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Exercise Zone : Limit Tak Hingga

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Limit Tak Hingga. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS


No.

\displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)=
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ALTERNATIF PENYELESAIAN
Misal x=p^2\to \sqrt{x}=p.

CARA BIASA

\begin{aligned} \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}}\\[3.7pt] &=\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}}\\[3.7pt] &=\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}}\\[3.7pt] &=\displaystyle\lim_{p\to\infty}\dfrac{2\sqrt{2013}p}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}\\[3.7pt] &=\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}}}\\[3.7pt] &=\displaystyle\lim_{p\to\infty}\dfrac{2\sqrt{2013}}{\sqrt{2013+\dfrac{\sqrt{2013}}p}+\sqrt{2013-\dfrac{\sqrt{2013}}p}}\\[3.7pt] &=\dfrac{2\sqrt{2013}}{\sqrt{2013+0}+\sqrt{2013-0}}\\[3.7pt] &=\dfrac{2\sqrt{2013}}{\sqrt{2013}+\sqrt{2013}}\\[3.7pt] &=\dfrac{2\sqrt{2013}}{2\sqrt{2013}}\\ &=\boxed{\boxed{1}} \end{aligned}

CARA CEPAT

\begin{aligned} \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}}\\[3.7pt] &=\dfrac{\sqrt{2013}+\sqrt{2013}}{2\sqrt{2013}}\\[3.7pt] &=\dfrac{2\sqrt{2013}}{2\sqrt{2013}}\\ &=\boxed{\boxed{1}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)=1.

No.

\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}=
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ALTERNATIF PENYELESAIAN

CARA BIASA

\begin{aligned} \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}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12-\dfrac4x+\dfrac{2013}{x^2}}+2}{1-\dfrac{2013}x}\\[3.7pt] &=\dfrac{\sqrt{12-0+0}+2}{1-0}\\[3.7pt] &=\dfrac{\sqrt{12}+2}1\\ &=\boxed{\boxed{2\sqrt3+2}} \end{aligned}

CARA CEPAT

\begin{aligned} \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\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{2\sqrt3x+2x}x\\[3.7pt] &=\displaystyle\lim_{x\to\infty}2\sqrt3+2\\ &=\boxed{\boxed{2\sqrt3+2}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}=2\sqrt3+2.

No.

\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)=
Matematika Zone Id
ALTERNATIF PENYELESAIAN
\begin{aligned} \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}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)=\dfrac52.

No.

\displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}=....
  1. 0
  2. 2
  3. 4
  1. 8
  2. ∞
SUMBER
ALTERNATIF PENYELESAIAN
\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)}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{8x^2}{2x^2}\\ &=\boxed{\boxed{4}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}=4.

No.

Nilai \displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x=
  1. −1
  2. 0
  3. -\dfrac12
  1. 2
  2. ∞
SUMBER
ALTERNATIF PENYELESAIAN

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}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{9x^2-3x-9x^2}{\sqrt{9x^2-3x}+3x}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{-3x}{\sqrt{9x^2-3x}+3x}\\[3.7pt] &=\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}\\[3.7pt] &=\dfrac{-3}{\sqrt{9-0}+3}\\[3.7pt] &=\dfrac{-3}{\sqrt{9}+3}\\[3.7pt] &=\dfrac{-3}{3+3}\\[3.7pt] &=\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}}\\[3.7pt] &=\dfrac{-3}{2(3)}\\[3.7pt] &=\dfrac{-3}6\\ &=\boxed{\boxed{-\dfrac12}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x=-\dfrac12.
JAWAB: C
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No.

Nilai dari \displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}= ....
  1. −2,4
  2. −2,5
  3. −2,6
  1. −2,7
  2. −2,8
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}&=\dfrac{-7-18}{2\sqrt{25}}\\[3.7pt] &=\dfrac{-25}{2(5)}\\[3.7pt] &=-\dfrac52\\ &=\boxed{\boxed{-2,5}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}=-2,5.
JAWAB: B

No.

Nilai \displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}=
  1. 10
  2. 3
  3. \dfrac54
  1. 8
  2. ∞
SUMBER
ALTERNATIF PENYELESAIAN
\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}}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{2x^2}{x}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}2x\\ &=2(\infty)\\ &=\boxed{\boxed{\infty}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}=\infty.
JAWAB: E

No.

\displaystyle\lim_{x\to\infty}\left(\dfrac{5-x^2}{4x^2+x}\right)\left(\dfrac{1-8x^3}{3x-2x^3}\right)^\frac32=
  1. -\dfrac14
  2. -\dfrac12
  3. −2
  1. 8
  2. 16
SUMBER
ALTERNATIF PENYELESAIAN
\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\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\left(\dfrac{-1}4\right)\left(\dfrac{-8}{-2}\right)^\frac32\\[3.7pt] &=\left(\dfrac{-1}4\right)(4)^\frac32\\[3.7pt] &=\left(\dfrac{-1}4\right)(8)\\[3.7pt] &=\boxed{\boxed{-2}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\dfrac{5-x^2}{4x^2+x}\right)\left(\dfrac{1-8x^3}{3x-2x^3}\right)^\frac32=-2.
JAWAB: C

No.

\displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x=....
Matematika Zone Id
ALTERNATIF PENYELESAIAN
\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}\\[3.7pt] &=\dfrac22\\ &=\boxed{\boxed{1}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x=1.

No.

Dengan menerapkan sifat limit tak hingga, {\displaystyle\lim_{x\to\infty}\dfrac{12x^5-6x^3+7x-1}{4x^5-8x+10}=}
  1. 0
  2. -\dfrac1{10}
  3. 3
  1. 12
  2. ∞
SUMBER
ALTERNATIF PENYELESAIAN

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}
Jadi, \displaystyle\lim_{x\to\infty}\dfrac{12x^5-6x^3+7x-1}{4x^5-8x+10}=3.
JAWAB: C


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