Berikut ini adalah kumpulan soal dan pembahasan mengenai limit tipe standar. Jika ingin bertanya soal, silahkan gabung ke grup
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No. 11
Nilai
\displaystyle\lim_{x\to0}\dfrac{2x}{x+2}= ...
\(\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{2x}{x+2}&=\dfrac{2(0)}{0+2}\\[8pt]
&=\dfrac02\\
&=\boxed{\boxed{0}}
\end{aligned}\)
No. 12
Jika
f(x)=\dfrac{3-\sqrt{2x+9}}x, maka
\displaystyle\lim_{x\to0}f(x)=
CARA 1 : KALI SEKAWAN
\(\begin{aligned}
\displaystyle\lim_{x\to0}f(x)&=\displaystyle\lim_{x\to0}\dfrac{3-\sqrt{2x+9}}x\cdot\dfrac{3+\sqrt{2x+9}}{3+\sqrt{2x+9}}\\[8pt]
&=\displaystyle\lim_{x\to0}\dfrac{9-(2x+9)}{x(3+\sqrt{2x+9})}\\[8pt]
&=\displaystyle\lim_{x\to0}\dfrac{9-2x-9}{x(3+\sqrt{2x+9})}\\[8pt]
&=\displaystyle\lim_{x\to0}\dfrac{-2x}{x(3+\sqrt{2x+9})}\\[8pt]
&=\displaystyle\lim_{x\to0}\dfrac{-2}{3+\sqrt{2x+9}}\\[8pt]
&=\dfrac{-2}{3+\sqrt{2(0)+9}}\\[8pt]
&=\dfrac{-2}{3+\sqrt{0+9}}\\[8pt]
&=\dfrac{-2}{3+\sqrt9}\\[8pt]
&=\dfrac{-2}{3+3}\\[8pt]
&=\dfrac{-2}6\\
&=\boxed{\boxed{-\dfrac13}}
\end{aligned}\)
CARA 2 : L'HOPITAL
\(\begin{aligned}
\displaystyle\lim_{x\to0}f(x)&=\displaystyle\lim_{x\to0}\dfrac{3-\sqrt{2x+9}}x\\[8pt]
&=\displaystyle\lim_{x\to0}\dfrac{0-\dfrac2{2\sqrt{2x+9}}}1\\[8pt]
&=\displaystyle\lim_{x\to0}-\dfrac1{\sqrt{2x+9}}\\[8pt]
&=-\dfrac1{\sqrt{2(0)+9}}\\[8pt]
&=-\dfrac1{\sqrt{0+9}}\\[8pt]
&=-\dfrac1{\sqrt9}\\
&=\boxed{\boxed{-\dfrac13}}
\end{aligned}\)
No. 13
Nilai dari
\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}} adalah
CARA 1
\(\begin{aligned}
\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}&=\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}\color{red}{\cdot\dfrac{\sqrt{2x-1}+\sqrt{x}}{\sqrt{2x-1}+\sqrt{x}}}\\[8pt]
&=\displaystyle\lim_{x\to1}\dfrac{(2x-2)\left(\sqrt{2x-1}+\sqrt{x}\right)}{2x-1-x}\\[8pt]
&=\displaystyle\lim_{x\to1}\dfrac{2\cancel{(x-1)}\left(\sqrt{2x-1}+\sqrt{x}\right)}{\cancel{x-1}}\\[8pt]
&=\displaystyle\lim_{x\to1}2\left(\sqrt{2x-1}+\sqrt{x}\right)\\
&=2\left(\sqrt{2(1)-1}+\sqrt{1}\right)\\
&=2\left(\sqrt{2-1}+1\right)\\
&=2\left(\sqrt1+1\right)\\
&=2\left(1+1\right)\\
&=2\left(2\right)\\
&=\boxed{\boxed{4}}
\end{aligned}\)
CARA 2
\(\begin{aligned}
\displaystyle\lim_{x\to1}\dfrac{2x-2}{\sqrt{2x-1}-\sqrt{x}}&=\displaystyle\lim_{x\to1}\dfrac2{\dfrac2{2\sqrt{2x-1}}-\dfrac1{2\sqrt{x}}}\\[21pt]
&=\dfrac2{\dfrac2{2\sqrt{2(1)-1}}-\dfrac1{2\sqrt1}}\\[21pt]
&=\dfrac2{\dfrac2{2\sqrt{2-1}}-\dfrac1{2(1)}}\\[21pt]
&=\dfrac2{\dfrac2{2\sqrt1}-\dfrac12}\\[21pt]
&=\dfrac2{\dfrac2{2(1)}-\dfrac12}\\[21pt]
