Exercise Zone : Limit Tak Hingga

Berikut ini adalah kumpulan soal mengenai limit tak hingga tingkat dasar. Ingin bertanya soal? silahkan gabung ke grup Facebook https://web.facebook.com/groups/matematikazoneid atau Telegram https://t.me/MatematikaZoneIdGrup.

Tipe:

Standar SBMPTN HOTS

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No. 1

\displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)=

Grup Telegram

Misal x=p^2\to \sqrt{x}=p.

CARA BIASA

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}})}& ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}(\sqrt{2013p^2+\sqrt{2013}p}-\sqrt{2013p^2-\sqrt{2013}p})\cdot {\displaystyle \frac{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}}}\\ & ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}{\displaystyle \frac{(2013p^2+\sqrt{2013}p)-(2013p^2-\sqrt{2013}p)}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}}}\\ & ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}{\displaystyle \frac{2013p^2+\sqrt{2013}p-2013p^2+\sqrt{2013}p}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}}}\\ & ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}{\displaystyle \frac{2\sqrt{2013}p}{\sqrt{2013p^2+\sqrt{2013}p}+\sqrt{2013p^2-\sqrt{2013}p}}}}\\ & ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{2\sqrt{2013}p}{p}}}{\sqrt{{\displaystyle \frac{2013p^2}{p^2}}+{\displaystyle \frac{\sqrt{2013}p}{p^2}}}+\sqrt{{\displaystyle \frac{2013p^2}{p^2}}-{\displaystyle \frac{\sqrt{2013}p}{p^2}}}}}}\\ & ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}{\displaystyle \frac{2\sqrt{2013}}{\sqrt{2013+{\displaystyle \frac{\sqrt{2013}}{p}}}+\sqrt{2013-{\displaystyle \frac{\sqrt{2013}}{p}}}}}}\\ & ={\displaystyle \frac{2\sqrt{2013}}{\sqrt{2013+0}+\sqrt{2013-0}}}\\ & ={\displaystyle \frac{2\sqrt{2013}}{\sqrt{2013}+\sqrt{2013}}}\\ & ={\displaystyle \frac{2\sqrt{2013}}{2\sqrt{2013}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

CARA CEPAT

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}})}& ={\displaystyle \underset{p\to \mathrm{\infty }}{lim}(\sqrt{2013p^2+\sqrt{2013}p}-\sqrt{2013p^2-\sqrt{2013}p})}\\ & ={\displaystyle \frac{\sqrt{2013}-(-\sqrt{2013})}{2\sqrt{2013}}}\\ & ={\displaystyle \frac{\sqrt{2013}+\sqrt{2013}}{2\sqrt{2013}}}\\ & ={\displaystyle \frac{2\sqrt{2013}}{2\sqrt{2013}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{2013x+\sqrt{2013x}}-\sqrt{2013x-\sqrt{2013x}}\right)=1.

No. 2

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

Grup Telegram

CARA BIASA

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{{\displaystyle \frac{12x^2}{x^2}}-{\displaystyle \frac{4x}{x^2}}+{\displaystyle \frac{2013}{x^2}}}+2\sqrt{{\displaystyle \frac{x^2}{x^2}}}}{{\displaystyle \frac{x}{x}}-{\displaystyle \frac{2013}{x}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{12-{\displaystyle \frac{4}{x}}+{\displaystyle \frac{2013}{x^2}}}+2}{1-{\displaystyle \frac{2013}{x}}}}}\\ & ={\displaystyle \frac{\sqrt{12-0+0}+2}{1-0}}\\ & ={\displaystyle \frac{\sqrt{12}+2}{1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\sqrt{3}+2}}}}\end{array}$

CARA CEPAT

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{12x^2}+2\sqrt{x^2}}{x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2\sqrt{3}x+2x}{x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}2\sqrt{3}+2}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\sqrt{3}+2}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{\sqrt{12x^2-4x+2013}+2\sqrt{x^2}}{x-2013}=2\sqrt3+2.

No. 3

\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)=

Telegram

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x)}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-x-2x)}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+2x-3}-x+\sqrt{4x^2+6x+3}-2x)}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+2x-3}-\sqrt{x^2}+\sqrt{4x^2+6x+3}-\sqrt{4x^2})}\\ & ={\displaystyle \frac{2-0}{2\sqrt{1}}}+{\displaystyle \frac{6-0}{2\sqrt{4}}}\\ & ={\displaystyle \frac{2}{2}}+{\displaystyle \frac{6}{4}}\\ & =1+{\displaystyle \frac{3}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{5}{2}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+2x-3}+\sqrt{4x^2+6x+3}-3x\right)=\dfrac52.

