Question

The value of ${lim}_{x\to 0}\frac{\sqrt[3]{31+{\sin }^{2}x}-\sqrt[4]{41-2\tan x}}{\sin x+{\tan }^{2}x}$ equals to

• A. $-1$
• B. $\frac{1}{2}$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{2}$

Using approximations, we have

$L={lim}_{x\to 0}\frac{\frac{{\sin }^{2}x}{3}+\frac{2\tan x}{4}}{\sin x+{\tan }^{2}x}={lim}_{x\to 0}\frac{\frac{x^2}{3}+\frac{2x}{4}}{x+x^2}$

$={lim}_{x\to 0}\frac{x(\frac{x}{3}+\frac{1}{2})}{x(1+x)}=\frac{1}{2}$