Question

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

Answer 1

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

Here $f\left(x\right)=e^x-e^{-x}$

$x\to$, $g\left(0\right)=\sin 0=0$

$\therefore \frac{f\left(0\right)}{g\left(0\right)}=\frac{0}{0}$

By usingL Hospitals rule

${lim}_{x\to 0}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}={lim}_{x\to 0}\frac{e^x+e^{-x}}{\cos x}=\frac{e^0+e^0}{\cos 0}=\frac{1+1}{1}=2$

${lim}_{x\to 0}\frac{e^x-e^{-x}}{\sin x}=2$