Question

Then value of $\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}$

• A. $e$
• B. $1$
• C. $e^2$
• D. $e^{-1}$

Answer 1

The correct option is D $e^{-1}$

Let $L=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}$

$L={lim}_{x\to 0}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{\frac{x}{\sin x}-1}}={\left(1\right)}^{\frac{1}{1-1}}\Rightarrow 1^{\infty }$form

$L=e^{{lim}_{x\to 0}\frac{\sin x}{x-\sin x}(\frac{\sin x}{x}-1)}$

$=e^{{lim}_{x\to 0}\frac{\sin x}{x-\sin x}\left(\frac{\sin x-x}{x}\right)}$

$L=e^{-{lim}_{x\to 0}\frac{\sin x}{x}}=e^{-1}$