No. 4

\displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}= ....

1. 0
2. 2
   
3. 4
   

4. 8
5. \infty
   

SUMBER

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{50+x+8x^2}{x(2x-5)}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{8x^2}{x(2x)}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{8x^2}{2x^2}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{50+x+8x^2}{x(2x-5)}=4.

No. 5

Nilai \displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x=

CARA 1

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2-3x}-3x}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2-3x}-3x\cdot {\displaystyle \frac{\sqrt{9x^2-3x}+3x}{\sqrt{9x^2-3x}+3x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{9x^2-3x-(3x{)}^2}{\sqrt{9x^2-3x}+3x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{9x^2-3x-9x^2}{\sqrt{9x^2-3x}+3x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-3x}{\sqrt{9x^2-3x}+3x}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-{\displaystyle \frac{3x}{x}}}{\sqrt{{\displaystyle \frac{9x^2}{x^2}}-{\displaystyle \frac{3x}{x^2}}}+{\displaystyle \frac{3x}{x}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-3}{\sqrt{9-{\displaystyle \frac{3}{x}}}+3}}}\\ & ={\displaystyle \frac{-3}{\sqrt{9-0}+3}}\\ & ={\displaystyle \frac{-3}{\sqrt{9}+3}}\\ & ={\displaystyle \frac{-3}{3+3}}\\ & ={\displaystyle \frac{-3}{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{2}}}}}}\end{array}$

CARA 2

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2-3x}-3x}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2-3x}-\sqrt{(3x{)}^2}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2-3x}-\sqrt{9x^2}}\\ & ={\displaystyle \frac{-3-0}{2\sqrt{9}}}\\ & ={\displaystyle \frac{-3}{2(3)}}\\ & ={\displaystyle \frac{-3}{6}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{2}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{9x^2-3x}-3x=.
JAWAB: C

No. 6

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
   

4. -2,7
5. -2,8
   

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}}& ={\displaystyle \frac{-7-18}{2\sqrt{25}}}\\ & ={\displaystyle \frac{-25}{2(5)}}\\ & =-{\displaystyle \frac{5}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2,5}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{25x^2-7x+16}-\sqrt{25x^2+18x+9}=-2,5.
JAWAB: B

No. 7

Nilai \displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}=

1. 10
2. 3
3. \dfrac54

4. 8
5. \infty

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

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{2x^2+3x}{\sqrt{x^2-x}}=\infty.
JAWAB: E

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=

1. -\dfrac14
2. -\dfrac12
3. -2

4. 8
5. 16

SUMBER

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\left({\displaystyle \frac{5-x^2}{4x^2+x}}\right){\left({\displaystyle \frac{1-8x^3}{3x-2x^3}}\right)}^{\frac{3}{2}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\left({\displaystyle \frac{-x^2}{4x^2}}\right){\left({\displaystyle \frac{-8x^3}{-2x^3}}\right)}^{\frac{3}{2}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\left({\displaystyle \frac{-1}{4}}\right){\left({\displaystyle \frac{-8}{-2}}\right)}^{\frac{3}{2}}}\\ & =\left({\displaystyle \frac{-1}{4}}\right)(4{)}^{\frac{3}{2}}\\ & =\left({\displaystyle \frac{-1}{4}}\right)(8)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2}}}}\end{array}$

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. 9

\displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x

SUMBER

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

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{x^2+2x}-x=1.

No. 10

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

4. 12
5. \infty

CARA 1

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{12x^5-6x^3+7x-1}{4x^5-8x+10}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{12x^5}{x^5}}-{\displaystyle \frac{6x^3}{x^5}}+{\displaystyle \frac{7x}{x^5}}-{\displaystyle \frac{1}{x^5}}}{{\displaystyle \frac{4x^5}{x^5}}-{\displaystyle \frac{8x}{x^5}}+{\displaystyle \frac{10}{x^5}}}}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{12-{\displaystyle \frac{6}{x^2}}+{\displaystyle \frac{7}{x^4}}-{\displaystyle \frac{1}{x^5}}}{4-{\displaystyle \frac{8}{x^4}}+{\displaystyle \frac{10}{x^5}}}}}\\ & ={\displaystyle \frac{12-0+0-0}{4-0+0}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

CARA 2

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{12x^5-6x^3+7x-1}{4x^5-8x+10}}}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{12\cancel{x^5}}{4\cancel{x^5}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

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

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