Exercise Zone : Limit

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

Tipe:

Standar SBMPTN HOTS

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No. 1

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

SUMBER

CARA 1

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

CARA 2

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

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(\sqrt{x-1}-\sqrt{5-x})(\sqrt{4x-3}-1)}{x^2-9}}}& ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(\sqrt{x-1}-\sqrt{5-x})(\sqrt{4x-3}-1)}{x^2-9}}\cdot {\displaystyle \frac{\sqrt{x-1}+\sqrt{5-x}}{\sqrt{x-1}+\sqrt{5-x}}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{((x-1)-(5-x))(\sqrt{4x-3}-1)}{(x^2-9)(\sqrt{x-1}+\sqrt{5-x})}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(x-1-5+x)(\sqrt{4x-3}-1)}{(x^2-9)(\sqrt{x-1}+\sqrt{5-x})}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(2x-6)(\sqrt{4x-3}-1)}{(x^2-9)(\sqrt{x-1}+\sqrt{5-x})}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{2(x-3)(\sqrt{4x-3}-1)}{(x+3)(x-3)(\sqrt{x-1}+\sqrt{5-x})}}}\\ & ={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{2(\sqrt{4x-3}-1)}{(x+3)(\sqrt{x-1}+\sqrt{5-x})}}}\\ & ={\displaystyle \frac{2(\sqrt{4(3)-3}-1)}{(3+3)(\sqrt{3-1}+\sqrt{5-3})}}\\ & ={\displaystyle \frac{2(\sqrt{9}-1)}{(3+3)(\sqrt{2}+\sqrt{2})}}\\ & ={\displaystyle \frac{2(3-1)}{(6)\left(2\sqrt{2}\right)}}\\ & ={\displaystyle \frac{2}{6\left(\sqrt{2}\right)}}\\ & ={\displaystyle \frac{1}{3\sqrt{2}}}\cdot {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}\\ & ={\displaystyle \frac{1}{3\cdot 2}}\sqrt{2}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{6}}\sqrt{2}}}}}\end{array}$$

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)= ....

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}({\displaystyle \frac{6}{x^2-x-2}}-{\displaystyle \frac{2}{x-2}})}& ={\displaystyle \underset{x\to 2}{lim}({\displaystyle \frac{6}{(x-2)(x+1)}}-{\displaystyle \frac{2}{x-2}})}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{6-2(x+1)}{(x-2)(x+1)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{6-2x-2}{(x-2)(x+1)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{-2x+4}{(x-2)(x+1)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{-2\cancel{(x-2)}}{\cancel{(x-2)}(x+1)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{-2}{x+1}}}\\ & ={\displaystyle \frac{-2}{2+1}}\\ & =-{\displaystyle \frac{2}{3}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to -2}{lim}{\displaystyle \frac{x^3-x^2-2x+8}{x^3+8}}}& ={\displaystyle \underset{x\to -2}{lim}{\displaystyle \frac{(x+2)(x^2-3x+4)}{(x+2)(x^2-2x+4)}}}\\ & ={\displaystyle \underset{x\to -2}{lim}{\displaystyle \frac{x^2-3x+4}{x^2-2x+4}}}\\ & ={\displaystyle \frac{(-2{)}^2-3(-2)+4}{(-2{)}^2-2(-2)+4}}\\ & ={\displaystyle \frac{4+6+4}{4+4+4}}\\ & ={\displaystyle \frac{14}{12}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{7}{6}}}}}}\end{array}$$

CARA 2 : L'HOPITAL

$$\begin{array}{rl}{\displaystyle \underset{x\to -2}{lim}{\displaystyle \frac{x^3-x^2-2x+8}{x^3+8}}}& ={\displaystyle \underset{x\to -2}{lim}{\displaystyle \frac{3x^2-2x-2}{3x^2}}}\\ & ={\displaystyle \frac{3(-2{)}^2-2(-2)-2}{3(-2{)}^2}}\\ & ={\displaystyle \frac{12+4-2}{12}}\\ & ={\displaystyle \frac{14}{12}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{7}{6}}}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}(g(x){\displaystyle \frac{x-2}{\sqrt{x}-\sqrt{2}}})}& ={\displaystyle \underset{x\to 2}{lim}(g(x){\displaystyle \frac{x-2}{\sqrt{x}-\sqrt{2}}}\cdot {\displaystyle \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}})}\\ & ={\displaystyle \underset{x\to 2}{lim}(g(x){\displaystyle \frac{(x-2)(\sqrt{x}+\sqrt{2})}{x-2}})}\\ & ={\displaystyle \underset{x\to 2}{lim}(g(x)(\sqrt{x}+\sqrt{2}))}\\ & =4(\sqrt{2}+\sqrt{2})\\ & =4\left(2\sqrt{2}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8\sqrt{2}}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{x^2-25}{\sqrt{x^2+24}-7}}}& ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{x^2-25}{\sqrt{x^2+24}-7}}\cdot {\displaystyle \frac{\sqrt{x^2+24}+7}{\sqrt{x^2+24}+7}}}\\ & ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{(x^2-25)(\sqrt{x^2+24}+7)}{x^2+24-49}}}\\ & ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{(x^2-25)(\sqrt{x^2+24}+7)}{x^2-25}}}\\ & ={\displaystyle \underset{x\to 5}{lim}(\sqrt{x^2+24}+7)}\\ & =\sqrt{5^2+24}+7\\ & =\sqrt{25+24}+7\\ & =\sqrt{49}+7\\ & =7+7\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 14}}}}\end{array}$$

