Exercise Zone : Limit

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

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No. 1

$\displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}=$ ....

SUMBER

CARA 1

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}&=\displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}\cdot\dfrac{3+\sqrt{2x+3}}{3+\sqrt{2x+3}}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{9-(2x+3)}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{9-2x-3}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{6-2x}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{-2(x-3)}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{-2}{3+\sqrt{2x+3}}\\[9pt] &=\dfrac{-2}{3+\sqrt{2(3)+3}}\\[9pt] &=\dfrac{-2}{3+\sqrt{6+3}}\\[9pt] &=\dfrac{-2}{3+\sqrt9}\\[9pt] &=\dfrac{-2}{3+3}\\[9pt] &=\dfrac{-2}6\\[9pt] &=\boxed{\boxed{-\dfrac13}} \end{aligned}$

CARA 2

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}&=\displaystyle\lim_{x\to3}\dfrac{0-\dfrac2{2\sqrt{2x+3}}}{1-0}\\[9pt] &=\displaystyle\lim_{x\to3}\left(-\dfrac1{\sqrt{2x+3}}\right)\\[9pt] &=-\dfrac1{\sqrt{2(3)+3}}\\[9pt] &=-\dfrac1{\sqrt{6+3}}\\[9pt] &=-\dfrac1{\sqrt9}\\[9pt] &=\boxed{\boxed{-\dfrac13}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}=-\dfrac13$.

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No. 2

$\displaystyle\lim_{x\to3}\dfrac{\left(\sqrt{x-1}-\sqrt{5-x}\right)\left(\sqrt{4x-3}-1\right)}{x^2-9}=$....

1. $-\dfrac16\sqrt2$
2. $-\dfrac14\sqrt2$
3. $-\dfrac12\sqrt2$

4. $\dfrac16\sqrt2$
5. $\dfrac12\sqrt2$

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{\left(\sqrt{x-1}-\sqrt{5-x}\right)\left(\sqrt{4x-3}-1\right)}{x^2-9}&=\displaystyle\lim_{x\to3}\dfrac{\left(\sqrt{x-1}-\sqrt{5-x}\right)\left(\sqrt{4x-3}-1\right)}{x^2-9}\cdot\dfrac{\sqrt{x-1}+\sqrt{5-x}}{\sqrt{x-1}+\sqrt{5-x}}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{\left((x-1)-(5-x)\right)\left(\sqrt{4x-3}-1\right)}{\left(x^2-9\right)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{(x-1-5+x)\left(\sqrt{4x-3}-1\right)}{\left(x^2-9\right)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{(2x-6)\left(\sqrt{4x-3}-1\right)}{\left(x^2-9\right)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{2(x-3)\left(\sqrt{4x-3}-1\right)}{(x+3)(x-3)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[9pt] &=\displaystyle\lim_{x\to3}\dfrac{2\left(\sqrt{4x-3}-1\right)}{(x+3)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[9pt] &=\dfrac{2\left(\sqrt{4(3)-3}-1\right)}{(3+3)\left(\sqrt{3-1}+\sqrt{5-3}\right)}\\[9pt] &=\dfrac{2\left(\sqrt9-1\right)}{(3+3)\left(\sqrt2+\sqrt2\right)}\\[9pt] &=\dfrac{2(3-1)}{(6)\left(2\sqrt2\right)}\\[9pt] &=\dfrac{2}{6\left(\sqrt2\right)}\\[9pt] &=\dfrac1{3\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\[9pt] &=\dfrac1{3\cdot2}\sqrt2\\[9pt] &=\boxed{\boxed{\dfrac16\sqrt2}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to3}\dfrac{\left(\sqrt{x-1}-\sqrt{5-x}\right)\left(\sqrt{4x-3}-1\right)}{x^2-9}=\dfrac16\sqrt2$.
JAWAB: D

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No. 3

$\displaystyle\lim_{x\to2}\left(\dfrac6{x^2-x-2}-\dfrac2{x-2}\right)=$ ....

