Answer
5
Work Step by Step
$\lim\limits_{x \to 0}\frac{\sin 5x}{x}=\lim\limits_{x \to 0}[\frac{\sin 5x}{5x}\cdot 5]$
$=5\cdot \lim\limits_{5x \to 0}[\frac{\sin 5x}{5x}]$ ( as x tends to 0, 5x also tends to 0 )
$= 5\cdot 1$ ( as $ \lim\limits_{x \to 0}\frac{\sin x}{x}=1$ )
= 5
