Evaluate the following limits:

$=\lim _{x \rightarrow 0} \frac{\left(2 \sin \frac{5 x-3 x}{2} \cos \frac{5 x+3 x}{2}\right)}{\sin x}[$ Applying $\sin C-\sin D$

$\left.=2 \sin \frac{C-D}{2} \cos \frac{C+D}{2}\right]$

$=\lim _{x \rightarrow 0} 2 \cos 4 x$

$=2 \times 1$

$\therefore \lim _{x \rightarrow 0} \frac{\sin 5 x-\sin 3 x}{\sin x}=2$