Exercise Zone : Limit Tak Hingga [2]

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

Tipe:

Standar SBMPTN HOTS

No.

\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4x^4-5x+1}}{5-2x^2}=

1. 0
2. -1
   
3. 1
   

4. 2
5. \infty
   

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}}\\[20pt] &=\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{4-\dfrac5{x^3}+\dfrac1{x^4}}}{\dfrac5{x^2}-2}\\[20pt] &=\dfrac{\sqrt{4-0+0}}{0-2}\\[8pt] &=\dfrac{\sqrt4}{-2}\\[8pt] &=\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}\\[8pt] &=\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

4. \dfrac14
5. \dfrac18

\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)\\[8pt] &=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2-\dfrac14x}-\sqrt{x^2}\right)\\[8pt] &=\dfrac{-\dfrac14-0}{2\sqrt1}\\[8pt] &=\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. 22
2. 21
   
3. 20
   

4. 19
5. 18
   

\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{4x^2+bx+5}-\sqrt{4x^2-6x}&=18\\[8pt] \dfrac{b-(-6)}{2\sqrt4}&=18\\[8pt] \dfrac{b+6}4&=18\\[8pt] b+6&=72\\ b&=\boxed{\boxed{66}} \end{aligned}

Jadi, b=66. JAWAB: -

No.

Nilai dari limit fungsi {\displaystyle\lim_{x\to\infty}\sqrt{(2x+2)(2x+4)}-2x+1} adalah

1. -2
2. 2
   
3. 0
   

4. 4
5. -4
   

\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)\\[8pt] &=\displaystyle\lim_{x\to\infty}\sqrt{4x^2+12x+8}-\sqrt{(2x+1)^2}\\[8pt] &=\displaystyle\lim_{x\to\infty}\sqrt{4x^2+12x+8}-\sqrt{4x^2+4x+1}\\[8pt] &=\dfrac{12-4}{2\sqrt4}\\[8pt] &=\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

4. -1
5. -\dfrac12

SKMP Juli 2020

\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

\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

CARA BIASA

CARA CEPAT

\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}

{2x\lt3x}, jadi hasilnya adalah -\infty.

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

CARA BIASA

CARA CEPAT

\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}

{3x\gt x}, jadi hasilnya adalah \infty.

Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{3x-10}-\sqrt{x+2}\right)=\infty.