Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{e^x-1+\sin x}{x}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{e^x-1+\sin x}{x}$

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

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

$=\log e+1$

$[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a,\underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=1+1=2$

Therefore,
$\underset{x\to 0}{lim}\frac{e^x-1+\sin x}{x}=2$