Question

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

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\sin 3x-\sin x}{\sin x}\right]$

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\frac{2\cos \left({\displaystyle \frac{3x+x}{2}}\right)\sin \left({\displaystyle \frac{3x-x}{2}}\right)}{\sin x}\right]\left[\because \sin \mathrm{C}-\sin \mathrm{D}=2\cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\cos 2x.\sin x}{\sin x}\right] \\ =2\cos 0 \\ =2 \end{gathered}$$