Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\tan 8x}{\sin 2x}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{\tan 8x}{\sin 2x}$

$=\underset{x\to 0}{lim}[\frac{\tan 8x}{8x}\times \frac{8x}{\frac{\sin 2x}{2x}\times 2x}]$

$=\underset{x\to 0}{lim}[\frac{\tan 8x}{8x}\times \frac{4}{\frac{\sin 2x}{2x}}][\because \underset{x\to 0}{lim}\frac{\tan 8x}{8x}=1,\underset{x\to 0}{lim}\frac{\sin 2x}{2x}=1]$

$=1\times \frac{4}{1}=4$

Therefore,
$li{m}_{x\to 0}\left[\frac{tan8x}{sin2x}\right]=4$