Exercise Zone : Limit [2]
Table of Contents
Tipe:
No.
Nilai- 0
- 1
- 3
- 9
- ∞
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{2x}{x+2}&=\dfrac{2(0)}{0+2}\\[3.7pt]
&=\dfrac02\\
&=\boxed{\boxed{0}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to0}\dfrac{2x}{x+2}=0 .
JAWAB: A
JAWAB: A
No.
Jika-\dfrac13 - 6
\dfrac23
- 12
\dfrac12
ALTERNATIF PENYELESAIAN
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}}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{9-(2x+9)}{x(3+\sqrt{2x+9})}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{9-2x-9}{x(3+\sqrt{2x+9})}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{-2x}{x(3+\sqrt{2x+9})}\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{-2}{3+\sqrt{2x+9}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt{2(0)+9}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt{0+9}}\\[3.7pt] &=\dfrac{-2}{3+\sqrt9}\\[3.7pt] &=\dfrac{-2}{3+3}\\[3.7pt] &=\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\\[3.7pt] &=\displaystyle\lim_{x\to0}\dfrac{0-\dfrac2{2\sqrt{2x+9}}}1\\[3.7pt] &=\displaystyle\lim_{x\to0}-\dfrac1{\sqrt{2x+9}}\\[3.7pt] &=-\dfrac1{\sqrt{2(0)+9}}\\[3.7pt] &=-\dfrac1{\sqrt{0+9}}\\[3.7pt] &=-\dfrac1{\sqrt9}\\ &=\boxed{\boxed{-\dfrac13}} \end{aligned}Jadi, \displaystyle\lim_{x\to0}f(x)=-\dfrac13 .
JAWAB: A
JAWAB: A
No.
Nilai dari- 1
- 2
- 3
- 4
- 5
ALTERNATIF PENYELESAIAN
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}}}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{(2x-2)\left(\sqrt{2x-1}+\sqrt{x}\right)}{2x-1-x}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{2\cancel{(x-1)}\left(\sqrt{2x-1}+\sqrt{x}\right)}{\cancel{x-1}}\\[3.7pt] &=\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}}}\\[3.7pt] &=\dfrac2{\dfrac2{2\sqrt{2(1)-1}}-\dfrac1{2\sqrt1}}\\[3.7pt] &=\dfrac2{\dfrac2{2\sqrt{2-1}}-\dfrac1{2(1)}}\\[3.7pt] &=\dfrac2{\dfrac2{2\sqrt1}-\dfrac12}\\[3.7pt] &=\dfrac2{\dfrac2{2(1)}-\dfrac12}\\[3.7pt] &=\dfrac2{\dfrac22-\dfrac12}\\[3.7pt] &=\dfrac2{\dfrac12}\\ &=\boxed{\boxed{4}} \end{aligned}Jadi, \displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}=4 .
JAWAB: D
JAWAB: D
No.
Diketahui- 0
\dfrac67 \dfrac98
\dfrac23 \dfrac54
ALTERNATIF PENYELESAIAN
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}\\[3.7pt] &=\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}}}\\[3.7pt] &=\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)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{4+2h^2-4+2h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{4h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac4{\sqrt{4+2h^2}+\sqrt{4-2h^2}}\\[3.7pt] &=\dfrac4{\sqrt{4+2(0)^2}+\sqrt{4-2(0)^2}}\\[3.7pt] &=\dfrac4{\sqrt{4+0}+\sqrt{4-0}}\\[3.7pt] &=\dfrac4{\sqrt4+\sqrt4}\\[3.7pt] &=\dfrac4{2+2}\\[3.7pt] &=\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}\\[3.7pt] &=\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)\\[3.7pt] &=\dfrac4{2\sqrt4}\\[3.7pt] &=\dfrac4{2(2)}\\[3.7pt] &=\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: TIDAK ADA
JAWAB: TIDAK ADA
No.
Nilai dari- 3
- −3
- −2
- 2
- 0
ALTERNATIF PENYELESAIAN
\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)}}\\[3.7pt]
&=\displaystyle\lim_{x\to3}\dfrac{x+6}x\\[3.7pt]
&=\dfrac{3+6}3\\[3.7pt]
&=\dfrac93\\
&=\boxed{\boxed{3}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x}=3 .
JAWAB: A
JAWAB: A
No.
Diketahui- 0
- −4
- 4
- −2
- 2
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}&=\dfrac14\\[3.7pt]
\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}\color{red}{\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2}}&=\dfrac14\\[3.7pt]
\displaystyle\lim_{x\to a}\dfrac{\cancel{x-4}}{\cancel{(x-4)}\left(\sqrt{x}+2\right)}&=\dfrac14\\[3.7pt]
\displaystyle\lim_{x\to a}\dfrac1{\sqrt{x}+2}&=\dfrac14\\[3.7pt]
\dfrac1{\sqrt{a}+2}&=\dfrac14\\[3.7pt]
\sqrt{a}+2&=4\\
\sqrt{a}&=2\\
a&=4
\end{aligned}
Jadi, a = 4 .
JAWAB: C
JAWAB: C
No.
JikaALTERNATIF PENYELESAIAN
\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}\\[3.7pt]
\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}\\[3.7pt]
\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}\\[3.7pt]
\dfrac{-2}2&\leq&\dfrac{\displaystyle\lim_{\theta\to-1}f(\theta)}2&\leq&\dfrac{-2}{2}&\quad\color{red}{\times2}\\[3.7pt]
-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 .
No.
Jika- −2
- −1
- 0
- 1
- 2
ALTERNATIF PENYELESAIAN
Jika f(x) = x2 − 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 .
JAWAB: E
JAWAB: E
No.
- 0
\dfrac12 - 1
- 2
- ∞
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}&=\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}\cdot{\color{red}{\dfrac{\sqrt{x}+x}{\sqrt{x}+x}}}\\[3.8pt]
&=\displaystyle\lim_{x\to0}\dfrac{x-x^2}{x+2x\sqrt{x}+x^2}\\[3.8pt]
&=\displaystyle\lim_{x\to0}\dfrac{x(1-x)}{x\left(1+2\sqrt{x}+x\right)}\\[3.8pt]
&=\displaystyle\lim_{x\to0}\dfrac{1-x}{1+2\sqrt{x}+x}\\[3.8pt]
&=\dfrac{1-0}{1+2\sqrt0+0}\\[3.8pt]
&=\dfrac11\\
&=\boxed{\boxed{1}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}=1 .
JAWAB: C
JAWAB: C
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}&=\displaystyle\lim_{x\to2}\dfrac{(x+7)(x-2)}{(2x+3)(x-2)}\\[3.8pt]
&=\displaystyle\lim_{x\to2}\dfrac{x+7}{2x+3}\\[3.8pt]
&=\dfrac{2+7}{2(2)+3}\\
&=\boxed{\boxed{\dfrac97}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}=\dfrac97 .
Post a Comment