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

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\left(\sqrt{4x+2}-\sqrt{4x}\right)\sqrt{x+1}&=\displaystyle\lim_{x\to\infty}\left(\left(\sqrt{4x+2}\right)\left(\sqrt{x+1}\right)-\left(\sqrt{4x}\right)\left(\sqrt{x+1}\right)\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{4x^2+6x+2}-\sqrt{4x^2+4x}\right)\\ &=\dfrac{6-4}{2\sqrt4}\\[9pt] &=\boxed{\boxed{\dfrac12}} \end{aligned}\)

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
   

\(\begin{aligned} &\displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}+\sqrt{x^2-x+2}-\sqrt{4x^2+8x-5}&=\displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}+\sqrt{x^2-x+2}-\sqrt{4\left(x^2+2x-\dfrac54\right)}\\ &=\displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}+\sqrt{x^2-x+2}-2\sqrt{x^2+2x-\dfrac54}\\ &=\displaystyle\lim_{x\to\infty}\sqrt{x^2+3x-5}-\sqrt{x^2+2x-\dfrac54}+\sqrt{x^2-x+2}-\sqrt{x^2+2x-\dfrac54}\\ &=\dfrac{3-2}{2\sqrt1}+\dfrac{-1-2}{2\sqrt1}\\ &=\dfrac12+\dfrac{-3}2\\ &=\boxed{\boxed{-1}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-3x-2&=\displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{4x^2-5x}-5x+2x-2\\ &=\displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-5x+2x-2-\sqrt{4x^2-5x}\\ &=\displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{(5x)^2}+\sqrt{(2x-2)^2}-\sqrt{4x^2-5x}\\ &=\displaystyle\lim_{x\to\infty}\sqrt{25x^2-100x}-\sqrt{25x^2}+\sqrt{4x^2-8x+4}-\sqrt{4x^2-5x}\\ &=\dfrac{-100-0}{2\sqrt{25}}+\dfrac{-8-(-5)}{2\sqrt4}\\[8pt] &=\dfrac{-100}{10}+\dfrac{-3}4\\[8pt] &=-10-\dfrac34\\ &=\boxed{\boxed{-\dfrac{43}4}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-5x&=\displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}+\sqrt{4x^2-3x+7}-3x-2x\\ &=\displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}-3x+\sqrt{4x^2-3x+7}-2x\\ &=\displaystyle\lim_{x\to\infty}\sqrt{9x^2+2}-\sqrt{9x^2}+\sqrt{4x^2-3x+7}-\sqrt{4x^2}\\ &=\dfrac{0-0}{2\sqrt9}+\dfrac{-3-0}{2\sqrt4}\\[8pt] &=0+\dfrac{-3}4\\ &=\boxed{\boxed{-\dfrac34}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to\infty}f(x)&=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+5x}-\sqrt{x^2-2x+1}\right)\\ &=\dfrac{5-(-2)}{2\sqrt1}\\[8pt] &=\dfrac72 \end{aligned}\)

\(\begin{aligned} \displaystyle\lim_{x\to\infty}g(x)&=\displaystyle\lim_{x\to\infty}\left(x-\sqrt{x^2+2x}\right)\\ &=\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2}-\sqrt{x^2+2x}\right)\\ &=\dfrac{0-2}{2\sqrt1}\\[8pt] &=\dfrac{-2}2\\ &=\boxed{\boxed{-1}} \end{aligned}\)

\(\begin{aligned} \displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}&=\dfrac{\displaystyle\lim_{x\to\infty}f(x)}{\displaystyle\lim_{x\to\infty}g(x)}\\[8pt] &=\dfrac{\dfrac72}{-1}\\ &=\boxed{\boxed{-\dfrac72}} \end{aligned}\)

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