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No. 1
\displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)=
Misal x=p^2\to \sqrt{x}=p.
CARA BIASA
\(\eqalign{
\displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)&=\displaystyle\lim_{p\to\infty}\left(\sqrt{2013p^2+\sqrt{2013}p}-\sqrt{2013p^2-\sqrt{2013}p}\right)\cdot\dfrac{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}\\
&=\displaystyle\lim_{p\to\infty}\dfrac{\left(2013p^2+\sqrt{2013}p\right)-\left(2013p^2-\sqrt{2013}p\right)}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}\\
&=\displaystyle\lim_{p\to\infty}\dfrac{2013p^2+\sqrt{2013}p-2013p^2+\sqrt{2013}p}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}\\
&=\displaystyle\lim_{p\to\infty}\dfrac{2\sqrt{2013}p}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}\\
&=\displaystyle\lim_{p\to\infty}\dfrac{\dfrac{2\sqrt{2013}p}p}{\sqrt{\dfrac{2013p^2}{p^2}+\dfrac{\sqrt{2013}p}{p^2}}+\sqrt{\dfrac{2013p^2}{p^2}-\dfrac{\sqrt{2013}p}{p^2}}}\\
&=\displaystyle\lim_{p\to\infty}\dfrac{2\sqrt{2013}}{\sqrt{2013+\dfrac{\sqrt{2013}}p}+\sqrt{2013-\dfrac{\sqrt{2013}}p}}\\
&=\dfrac{2\sqrt{2013}}{\sqrt{2013+0}+\sqrt{2013-0}}\\
&=\dfrac{2\sqrt{2013}}{\sqrt{2013}+\sqrt{2013}}\\
&=\dfrac{2\sqrt{2013}}{2\sqrt{2013}}\\
&=\boxed{\boxed{1}}
}\)
CARA CEPAT
\(\eqalign{
\displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)&=\displaystyle\lim_{p\to\infty}\left(\sqrt{2013p^2+\sqrt{2013}p}-\sqrt{2013p^2-\sqrt{2013}p}\right)\\
&=\dfrac{\sqrt{2013}-\left(-\sqrt{2013}\right)}{2\sqrt{2013}}\\
&=\dfrac{\sqrt{2013}+\sqrt{2013}}{2\sqrt{2013}}\\
&=\dfrac{2\sqrt{2013}}{2\sqrt{2013}}\\
&=\boxed{\boxed{1}}
}\)
No. 2
\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}=
CARA BIASA
\(\eqalign{
\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}&=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{\dfrac{12x^2}{x^2}-\dfrac{4x}{x^2}+\dfrac{2013}{x^2}}+2\sqrt{\dfrac{x^2}{x^2}}}{\dfrac{x}x-\dfrac{2013}x}\\
&=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12-\dfrac4x+\dfrac{2013}{x^2}}+2}{1-\dfrac{2013}x}\\
&=\dfrac{\sqrt{12-0+0}+2}{1-0}\\
&=\dfrac{\sqrt{12}+2}1\\
&=\boxed{\boxed{2\sqrt3+2}}
}\)
CARA CEPAT
\(\eqalign{
\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}&=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2}+2\sqrt{x^2}}x\\
&=\displaystyle\lim_{x\to\infty}\dfrac{2\sqrt3x+2x}x\\
&=\displaystyle\lim_{x\to\infty}2\sqrt3+2\\
&=\boxed{\boxed{2\sqrt3+2}}
}\)
No. 3
\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)=
\(\eqalign{
\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-x-2x\right)\\
&=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}-x+\sqrt{4x^2+6x+3}-2x\right)\\
&=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}-\sqrt{x^2}+\sqrt{4x^2+6x+3}-\sqrt{4x^2}\right)\\
&=\dfrac{2-0}{2\sqrt1}+\dfrac{6-0}{2\sqrt4}\\
&=\dfrac22+\dfrac64\\
&=1+\dfrac32\\
&=\boxed{\boxed{\dfrac52}}
}\)
No. 4
\displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}= ....
