\sqrt{x}-2}{x-4}}\cdot {\displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2}}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \underset{x\to a}{lim}{\displaystyle \frac{\cancel{x-4}}{\cancel{(x-4)}(\sqrt{x}+2)}}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \underset{x\to a}{lim}{\displaystyle \frac{1}{\sqrt{x}+2}}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{\sqrt{a}+2}}& ={\displaystyle \frac{1}{4}}\\ \sqrt{a}+2& =4\\ \sqrt{a}& =2\\ a& =4\end{array}$

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)!

$\begin{array}{rcccl}{\displaystyle \frac{{\theta }^2+\theta -2}{\theta +3}}& \le & {\displaystyle \frac{f(\theta )}{\theta +3}}& \le & {\displaystyle \frac{{\theta }^2+2\theta -1}{\theta +3}}\\ {\displaystyle \underset{\theta \to -1}{lim}{\displaystyle \frac{{\theta }^2+\theta -2}{\theta +3}}}& \le & {\displaystyle \underset{\theta \to -1}{lim}{\displaystyle \frac{f(\theta )}{\theta +3}}}& \le & {\displaystyle \underset{\theta \to -1}{lim}{\displaystyle \frac{{\theta }^2+2\theta -1}{\theta +3}}}\\ {\displaystyle \frac{(-1{)}^2+(-1)-2}{-1+3}}& \le & {\displaystyle \frac{{\displaystyle \underset{\theta \to -1}{lim}f(\theta )}}{-1+3}}& \le & {\displaystyle \frac{(-1{)}^2+2(-1)-1}{-1+3}}\\ {\displaystyle \frac{-2}{2}}& \le & {\displaystyle \frac{{\displaystyle \underset{\theta \to -1}{lim}f(\theta )}}{2}}& \le & {\displaystyle \frac{-2}{2}}& \times 2\\ -2& \le & {\displaystyle \underset{\theta \to -1}{lim}f(\theta )}& \le & -2\end{array}$
\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}.
$\begin{array}{rl}{2}^2-3(2)+a& =0\\ 4-6+a& =0\\ -2+a& =0\\ a& =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, a=2.

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No. 19

{\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}=} ....

1. 0
2. \dfrac12
   
3. 1
   

4. 2
5. \infty
   

dari Dyah Mesha

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{x}-x}{\sqrt{x}+x}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{x}-x}{\sqrt{x}+x}}\cdot {\displaystyle \frac{\sqrt{x}+x}{\sqrt{x}+x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x-x^2}{x+2x\sqrt{x}+x^2}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x(1-x)}{x(1+2\sqrt{x}+x)}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1-x}{1+2\sqrt{x}+x}}}\\ & ={\displaystyle \frac{1-0}{1+2\sqrt{0}+0}}\\ & ={\displaystyle \frac{1}{1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}=1.
JAWAB: C

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No. 20

\displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}=

Telegram

$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^2+5x-14}{2x^2-x-6}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x+7)(x-2)}{(2x+3)(x-2)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x+7}{2x+3}}}\\ & ={\displaystyle \frac{2+7}{2(2)+3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{9}{7}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}=\dfrac97.

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