Question

Find the following limit:
$\underset{x\to 0}{lim}\frac{\sin 5x-\sin 3x}{\sin x}.$

Answer 1

$=\underset{x\to 0}{lim}\frac{\sin 5x-\sin 3x}{\sin x}$

$=\underset{x\to 0}{lim}\frac{2\cos \left(\frac{5x+3x}{2}\right)\sin \left(\frac{5x-3x}{2}\right)}{\sin x}$

$=\underset{x\to 0}{lim}\frac{2\cos 4x\sin x}{\sin x}[\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$=\underset{x\to 0}{lim}2\cos 4x$

$=2\cos 4\left(0\right)$

$=2\times 1$.