Question

Solve
$\underset{x\to 0}{lim}\frac{\tan 8x}{\sin 2x}$

Answer 1

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

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

We know that,

$\underset{x\to 0}{lim}\frac{\tan 8x}{8x}=1$

$\underset{x\to 0}{lim}\frac{2x}{\sin 2x}=1$

$\therefore \underset{x\to 0}{lim}\frac{\tan 8x}{8x}\times \frac{2x}{\sin 2x}\times \frac{8x}{2x}=\underset{x\to 0}{lim}1\times 1\times \frac{8}{2}$

Hence, $\underset{x\to 0}{lim}\frac{\tan 8x}{\sin 2x}=4$