Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}$

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

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

Put $y=x-\sin x$
As $x\to 0\Rightarrow x-\sin x\to 0$,
$\therefore y\to 0$

$=e^{\sin 0}\underset{y\to 0}{lim}\frac{e^y-1}{y}$

$=1\times \log e=1$
$[\because {lim}_{x\to 0}\left(\frac{a^x-1}{x}\right)=\log a]$

Therefore,
$\underset{x\to 0}{lim}\frac{e^x-e^{\sin x}}{x-\sin x}=1$