$\begin{aligned} \displaystyle\lim_{x\to2}\left(\dfrac6{x^2-x-2}-\dfrac2{x-2}\right)&=\displaystyle\lim_{x\to2}\left(\dfrac6{(x-2)(x+1)}-\dfrac2{x-2}\right)\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{6-2(x+1)}{(x-2)(x+1)}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{6-2x-2}{(x-2)(x+1)}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{-2x+4}{(x-2)(x+1)}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{-2\cancel{(x-2)}}{\cancel{(x-2)}(x+1)}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{-2}{x+1}\\[8pt] &=\dfrac{-2}{2+1}\\[8pt] &=-\dfrac23 \end{aligned}$

Jadi, $\displaystyle\lim_{x\to2}\left(\dfrac6{x^2-x-2}-\dfrac2{x-2}\right)=-\dfrac23$.

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No. 4

$\displaystyle\lim_{x\to-2}\dfrac{x^3-x^2-2x+8}{x^3+8}=$ ....

Ganesha Operation

CARA 1 : PEMFAKTORAN

$\begin{aligned} \displaystyle\lim_{x\to-2}\dfrac{x^3-x^2-2x+8}{x^3+8}&=\displaystyle\lim_{x\to-2}\dfrac{(x+2)\left(x^2-3x+4\right)}{(x+2)\left(x^2-2x+4\right)}\\[8pt] &=\displaystyle\lim_{x\to-2}\dfrac{x^2-3x+4}{x^2-2x+4}\\[8pt] &=\dfrac{(-2)^2-3(-2)+4}{(-2)^2-2(-2)+4}\\[8pt] &=\dfrac{4+6+4}{4+4+4}\\[8pt] &=\dfrac{14}{12}\\[8pt] &=\boxed{\boxed{\dfrac76}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} \displaystyle\lim_{x\to-2}\dfrac{x^3-x^2-2x+8}{x^3+8}&=\displaystyle\lim_{x\to-2}\dfrac{3x^2-2x-2}{3x^2}\\[8pt] &=\dfrac{3(-2)^2-2(-2)-2}{3(-2)^2}\\[8pt] &=\dfrac{12+4-2}{12}\\[8pt] &=\dfrac{14}{12}\\[8pt] &=\boxed{\boxed{\dfrac76}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to-2}\dfrac{x^3-x^2-2x+8}{x^3+8}=\dfrac76$.

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No. 5

Diketahui fungsi $g$ kontinu di $x=2$ dan $\displaystyle\lim_{x\to2}g(x)=4$. Nilai $\displaystyle\lim_{x\to2}\left(g(x)\dfrac{x-2}{\sqrt{x}-\sqrt2}\right)$ adalah ....

Ganesha Operation

$\begin{aligned} \displaystyle\lim_{x\to2}\left(g(x)\dfrac{x-2}{\sqrt{x}-\sqrt2}\right)&=\displaystyle\lim_{x\to2}\left(g(x)\dfrac{x-2}{\sqrt{x}-\sqrt2}\cdot\dfrac{\sqrt{x}+\sqrt2}{\sqrt{x}+\sqrt2}\right)\\[8pt] &=\displaystyle\lim_{x\to2}\left(g(x)\dfrac{(x-2)\left(\sqrt{x}+\sqrt2\right)}{x-2}\right)\\[8pt] &=\displaystyle\lim_{x\to2}\left(g(x)\left(\sqrt{x}+\sqrt2\right)\right)\\ &=4\left(\sqrt2+\sqrt2\right)\\ &=4\left(2\sqrt2\right)\\ &=\boxed{\boxed{8\sqrt2}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to2}\left(g(x)\dfrac{x-2}{\sqrt{x}-\sqrt2}\right)=8\sqrt2$.

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No. 5

Tentukan nilai dari $\displaystyle\lim_{x\to5}\dfrac{x^2-25}{\sqrt{x^2+24}-7}$!