&=\dfrac2{\dfrac22-\dfrac12}\\[21pt]
&=\dfrac2{\dfrac12}\\
&=\boxed{\boxed{4}}
\end{aligned}\)
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
CARA 1
\(\begin{aligned}
\displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\sqrt{1+3+2h^2}-\sqrt{1+3-2h^2}}{h^2}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{\sqrt{4+2h^2}-\sqrt{4-2h^2}}{h^2}\color{red}{\cdot\dfrac{\sqrt{4+2h^2}+\sqrt{4-2h^2}}{\sqrt{4+2h^2}+\sqrt{4-2h^2}}}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{\left(4+2h^2\right)-\left(4-2h^2\right)}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{4+2h^2-4+2h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac{4h^2}{h^2\left(\sqrt{4+2h^2}+\sqrt{4-2h^2}\right)}\\[8pt]
&=\displaystyle\lim_{h\to0}\dfrac4{\sqrt{4+2h^2}+\sqrt{4-2h^2}}\\[8pt]
&=\dfrac4{\sqrt{4+2(0)^2}+\sqrt{4-2(0)^2}}\\[8pt]
&=\dfrac4{\sqrt{4+0}+\sqrt{4-0}}\\[8pt]
&=\dfrac4{\sqrt4+\sqrt4}\\[8pt]
&=\dfrac4{2+2}\\[8pt]
&=\dfrac44\\
&=\boxed{\boxed{1}}
\end{aligned}\)
CARA 2 : L'HOPITAL
\(\begin{aligned}
f(x)&=\sqrt{1+x}\\
f'(x)&=\dfrac1{2\sqrt{1+x}}
\end{aligned}\)
\(\begin{aligned}
\displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-2h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{4h\ f'\left(3+2h^2\right)-(-4h)f'\left(3-2h^2\right)}{2h}\\[8pt]
&=\displaystyle\lim_{h\to0}\left(2f'\left(3+2h^2\right)+2f'\left(3-2h^2\right)\right)\\
&=2f'\left(3+2(0)^2\right)+2f'\left(3-2(0)^2\right)\\
&=2f'(3)+2f'(3)\\
&=4f'(3)\\
&=4\left(\dfrac1{2\sqrt{1+3}}\right)\\[8pt]
&=\dfrac4{2\sqrt4}\\[8pt]
&=\dfrac4{2(2)}\\[8pt]
&=\dfrac44\\
&=\boxed{\boxed{1}}
\end{aligned}\)
No. 15
Nilai dari
\displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x} adalah
\(\begin{aligned}
\displaystyle\lim_{x\to3}\dfrac{x^2+3x-18}{x^2-3x}&=\displaystyle\lim_{x\to3}\dfrac{(x+6)\cancel{(x-3)}}{x\cancel{(x-3)}}\\[8pt]
&=\displaystyle\lim_{x\to3}\dfrac{x+6}x\\[8pt]
&=\dfrac{3+6}3\\[8pt]
&=\dfrac93\\
&=\boxed{\boxed{3}}
\end{aligned}\)
No. 16
Diketahui
{\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}=\dfrac14}, nilai
a adalah
\(\begin{aligned}
\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}&=\dfrac14\\[8pt]
\displaystyle\lim_{x\to a}\dfrac{\sqrt{x}-2}{x-4}\color{red}{\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2}}&=\dfrac14\\[8pt]
\displaystyle\lim_{x\to a}\dfrac{\cancel{x-4}}{\cancel{(x-4)}\left(\sqrt{x}+2\right)}&=\dfrac14\\[8pt]
\displaystyle\lim_{x\to a}\dfrac1{\sqrt{x}+2}&=\dfrac14\\[8pt]
\dfrac1{\sqrt{a}+2}&=\dfrac14\\[8pt]
\sqrt{a}+2&=4\\
\sqrt{a}&=2\\
a&=4
\end{aligned}\)
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)!
