SBMPTN Zone : Limit Tak HIngga

Berikut ini adalah kumpulan soal mengenai limit tak hingga tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Tentukan nilai dari \displaystyle\lim_{x\to\infty}\left(\sqrt{4x+2}-\sqrt{4x}\right)\sqrt{x+1}!

Ganesha Operation

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

Jadi, \displaystyle\lim_{x\to\infty}\left(\sqrt{4x+2}-\sqrt{4x}\right)\sqrt{x+1}=\dfrac12.

No. 2

\displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}+\sqrt{x^2-x+2}-\sqrt{4x^2+8x-5}= ....

1. \dfrac32
2. 1
   
3. \dfrac12
   

4. 0
5. -1
   

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

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}+\sqrt{x^2-x+2}-\sqrt{4x^2+8x-5}=-1.

No. 3

Tentukan nilai \displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-3x-2?

SUMBER

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-3x-2}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-5x+2x-2}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-100x}-5x+2x-2-\sqrt{4x^2-5x}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-100x}-\sqrt{(5x{)}^2}+\sqrt{(2x-2{)}^2}-\sqrt{4x^2-5x}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{25x^2-100x}-\sqrt{25x^2}+\sqrt{4x^2-8x+4}-\sqrt{4x^2-5x}}\\ & ={\displaystyle \frac{-100-0}{2\sqrt{25}}}+{\displaystyle \frac{-8-(-5)}{2\sqrt{4}}}\\ & ={\displaystyle \frac{-100}{10}}+{\displaystyle \frac{-3}{4}}\\ & =-10-{\displaystyle \frac{3}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{43}{4}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-3x-2=-\dfrac{43}4.

No. 4

Tentukan nilai \displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-5x?

SUMBER

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-5x}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-3x-2x}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+2}-3x+\sqrt{4x^2-3x+7}-2x}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+2}-\sqrt{9x^2}+\sqrt{4x^2-3x+7}-\sqrt{4x^2}}\\ & ={\displaystyle \frac{0-0}{2\sqrt{9}}}+{\displaystyle \frac{-3-0}{2\sqrt{4}}}\\ & =0+{\displaystyle \frac{-3}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{3}{4}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-5x=-\dfrac34.

No. 5

Jika diketahui dua fungsi f(x) dan g(x) dengan {f(x)=\sqrt{x^2+5x}-\sqrt{x^2-2x+1}} dan {g(x)=x-\sqrt{x^2+2x}}, maka nilai dari {\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=}

1. \dfrac72
2. \dfrac32
3. 0

4. -\dfrac32
5. -\dfrac72

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)}& ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+5x}-\sqrt{x^2-2x+1})}\\ & ={\displaystyle \frac{5-(-2)}{2\sqrt{1}}}\\ & ={\displaystyle \frac{7}{2}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{f(x)}{g(x)}}}& ={\displaystyle \frac{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}f(x)}}{{\displaystyle \underset{x\to \mathrm{\infty }}{lim}g(x)}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{7}{2}}}{-1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{7}{2}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=-\dfrac72.
JAWAB: E