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Exercise Zone : Limit Tak Hingga [2]

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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}\dfrac{\sqrt{4x^4-5x+1}}{5-2x^2}=
  1. 0
  2. −1
  3. 1
  1. 2
  2. ∞
SUMBER
ALTERNATIF PENYELESAIAN

CARA 1

\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4x^4-5x+1}}{5-2x^2}&=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{\dfrac{4x^4}{\color{red}x^4}-\dfrac{5x}{\color{red}x^4}+\dfrac1{\color{red}x^4}}}{\dfrac5{\color{red}x^2}-\dfrac{2x^2}{\color{red}x^2}}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4-\dfrac5{x^3}+\dfrac1{x^4}}}{\dfrac5{x^2}-2}\\[3.7pt] &=\dfrac{\sqrt{4-0+0}}{0-2}\\[3.7pt] &=\dfrac{\sqrt4}{-2}\\[3.7pt] &=\dfrac2{-2}\\ &=\boxed{\boxed{-1}} \end{aligned}

CARA 2

\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4x^4-5x+1}}{5-2x^2}&=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4x^4}}{-2x^2}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\dfrac{2{\color{red}{\cancel{\color{black}x^2}}}}{-2{\color{red}{\cancel{\color{black}x^2}}}}\\ &=\boxed{\boxed{-1}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4x^4-5x+1}}{5-2x^2}=-1.
JAWAB: B

No.

Nilai dari {\displaystyle\lim_{x\to\infty}\left(\dfrac12\sqrt{4x^2-x}-x\right)=}
  1. -\dfrac18
  2. -\dfrac14
  3. -\dfrac12
  1. \dfrac14
  2. \dfrac18
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\dfrac12\sqrt{4x^2-x}-x\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{\dfrac14\left(4x^2-x\right)}-\sqrt{x^2}\right)\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2-\dfrac14x}-\sqrt{x^2}\right)\\[3.7pt] &=\dfrac{-\dfrac14-0}{2\sqrt1}\\[3.7pt] &=\dfrac{-\dfrac14}2\\ &=\boxed{\boxed{-\dfrac18}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\dfrac12\sqrt{4x^2-x}-x\right)=-\dfrac18.
JAWAB: A

No.

Nilai b yang memenuhi {\displaystyle\lim_{x\to\infty}\sqrt{4x^2+bx+5}-\sqrt{4x^2-6x}=18} adalah
  1. 65
  2. 66
  3. 67
  1. 68
  2. 69
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{4x^2+bx+5}-\sqrt{4x^2-6x}&=18\\[3.7pt] \dfrac{b-(-6)}{2\sqrt4}&=18\\[3.7pt] \dfrac{b+6}4&=18\\[3.7pt] b+6&=72\\ b&=\boxed{\boxed{66}} \end{aligned}
Jadi, b = 66.
JAWAB: B

No.

Nilai dari limit fungsi {\displaystyle\lim_{x\to\infty}\sqrt{(2x+2)(2x+4)}-2x+1} adalah
  1. −2
  2. 2
  3. 0
  1. 4
  2. −4
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{(2x+2)(2x+4)}-2x+1&=\displaystyle\lim_{x\to\infty}\sqrt{4x^2+8x+4x+8}-(2x+1)\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\sqrt{4x^2+12x+8}-\sqrt{(2x+1)^2}\\[3.7pt] &=\displaystyle\lim_{x\to\infty}\sqrt{4x^2+12x+8}-\sqrt{4x^2+4x+1}\\[3.7pt] &=\dfrac{12-4}{2\sqrt4}\\[3.7pt] &=\dfrac84\\ &=\boxed{\boxed{2}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\sqrt{(2x+2)(2x+4)}-2x+1.
JAWAB: B

No.

\displaystyle\lim_{x\to\infty}\left(\sqrt{(x+2)(x-3)}-x+1\right)=
  1. \dfrac12
  2. 1
  3. 3
  1. −1
  2. -\dfrac12
SKMP Juli 2020
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\sqrt{(x+2)(x-3)}-x+1\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2-x-6}-(x-1)\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2-x-6}-\sqrt{(x-1)^2}\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2-x-6}-\sqrt{x^2-2x+1}\right)\\ &=\dfrac{-1-(-2)}{2\sqrt1}\\ &=\boxed{\boxed{\dfrac12}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{(x+2)(x-3)}-x+1\right)=\dfrac12.
JAWAB: A

No.

Nilai \displaystyle\lim_{x\to\infty}\left(3x-\sqrt{9x^2+27}\right)= ....
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\left(3x-\sqrt{9x^2+27}\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{9x^2}-\sqrt{9x^2+27}\right)\\ &=\dfrac{0-0}{2\sqrt9}\\ &=\boxed{\boxed{0}} \end{aligned}
Jadi, \displaystyle\lim_{x\to\infty}\left(3x-\sqrt{9x^2+27}\right)=0.