Ganesha Operation

CARA 1 : KALIKAN DENGAN SEKAWAN

$\begin{aligned} \displaystyle\lim_{x\to5}\dfrac{x^2-25}{\sqrt{x^2+24}-7}&=\displaystyle\lim_{x\to5}\dfrac{x^2-25}{\sqrt{x^2+24}-7}\cdot\dfrac{\sqrt{x^2+24}+7}{\sqrt{x^2+24}+7}\\[9pt] &=\displaystyle\lim_{x\to5}\dfrac{\left(x^2-25\right)\left(\sqrt{x^2+24}+7\right)}{x^2+24-49}\\[9pt] &=\displaystyle\lim_{x\to5}\dfrac{\left(x^2-25\right)\left(\sqrt{x^2+24}+7\right)}{x^2-25}\\[9pt] &=\displaystyle\lim_{x\to5}\left(\sqrt{x^2+24}+7\right)\\ &=\sqrt{5^2+24}+7\\ &=\sqrt{25+24}+7\\ &=\sqrt{49}+7\\ &=7+7\\ &=\boxed{\boxed{14}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} \displaystyle\lim_{x\to5}\dfrac{x^2-25}{\sqrt{x^2+24}-7}&=\displaystyle\lim_{x\to5}\dfrac{2x}{\dfrac{2x}{2\sqrt{x^2+24}}}\\[9pt] &=\displaystyle\lim_{x\to5}2\sqrt{x^2+24}\\ &=2\sqrt{5^2+24}\\ &=2\sqrt{25+24}\\ &=2\sqrt{49}\\ &=2(7)\\ &=\boxed{\boxed{14}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to5}\dfrac{x^2-25}{\sqrt{x^2+24}-7}=14$.

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No. 6

Tentukan nilai dari $\displaystyle\lim_{x\to2}\dfrac{x^3+2x^2-4x-8}{x^3-8}$!

Ganesha Operation

CARA 1 : PEMFAKTORAN

2	1	2	-4	-8
		2	8	8
	1	4	4	0

2	1	0	0	-8
		2	4	8
	1	2	4	0

$\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{x^3+2x^2-4x-8}{x^3-8}&=\displaystyle\lim_{x\to2}\dfrac{\cancel{(x-2)}\left(x^2+4x+4\right)}{\cancel{(x-2)}\left(x^2+2x+4\right)}\\[9pt] &=\dfrac{2^2+4(2)+4}{2^2+2(2)+4}\\[9pt] &=\dfrac{4+8+4}{4+4+4}\\[9pt] &=\dfrac{16}{12}\\[9pt] &=\boxed{\boxed{\dfrac43}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{x^3+2x^2-4x-8}{x^3-8}&=\displaystyle\lim_{x\to2}\dfrac{3x^2+4x-4}{3x^2}\\[9pt] &=\dfrac{3(2)^2+4(2)-4}{3(2)^2}\\[9pt] &=\dfrac{3(4)+8-4}{3(4)}\\[9pt] &=\dfrac{12+4}{12}\\[9pt] &=\dfrac{16}{12}\\[9pt] &=\boxed{\boxed{\dfrac43}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to2}\dfrac{x^3+2x^2-4x-8}{x^3-8}=\dfrac43$.

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No. 7

Jika $f(x)=\cos\left(\dfrac12x\right)$, maka $\displaystyle\lim_{h\to0}\dfrac{f\left(x+\dfrac{h}2\right)-2f(x)+f\left(x-\dfrac{h}2\right)}{h^2}=$ ....

1. $\dfrac12\sin2x$
2. $-\dfrac1{16}\cos\dfrac12x$
3. $-\dfrac1{16}\sin\dfrac12x$

