# Solve this following

Question:

The value of $\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)$ is equal

to:

1. 0

2. 4

3. $-4$

4. $-1$

Correct Option: 3,

Solution:

$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)$

$\left(\frac{(\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x})}{\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x}}\right)$

$\left(\frac{(\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x})}{\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x}}\right)$

$\left(\frac{(\sqrt[2]{1-\sin x}+\sqrt[2]{1+\sin x})}{\sqrt[2]{1-\sin x}+\sqrt[2]{1+\sin x}}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{x}{1-\sin x-(1+\sin x)}\right)$

$(\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x})(\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x})$

$(\sqrt[2]{1-\sin x}+\sqrt[2]{1+\sin x})$

$=\lim _{x \rightarrow 0} \frac{x}{(-2 \sin x)}(\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x})$

$(\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x})(\sqrt[2]{1-\sin x}+\sqrt[2]{1+\sin x})$

$=\lim _{x \rightarrow 0}\left(-\frac{1}{2}\right)(2)(2)(2)\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}=-4$