Question

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

Answer 1

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

Dividing the Numerator and Denominator by $x$,
$=\underset{x\to 0}{lim}\frac{\frac{e^{2x}-e^x+1-1}{x}}{2\times \frac{\sin 2x}{2x}}$

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

$=\frac{2\log e-\log e}{2\times 1}$
$[\because \underset{x\to 0}{lim}\frac{a^x-1}{x}=\log a,\underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=\frac{\log e}{2}=\frac{1}{2}$

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