Question

Evaluate :
$\underset{x\to 0}{lim}\frac{\sin 3x-sinx}{\sin x}.$

Answer 1

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

it is of the form $\frac{0}{0}$

So, we will apply L Hospital rule

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}(\sin 3x-sinx)}{\frac{d}{dx}\sin x}=$ $\underset{x\to 0}{lim}\frac{3\cos 3x-\cos x}{\cos x}=\frac{3-1}{1}=2$