Exercise Zone : Limit
Table of Contents
Tipe:
No.
ALTERNATIF PENYELESAIAN
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}}\\[3.7pt] &=\displaystyle\lim_{x\to3}\dfrac{9-(2x+3)}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[3.7pt] &=\displaystyle\lim_{x\to3}\dfrac{9-2x-3}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[3.7pt] &=\displaystyle\lim_{x\to3}\dfrac{6-2x}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[3.7pt] &=\displaystyle\lim_{x\to3}\dfrac{-2(x-3)}{(x-3)\left(3+\sqrt{2x+3}\right)}\\[3.7pt] &=\displaystyle\lim_{x\to3}\dfrac{-2}{3+\sqrt{2x+3}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt{2(3)+3}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt{6+3}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt9}\\[3.7pt] &=\dfrac{-2}{3+3}\\[3.7pt] &=\dfrac{-2}6\\[3.7pt] &=\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}\\[3.7pt] &=\displaystyle\lim_{x\to3}\left(-\dfrac1{\sqrt{2x+3}}\right)\\[3.7pt] &=-\dfrac1{\sqrt{2(3)+3}}\\[3.7pt] &=-\dfrac1{\sqrt{6+3}}\\[3.7pt] &=-\dfrac1{\sqrt9}\\[3.7pt] &=\boxed{\boxed{-\dfrac13}} \end{aligned}Jadi, \displaystyle\lim_{x\to3}\dfrac{3-\sqrt{2x+3}}{x-3}=-\dfrac13 .
No.
-\dfrac16\sqrt2 -\dfrac14\sqrt2 -\dfrac12\sqrt2
\dfrac16\sqrt2 \dfrac12\sqrt2
ALTERNATIF PENYELESAIAN
\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}}\\[3.7pt]
&=\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)}\\[3.7pt]
&=\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)}\\[3.7pt]
&=\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)}\\[3.7pt]
&=\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)}\\[3.7pt]
&=\displaystyle\lim_{x\to3}\dfrac{2\left(\sqrt{4x-3}-1\right)}{(x+3)\left(\sqrt{x-1}+\sqrt{5-x}\right)}\\[3.7pt]
&=\dfrac{2\left(\sqrt{4(3)-3}-1\right)}{(3+3)\left(\sqrt{3-1}+\sqrt{5-3}\right)}\\[3.7pt]
&=\dfrac{2\left(\sqrt9-1\right)}{(3+3)\left(\sqrt2+\sqrt2\right)}\\[3.7pt]
&=\dfrac{2(3-1)}{(6)\left(2\sqrt2\right)}\\[3.7pt]
&=\dfrac{2}{6\left(\sqrt2\right)}\\[3.7pt]
&=\dfrac1{3\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\[3.7pt]
&=\dfrac1{3\cdot2}\sqrt2\\[3.7pt]
&=\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
JAWAB: D
No.
ALTERNATIF PENYELESAIAN
\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)\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{6-2(x+1)}{(x-2)(x+1)}\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{6-2x-2}{(x-2)(x+1)}\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{-2x+4}{(x-2)(x+1)}\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{-2\cancel{(x-2)}}{\cancel{(x-2)}(x+1)}\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{-2}{x+1}\\[3.7pt]
&=\dfrac{-2}{2+1}\\[3.7pt]
&=-\dfrac23
\end{aligned}
Jadi, \displaystyle\lim_{x\to2}\left(\dfrac6{x^2-x-2}-\dfrac2{x-2}\right)=-\dfrac23 .
No.
ALTERNATIF PENYELESAIAN
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)}\\[3.7pt] &=\displaystyle\lim_{x\to-2}\dfrac{x^2-3x+4}{x^2-2x+4}\\[3.7pt] &=\dfrac{(-2)^2-3(-2)+4}{(-2)^2-2(-2)+4}\\[3.7pt] &=\dfrac{4+6+4}{4+4+4}\\[3.7pt] &=\dfrac{14}{12}\\[3.7pt] &=\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}\\[3.7pt] &=\dfrac{3(-2)^2-2(-2)-2}{3(-2)^2}\\[3.7pt] &=\dfrac{12+4-2}{12}\\[3.7pt] &=\dfrac{14}{12}\\[3.7pt] &=\boxed{\boxed{\dfrac76}} \end{aligned}Jadi, \displaystyle\lim_{x\to-2}\dfrac{x^3-x^2-2x+8}{x^3+8}=\dfrac76 .
No.
Diketahui fungsi g kontinu diALTERNATIF PENYELESAIAN
\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)\\[3.7pt]
&=\displaystyle\lim_{x\to2}\left(g(x)\dfrac{(x-2)\left(\sqrt{x}+\sqrt2\right)}{x-2}\right)\\[3.7pt]
&=\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 .
No.
