Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\sin 2x}{e^x-1}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{\sin 2x}{e^x-1}$

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

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

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

$=\frac{2\times 1}{1}=2$

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