Question

For $x>0$, the value of ${lim}_{x\to 0}\{{\left(\sin x\right)}^{1/x}+{\left(\frac{1}{x}\right)}^{\sin x}\}$ is?

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$\underset{x\to 0}{lim}{\left(\sin x\right)}^{\frac{1}{x}}+\underset{x\to 0}{lim}{\left(\frac{1}{x}\right)}^{\sin x}$
$=0+\underset{x\to 0}{lim}{e}^{\log {\left(\frac{1}{x}\right)}^{\sin x}}=e\underset{x\to 0}{lim}\frac{\log \left(\frac{1}{x}\right)}{\csc x}$
$=\exp \left(\underset{x\to 0}{lim}\frac{x(-\frac{1}{x^2})}{-\csc x\cot x}\right)=\exp \left(\underset{x\to 0}{lim}\frac{\sin x}{x}\tan x\right)=e^0=1$