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}&=\displaystyle\lim_{x\to\infty}\dfrac{8x^2}{x(2x)}\\[8pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{8x^2}{2x^2}\\[6pt]
&=\boxed{\boxed{4}}
\end{aligned}\)
No. 5
Nilai
\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x=
CARA 1
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x&=\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x\cdot\dfrac{\sqrt{9x^2-3x}+3x}{\sqrt{9x^2-3x}+3x}\\
&=\displaystyle\lim_{x\to\infty}\dfrac{9x^2-3x-(3x)^2}{\sqrt{9x^2-3x}+3x}\\[8pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{9x^2-3x-9x^2}{\sqrt{9x^2-3x}+3x}\\[8pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{-3x}{\sqrt{9x^2-3x}+3x}\\[8pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{-\dfrac{3x}x}{\sqrt{\dfrac{9x^2}{x^2}-\dfrac{3x}{x^2}}+\dfrac{3x}x}\\[15pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{-3}{\sqrt{9-\dfrac3x}+3}\\[8pt]
&=\dfrac{-3}{\sqrt{9-0}+3}\\[8pt]
&=\dfrac{-3}{\sqrt{9}+3}\\[8pt]
&=\dfrac{-3}{3+3}\\[8pt]
&=\dfrac{-3}6\\
&=\boxed{\boxed{-\dfrac12}}
\end{aligned}\)
CARA 2
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x&=\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-\sqrt{(3x)^2}\\
&=\displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-\sqrt{9x^2}\\
&=\dfrac{-3-0}{2\sqrt{9}}\\[8pt]
&=\dfrac{-3}{2(3)}\\[8pt]
&=\dfrac{-3}6\\
&=\boxed{\boxed{-\dfrac12}}
\end{aligned}\)No. 6
Nilai dari
\displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}= ....
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}&=\dfrac{-7-18}{2\sqrt{25}}\\[8pt]
&=\dfrac{-25}{2(5)}\\[8pt]
&=-\dfrac52\\
&=\boxed{\boxed{-2,5}}
\end{aligned}\)
No. 7
Nilai
\displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}=
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}&=\displaystyle\lim_{x\to\infty}\dfrac{2x^2}{\sqrt{x^2}}\\[8pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{2x^2}{x}\\[8pt]
&=\displaystyle\lim_{x\to\infty}2x\\
&=2(\infty)\\
&=\boxed{\boxed{\infty}}
\end{aligned}\)
No. 8
\displaystyle\lim_{x\to\infty}\left(\dfrac{5-x^2}{4x^2+x}\right)\left(\dfrac{1-8x^3}{3x-2x^3}\right)^\frac32=
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\left(\dfrac{5-x^2}{4x^2+x}\right)\left(\dfrac{1-8x^3}{3x-2x^3}\right)^\frac32&=\displaystyle\lim_{x\to\infty}\left(\dfrac{-x^2}{4x^2}\right)\left(\dfrac{-8x^3}{-2x^3}\right)^\frac32\\[8pt]
&=\displaystyle\lim_{x\to\infty}\left(\dfrac{-1}4\right)\left(\dfrac{-8}{-2}\right)^\frac32\\[8pt]
&=\left(\dfrac{-1}4\right)(4)^\frac32\\[8pt]
&=\left(\dfrac{-1}4\right)(8)\\[8pt]
&=\boxed{\boxed{-2}}
\end{aligned}\)
No. 9
\displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x&=\displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-\sqrt{x^2}\\
&=\dfrac{2-0}{2\sqrt1}\\[8pt]
&=\dfrac22\\
&=\boxed{\boxed{1}}
\end{aligned}\)
No. 10
Dengan menerapkan sifat limit tak hingga,
{\displaystyle\lim_{x\to\infty}\dfrac{12x^5-6x^3+7x-1}{4x^5-8x+10}=}
CARA 1
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\dfrac{12x^5-6x^3+7x-1}{4x^5-8x+10}&=\displaystyle\lim_{x\to\infty}\dfrac{\dfrac{12x^5}{\color{red}{x^5}}-\dfrac{6x^3}{\color{red}{x^5}}+\dfrac{7x}{\color{red}{x^5}}-\dfrac1{\color{red}{x^5}}}{\dfrac{4x^5}{\color{red}{x^5}}-\dfrac{8x}{\color{red}{x^5}}+\dfrac{10}{\color{red}{x^5}}}\\[20pt]
&=\displaystyle\lim_{x\to\infty}\dfrac{12-\dfrac6{x^2}+\dfrac7{x^4}-\dfrac1{x^5}}{4-\dfrac8{x^4}+\dfrac{10}{x^5}}\\[20pt]
&=\dfrac{12-0+0-0}{4-0+0}\\
&=\boxed{\boxed{3}}
\end{aligned}\)
CARA 2
\(\begin{aligned}
\displaystyle\lim_{x\to\infty}\dfrac{12x^5-6x^3+7x-1}{4x^5-8x+10}&=\displaystyle\lim_{x\to\infty}\dfrac{12{\color{red}{\cancel{\color{black}{x^5}}}}}{4{\color{red}{\cancel{\color{black}{x^5}}}}}\\
&=\boxed{\boxed{3}}
\end{aligned}\)
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