4. $\dfrac1{16}\cos2x$
5. $-\dfrac1{16}\sin2x$

CARA 1

$\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(x+\dfrac{h}2\right)-2f(x)+f\left(x-\dfrac{h}2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\cos\dfrac12\left(x+\dfrac{h}2\right)-2\cos\dfrac12x+\cos\dfrac12\left(x-\dfrac{h}2\right)}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\cos\left(\dfrac12x+\dfrac{h}4\right)-2\cos\dfrac12x+\cos\left(\dfrac12x-\dfrac{h}4\right)}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\cos\dfrac12x\cos\dfrac{h}4-\sin\dfrac12x\sin\dfrac{h}4-2\cos\dfrac12x+\cos\dfrac12x\cos\dfrac{h}4+\sin\dfrac12x\sin\dfrac{h}4}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\cos\dfrac{h}4-2\cos\dfrac12x+\cos\dfrac12x\cos\dfrac{h}4}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\left(\cos\dfrac{h}4-1\right)}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\left(1-2\sin^2\dfrac{h}8-1\right)}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{-4\cos\dfrac12x\sin^2\dfrac{h}8}{h^2}\\[8pt] &=\displaystyle\lim_{h\to0}\left(-4\cos\dfrac12x\right)\cdot\dfrac{\sin\dfrac{h}8}h\cdot\dfrac{\sin\dfrac{h}8}h\\[8pt] &=-4\cos\dfrac12x\cdot\dfrac18\cdot\dfrac18\\ &=\boxed{\boxed{-\dfrac1{16}\cos\dfrac12x}} \end{aligned}$

CARA 2 : L'HOPITAL

$\begin{aligned} f'(x)&=-\dfrac12\sin\dfrac12x\\[8pt] f''(x)&=-\dfrac14\cos\dfrac12x \end{aligned}$

$\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(x+\dfrac{h}2\right)-2f(x)+f\left(x-\dfrac{h}2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\dfrac12f'\left(x+\dfrac{h}2\right)-0+\left(-\dfrac12\right)f'\left(x-\dfrac{h}2\right)}{2h}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\dfrac12f'\left(x+\dfrac{h}2\right)-\dfrac12f'\left(x-\dfrac{h}2\right)}{2h}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{\dfrac14f''\left(x+\dfrac{h}2\right)+\dfrac14f''\left(x-\dfrac{h}2\right)}2\\[8pt] &=\dfrac{\dfrac14f''\left(x+\dfrac02\right)+\dfrac14f''\left(x-\dfrac02\right)}2\\[8pt] &=\dfrac{\dfrac14f''\left(x\right)+\dfrac14f''\left(x\right)}2\\[8pt] &=\dfrac{\dfrac12f''\left(x\right)}2\\[8pt] &=\dfrac14f''\left(x\right)\\[8pt] &=\dfrac14\left(-\dfrac14\cos\dfrac12x\right)\\ &=\boxed{\boxed{-\dfrac1{16}\cos\dfrac12x}} \end{aligned}$

Jadi, $\displaystyle\lim_{h\to0}\dfrac{f\left(x+\dfrac{h}2\right)-2f(x)+f\left(x-\dfrac{h}2\right)}{h^2}=-\dfrac1{16}\cos\dfrac12x$.
JAWAB: B

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No. 8

$f(x)=x^2+2x$
$\displaystyle\lim_{h\to0}\dfrac{f(x-2h)-f(x)}{4h}=$

$\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f(x-2h)-f(x)}{4h}&=\displaystyle\lim_{h\to0}\dfrac{(x-2h)^2+2(x-2h)-\left(x^2+2x\right)}{4h}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{x^2-4xh+4h^2+2x-4h-x^2-2x}{4h}\\[8pt] &=\displaystyle\lim_{h\to0}\dfrac{-4xh+4h^2+-4h}{4h}\\[8pt] &=\displaystyle\lim_{h\to0}(-x+h+-1)\\ &=-x+0-1\\ &=-x-1 \end{aligned}$

Jadi,

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No. 9

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

1. $0$
2. $-\dfrac12$
3. $-1$

4. $-2$
5. $\infty$

CARA BIASA : KALI SEKAWAN

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

CARA L'HOPITAL

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

Jadi, $\displaystyle\lim_{x\to0}f(x)=-1$.
JAWAB: C

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No. 10

$\displaystyle\lim_{x\to2}(3x+2)=$ ....

1. $4$
2. $5$
   
3. $6$
   

4. $7$
5. $8$
   

$\begin{aligned} \displaystyle\lim_{x\to2}(3x+2)&=3(2)+2\\ &=6+2\\ &=\boxed{\boxed{8}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to2}(3x+2)=8$.
JAWAB: E

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