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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{4x^4-5x+1}}{5-2x^2}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{{\displaystyle \frac{4x^4}{x^4}}-{\displaystyle \frac{5x}{x^4}}+{\displaystyle \frac{1}{x^4}}}}{{\displaystyle \frac{5}{x^2}}-{\displaystyle \frac{2x^2}{x^2}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{4-{\displaystyle \frac{5}{x^3}}+{\displaystyle \frac{1}{x^4}}}}{{\displaystyle \frac{5}{x^2}}-2}}}\\ & ={\displaystyle \frac{\sqrt{4-0+0}}{0-2}}\\ & ={\displaystyle \frac{\sqrt{4}}{-2}}\\ & ={\displaystyle \frac{2}{-2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$$

CARA 2

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{4x^4-5x+1}}{5-2x^2}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{4x^4}}{-2x^2}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2\cancel{x^2}}{-2\cancel{x^2}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}({\displaystyle \frac{1}{2}}\sqrt{4x^2-x}-x)}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{{\displaystyle \frac{1}{4}}(4x^2-x)}-\sqrt{x^2})}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2-{\displaystyle \frac{1}{4}}x}-\sqrt{x^2})}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{4}}-0}{2\sqrt{1}}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{4}}}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{8}}}}}}\end{array}$$

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{4x^2+bx+5}-\sqrt{4x^2-6x}}& =18\\ {\displaystyle \frac{b-(-6)}{2\sqrt{4}}}& =18\\ {\displaystyle \frac{b+6}{4}}& =18\\ b+6& =72\\ b& =\boxed{{\displaystyle \boxed{{\displaystyle 66}}}}\end{array}$$

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{(2x+2)(2x+4)}-2x+1}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{4x^2+8x+4x+8}-(2x+1)}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{4x^2+12x+8}-\sqrt{(2x+1{)}^2}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{4x^2+12x+8}-\sqrt{4x^2+4x+1}}\\ & ={\displaystyle \frac{12-4}{2\sqrt{4}}}\\ & ={\displaystyle \frac{8}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{(x+2)(x-3)}-x+1)}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2-x-6}-(x-1))}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2-x-6}-\sqrt{(x-1{)}^2})}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2-x-6}-\sqrt{x^2-2x+1})}\\ & ={\displaystyle \frac{-1-(-2)}{2\sqrt{1}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3x-\sqrt{9x^2+27})}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{9x^2}-\sqrt{9x^2+27})}\\ & ={\displaystyle \frac{0-0}{2\sqrt{9}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 0}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{2x+3}-\sqrt{3x-2})}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{2x+3}-\sqrt{3x-2})\cdot {\displaystyle \frac{\sqrt{2x+3}+\sqrt{3x-2}}{\sqrt{2x+3}+\sqrt{3x-2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(2x+3)-(3x-2)}{\sqrt{2x+3}+\sqrt{3x-2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x+3-3x+2}{\sqrt{2x+3}+\sqrt{3x-2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-x+5}{\sqrt{2x+3}+\sqrt{3x-2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-{\displaystyle \frac{x}{x}}+{\displaystyle \frac{5}{x}}}{\sqrt{{\displaystyle \frac{2x}{x^2}}+{\displaystyle \frac{3}{x^2}}}+\sqrt{{\displaystyle \frac{3x}{x^2}}-{\displaystyle \frac{2}{x^2}}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-1+{\displaystyle \frac{5}{x}}}{\sqrt{{\displaystyle \frac{2}{x}}+{\displaystyle \frac{3}{x^2}}}+\sqrt{{\displaystyle \frac{3}{x}}-{\displaystyle \frac{2}{x^2}}}}}}\\ & ={\displaystyle \frac{-1+0}{\sqrt{0+0}+\sqrt{0-0}}}\\ & ={\displaystyle \frac{-1}{0}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -\mathrm{\infty }}}}}\end{array}$$

{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

$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{3x-10}-\sqrt{x+2})}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{3x-10}-\sqrt{x+2})\cdot {\displaystyle \frac{\sqrt{3x-10}+\sqrt{x+2}}{\sqrt{3x-10}+\sqrt{x+2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(3x-10)-(x+2)}{\sqrt{3x-10}+\sqrt{x+2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3x-10-x-2}{\sqrt{3x-10}+\sqrt{x+2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x-12}{\sqrt{3x-10}+\sqrt{x+2}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{2x}{x}}-{\displaystyle \frac{12}{x}}}{\sqrt{{\displaystyle \frac{3x}{x^2}}-{\displaystyle \frac{10}{x^2}}}+\sqrt{{\displaystyle \frac{x}{x^2}}+{\displaystyle \frac{2}{x^2}}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2-{\displaystyle \frac{12}{x}}}{\sqrt{{\displaystyle \frac{3}{x}}-{\displaystyle \frac{10}{x^2}}}+\sqrt{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^2}}}}}}\\ & ={\displaystyle \frac{2-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{2}{0}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \mathrm{\infty }}}}}\end{array}$$

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

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