No.

Tentukan nilai dari {\displaystyle\lim_{x\to\infty}\left(\sqrt{2x+3}-\sqrt{3x-2}\right)}
Brainly.co.id
ALTERNATIF PENYELESAIAN

CARA BIASA

\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\sqrt{2x+3}-\sqrt{3x-2}\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{2x+3}-\sqrt{3x-2}\right){\color{red}\cdot\dfrac{\sqrt{2x+3}+\sqrt{3x-2}}{\sqrt{2x+3}+\sqrt{3x-2}}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{(2x+3)-(3x-2)}{\sqrt{2x+3}+\sqrt{3x-2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{2x+3-3x+2}{\sqrt{2x+3}+\sqrt{3x-2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{-x+5}{\sqrt{2x+3}+\sqrt{3x-2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{-\dfrac{x}x+\dfrac5x}{\sqrt{\dfrac{2x}{x^2}+\dfrac3{x^2}}+\sqrt{\dfrac{3x}{x^2}-\dfrac2{x^2}}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{-1+\dfrac5x}{\sqrt{\dfrac2x+\dfrac3{x^2}}+\sqrt{\dfrac3x-\dfrac2{x^2}}}\\ &=\dfrac{-1+0}{\sqrt{0+0}+\sqrt{0-0}}\\ &=\dfrac{-1}0\\ &=\boxed{\boxed{-\infty}} \end{aligned}

CARA CEPAT

2x <3x, jadi hasilnya adalah −∞.
Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{2x+3}-\sqrt{3x-2}\right)=-\infty.

No.

Tentukan nilai dari {\displaystyle\lim_{x\to\infty}\left(\sqrt{3x-10}-\sqrt{x+2}\right)}
Brainly.co.id
ALTERNATIF PENYELESAIAN

CARA BIASA

\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\sqrt{3x-10}-\sqrt{x+2}\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{3x-10}-\sqrt{x+2}\right){\color{red}\cdot\dfrac{\sqrt{3x-10}+\sqrt{x+2}}{\sqrt{3x-10}+\sqrt{x+2}}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{(3x-10)-(x+2)}{\sqrt{3x-10}+\sqrt{x+2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{3x-10-x-2}{\sqrt{3x-10}+\sqrt{x+2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{2x-12}{\sqrt{3x-10}+\sqrt{x+2}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{\dfrac{2x}x-\dfrac{12}x}{\sqrt{\dfrac{3x}{x^2}-\dfrac{10}{x^2}}+\sqrt{\dfrac{x}{x^2}+\dfrac2{x^2}}}\\ &=\displaystyle\lim_{x\to\infty}\dfrac{2-\dfrac{12}x}{\sqrt{\dfrac3x-\dfrac{10}{x^2}}+\sqrt{\dfrac1x+\dfrac2{x^2}}}\\ &=\dfrac{2-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\dfrac20\\ &=\boxed{\boxed{\infty}} \end{aligned}

CARA CEPAT

3x > x, jadi hasilnya adalah ∞.
Jadi,
JAWAB:

No.

Nilai dari \displaystyle\lim_{x\to\infty}\left((3x-2)-\sqrt{9x^2-2x+5}\right) adalah ....
latiseducation.com
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\left((3x-2)-\sqrt{9x^2-2x+5}\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{9x^2-12x+4}-\sqrt{9x^2-2x+5}\right)\\ &=\frac{b-q}{2\sqrt{a}}\\[3.5pt] &=\frac{-12-(-2)}{2\sqrt9}\\[3.5pt] &=\frac{-10}{2(3)}\\[3.5pt] &=\frac{-10}6\\ &=\boxed{\boxed{-\frac53}} \end{aligned}
Jadi, nilai dari \displaystyle\lim_{x\to\infty}\left((3x-2)-\sqrt{9x^2-2x+5}\right) adalah -\dfrac53.

No.

Nilai dari \displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-5x+3\right) adalah ....
latiseducation.com
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-5x+3\right)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-(5x-3)\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-\sqrt{(5x-3)^2}\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-\sqrt{25x^2-30x+9}\right)\\ &=\frac{b-q}{2\sqrt{a}}\\[3.5pt] &=\frac{-9-(-30)}{2\sqrt{25}}\\[3.5pt] &=\frac{21}{2(5)}\\ &=\boxed{\boxed{\frac{21}{10}}} \end{aligned}
Jadi, nilai dari \displaystyle\lim_{x\to\infty}\left(\sqrt{25x^2-9x-6}-5x+3\right) adalah \dfrac{21}{10}.


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