CARA 2 : L'HOPITAL

$$\begin{array}{rl}{\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{x^2-25}{\sqrt{x^2+24}-7}}}& ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{2x}{{\displaystyle \frac{2x}{2\sqrt{x^2+24}}}}}}\\ & ={\displaystyle \underset{x\to 5}{lim}2\sqrt{x^2+24}}\\ & =2\sqrt{5^2+24}\\ & =2\sqrt{25+24}\\ & =2\sqrt{49}\\ & =2(7)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 14}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^3+2x^2-4x-8}{x^3-8}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\cancel{(x-2)}(x^2+4x+4)}{\cancel{(x-2)}(x^2+2x+4)}}}\\ & ={\displaystyle \frac{2^2+4(2)+4}{2^2+2(2)+4}}\\ & ={\displaystyle \frac{4+8+4}{4+4+4}}\\ & ={\displaystyle \frac{16}{12}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$$

CARA 2 : L'HOPITAL

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^3+2x^2-4x-8}{x^3-8}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{3x^2+4x-4}{3x^2}}}\\ & ={\displaystyle \frac{3(2{)}^2+4(2)-4}{3(2{)}^2}}\\ & ={\displaystyle \frac{3(4)+8-4}{3(4)}}\\ & ={\displaystyle \frac{12+4}{12}}\\ & ={\displaystyle \frac{16}{12}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$$

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{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(x+{\displaystyle \frac{h}{2}})-2f(x)+f(x-{\displaystyle \frac{h}{2}})}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\cos {\displaystyle \frac{1}{2}}(x+{\displaystyle \frac{h}{2}})-2\cos {\displaystyle \frac{1}{2}}x+\cos {\displaystyle \frac{1}{2}}(x-{\displaystyle \frac{h}{2}})}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\cos ({\displaystyle \frac{1}{2}}x+{\displaystyle \frac{h}{4}})-2\cos {\displaystyle \frac{1}{2}}x+\cos ({\displaystyle \frac{1}{2}}x-{\displaystyle \frac{h}{4}})}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\cos {\displaystyle \frac{1}{2}}x\cos {\displaystyle \frac{h}{4}}-\sin {\displaystyle \frac{1}{2}}x\sin {\displaystyle \frac{h}{4}}-2\cos {\displaystyle \frac{1}{2}}x+\cos {\displaystyle \frac{1}{2}}x\cos {\displaystyle \frac{h}{4}}+\sin {\displaystyle \frac{1}{2}}x\sin {\displaystyle \frac{h}{4}}}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}}x\cos {\displaystyle \frac{h}{4}}-2\cos {\displaystyle \frac{1}{2}}x+\cos {\displaystyle \frac{1}{2}}x\cos {\displaystyle \frac{h}{4}}}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}}x(\cos {\displaystyle \frac{h}{4}}-1)}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}}x(1-2{\sin }^2{\displaystyle \frac{h}{8}}-1)}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{-4\cos {\displaystyle \frac{1}{2}}x{\sin }^2{\displaystyle \frac{h}{8}}}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}(-4\cos {\displaystyle \frac{1}{2}}x)\cdot {\displaystyle \frac{\sin {\displaystyle \frac{h}{8}}}{h}}\cdot {\displaystyle \frac{\sin {\displaystyle \frac{h}{8}}}{h}}}\\ & =-4\cos {\displaystyle \frac{1}{2}}x\cdot {\displaystyle \frac{1}{8}}\cdot {\displaystyle \frac{1}{8}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{16}}\cos {\displaystyle \frac{1}{2}}x}}}}\end{array}$$

CARA 2 : L'HOPITAL

$$\begin{array}{rl}{f}^{\prime }(x)& =-{\displaystyle \frac{1}{2}}\sin {\displaystyle \frac{1}{2}}x\\ f^{″}(x)& =-{\displaystyle \frac{1}{4}}\cos {\displaystyle \frac{1}{2}}x\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(x+{\displaystyle \frac{h}{2}})-2f(x)+f(x-{\displaystyle \frac{h}{2}})}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{2}}{f}^{\prime }(x+{\displaystyle \frac{h}{2}})-0+(-{\displaystyle \frac{1}{2}})f^{\prime }(x-{\displaystyle \frac{h}{2}})}{2h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{2}}{f}^{\prime }(x+{\displaystyle \frac{h}{2}})-{\displaystyle \frac{1}{2}}{f}^{\prime }(x-{\displaystyle \frac{h}{2}})}{2h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{4}}{f}^{″}(x+{\displaystyle \frac{h}{2}})+{\displaystyle \frac{1}{4}}{f}^{″}(x-{\displaystyle \frac{h}{2}})}{2}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{4}}{f}^{″}(x+{\displaystyle \frac{0}{2}})+{\displaystyle \frac{1}{4}}{f}^{″}(x-{\displaystyle \frac{0}{2}})}{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{4}}{f}^{″}\left(x\right)+{\displaystyle \frac{1}{4}}{f}^{″}\left(x\right)}{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}}{f}^{″}\left(x\right)}{2}}\\ & ={\displaystyle \frac{1}{4}}{f}^{″}\left(x\right)\\ & ={\displaystyle \frac{1}{4}}(-{\displaystyle \frac{1}{4}}\cos {\displaystyle \frac{1}{2}}x)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{16}}\cos {\displaystyle \frac{1}{2}}x}}}}\end{array}$$

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}=

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

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

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

CARA L'HOPITAL

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

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}(3x+2)}& =3(2)+2\\ & =6+2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$$

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

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