Question

$e^m={\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}$, then the value of $m$ is

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

The correct option is B $-1$

${\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}=exp\left(\underset{x\to 0}{lim}\frac{\sin x}{x-\sin x}(\frac{\sin x}{x}-1)\right)$
$[$ Using ${lim}_{x\to a}{\left[f\left(x\right)\right]}^{g\left(x\right)}=exp\left({lim}_{x\to a}g\left(x\right)[f\left(x\right)-1]\right)]$
as $f\left(x\right)=\frac{\sin x}{x}\to 1$ and $g\left(x\right)=\frac{\sin x}{x-\sin x}=\frac{\frac{\sin x}{x}}{1-\frac{\sin x}{x}}\to \infty$ $[$ as $x\to 0]$
$=exp\left(\underset{x\to 0}{lim}\frac{\sin x}{x}\right)=e^{-1}$