Question

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

• A. $e$
• B. $e^2$
• C. $e^3$
• D. $\frac{1}{e}$

Answer 1

The correct option is D $\frac{1}{e}$

$\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}=\underset{x\to 0}{lim}{(1+\frac{\sin x-x}{x})}^{\frac{\sin x}{x-\sin x}}$

$\underset{x\to 0}{lim}\left(\right(1+x{)}^{\frac{1}{x}})=e$

Using above formula we get,
$=\underset{x\to 0}{lim}{(1+\frac{\sin x-x}{x})}^{\frac{\sin x}{\sin x-x}.(-1)}=e^{-1}=\frac{1}{e}$