\(\begin{array}{rcccll}
\dfrac{\theta^2+\theta-2}{\theta+3}&\leq&\dfrac{f(\theta)}{\theta+3}&\leq&\dfrac{\theta^2+2\theta-1}{\theta+3}\\[8pt]
\displaystyle\lim_{\theta\to-1}\dfrac{\theta^2+\theta-2}{\theta+3}&\leq&\displaystyle\lim_{\theta\to-1}\dfrac{f(\theta)}{\theta+3}&\leq&\displaystyle\lim_{\theta\to-1}\dfrac{\theta^2+2\theta-1}{\theta+3}\\[8pt]
\dfrac{(-1)^2+(-1)-2}{-1+3}&\leq&\dfrac{\displaystyle\lim_{\theta\to-1}f(\theta)}{-1+3}&\leq&\dfrac{(-1)^2+2(-1)-1}{-1+3}\\[8pt]
\dfrac{-2}2&\leq&\dfrac{\displaystyle\lim_{\theta\to-1}f(\theta)}2&\leq&\dfrac{-2}{2}&\quad\color{red}{\times2}\\[8pt]
-2&\leq&\displaystyle\lim_{\theta\to-1}f(\theta)&\leq&-2
\end{array}\)
\displaystyle\lim_{\theta\to-1}f(\theta)=-2
No. 18
Jika
{\displaystyle\lim_{x\to2}\dfrac{x^2-3x+a}{x-2}=1} maka nilai
a adalah ....
Jika {f(x)=x^2-3x+a}, maka {f(2)=0}.
\(\begin{aligned}
2^2-3(2)+a&=0\\
4-6+a&=0\\
-2+a&=0\\
a&=\boxed{\boxed{2}}
\end{aligned}\)
No. 19
{\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}=} ....
\(\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}&=\displaystyle\lim_{x\to0}\dfrac{\sqrt{x}-x}{\sqrt{x}+x}\cdot{\color{red}{\dfrac{\sqrt{x}+x}{\sqrt{x}+x}}}\\[10pt]
&=\displaystyle\lim_{x\to0}\dfrac{x-x^2}{x+2x\sqrt{x}+x^2}\\[10pt]
&=\displaystyle\lim_{x\to0}\dfrac{x(1-x)}{x\left(1+2\sqrt{x}+x\right)}\\[10pt]
&=\displaystyle\lim_{x\to0}\dfrac{1-x}{1+2\sqrt{x}+x}\\[10pt]
&=\dfrac{1-0}{1+2\sqrt0+0}\\[10pt]
&=\dfrac11\\
&=\boxed{\boxed{1}}
\end{aligned}\)
No. 20
\displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}=
\(\eqalign{
\displaystyle\lim_{x\to2}\dfrac{x^2+5x-14}{2x^2-x-6}&=\displaystyle\lim_{x\to2}\dfrac{(x+7)(x-2)}{(2x+3)(x-2)}\\
&=\displaystyle\lim_{x\to2}\dfrac{x+7}{2x+3}\\
&=\dfrac{2+7}{2(2)+3}\\
&=\boxed{\boxed{\dfrac97}}
}\)
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