Tentukan nilai dariALTERNATIF PENYELESAIAN
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}\\[3.7pt] &=\displaystyle\lim_{x\to5}\dfrac{\left(x^2-25\right)\left(\sqrt{x^2+24}+7\right)}{x^2+24-49}\\[3.7pt] &=\displaystyle\lim_{x\to5}\dfrac{\left(x^2-25\right)\left(\sqrt{x^2+24}+7\right)}{x^2-25}\\[3.7pt] &=\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}}}\\[3.7pt] &=\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 .
No.
Tentukan nilai dariALTERNATIF PENYELESAIAN
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 |
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}\\[3.7pt] &=\dfrac{3(2)^2+4(2)-4}{3(2)^2}\\[3.7pt] &=\dfrac{3(4)+8-4}{3(4)}\\[3.7pt] &=\dfrac{12+4}{12}\\[3.7pt] &=\dfrac{16}{12}\\[3.7pt] &=\boxed{\boxed{\dfrac43}} \end{aligned}Jadi, \displaystyle\lim_{x\to2}\dfrac{x^3+2x^2-4x-8}{x^3-8}=\dfrac43 .
No.
Jika\dfrac12\sin2x -\dfrac1{16}\cos\dfrac12x -\dfrac1{16}\sin\dfrac12x
\dfrac1{16}\cos2x -\dfrac1{16}\sin2x
ALTERNATIF PENYELESAIAN
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}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{\cos\left(\dfrac12x+\dfrac{h}4\right)-2\cos\dfrac12x+\cos\left(\dfrac12x-\dfrac{h}4\right)}{h^2}\\[3.7pt] &=\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}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\cos\dfrac{h}4-2\cos\dfrac12x+\cos\dfrac12x\cos\dfrac{h}4}{h^2}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\left(\cos\dfrac{h}4-1\right)}{h^2}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{2\cos\dfrac12x\left(1-2\sin^2\dfrac{h}8-1\right)}{h^2}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{-4\cos\dfrac12x\sin^2\dfrac{h}8}{h^2}\\[3.7pt] &=\displaystyle\lim_{h\to0}\left(-4\cos\dfrac12x\right)\cdot\dfrac{\sin\dfrac{h}8}h\cdot\dfrac{\sin\dfrac{h}8}h\\[3.7pt] &=-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\\[3.7pt] 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}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{\dfrac12f'\left(x+\dfrac{h}2\right)-\dfrac12f'\left(x-\dfrac{h}2\right)}{2h}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{\dfrac14f''\left(x+\dfrac{h}2\right)+\dfrac14f''\left(x-\dfrac{h}2\right)}2\\[3.7pt] &=\dfrac{\dfrac14f''\left(x+\dfrac02\right)+\dfrac14f''\left(x-\dfrac02\right)}2\\[3.7pt] &=\dfrac{\dfrac14f''\left(x\right)+\dfrac14f''\left(x\right)}2\\[3.7pt] &=\dfrac{\dfrac12f''\left(x\right)}2\\[3.7pt] &=\dfrac14f''\left(x\right)\\[3.7pt] &=\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
JAWAB: B
No.
ALTERNATIF PENYELESAIAN
\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}\\[3.7pt]
&=\displaystyle\lim_{h\to0}\dfrac{x^2-4xh+4h^2+2x-4h-x^2-2x}{4h}\\[3.7pt]
&=\displaystyle\lim_{h\to0}\dfrac{-4xh+4h^2+-4h}{4h}\\[3.7pt]
&=\displaystyle\lim_{h\to0}(-x+h+-1)\\
&=-x+0-1\\
&=-x-1
\end{aligned}
Jadi, \displaystyle\lim_{h\to0}\dfrac{f(x-2h)-f(x)}{4h}=-x-1 .
No.
Jika- 0
-\dfrac12 - −1
- −2
- ∞
ALTERNATIF PENYELESAIAN
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}}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{x^2-2x\sqrt{x}+x}{x^2-x}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{x(x-2\sqrt{x}+1)}{x(x-1)}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{x-2\sqrt{x}+1}{x-1}\\[3.7pt] &=\dfrac{0-2\sqrt{0}+1}{0-1}\\[3.7pt] &=\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}}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{1-\dfrac1{2\sqrt{x}}}{1+\dfrac1{2\sqrt{x}}}\cdot\dfrac{2\sqrt{x}}{2\sqrt{x}}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{2\sqrt{x}-1}{2\sqrt{x}+1}\\[3.7pt] &=\dfrac{2\sqrt{0}-1}{2\sqrt{0}+1}\\[3.7pt] &=\dfrac{-1}1\\ &=\boxed{\boxed{-1}} \end{aligned}Jadi, \displaystyle\lim_{x\to0}f(x)=-1 .
JAWAB: C
JAWAB: C
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