Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{e^{x+2}-e^2}{x}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{e^{x+2}-e^2}{x}$

$=\underset{x\to 0}{lim}\frac{e^2(e^x-1)}{x}$

$=e^2\underset{x\to 0}{lim}\frac{(e^x-1)}{x}$

$=e^2\times \log e[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

$=e^2\times 1$
$=e^2$

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