## Calculus (3rd Edition)

$\lim\limits_{x \to 0}\frac{\sin 5x}{x}=\lim\limits_{x \to 0}[\frac{\sin 5x}{5x}\cdot 5]$ $=5\cdot \lim\limits_{5x \to 0}[\frac{\sin 5x}{5x}]$ ( 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