Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\sin 3x}{5x}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{\sin 3x}{5x}$

$=\frac{1}{5}\underset{x\to 0}{lim}\frac{3\times \sin 3x}{3x}$

$=\frac{1}{5}\times 3\times \underset{x\to 0}{lim}\frac{\sin 3x}{3x}$

$=\frac{1}{5}\times 3\times 1=\frac{3}{5}[\because \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1]$

Therefore,
$\underset{x\to 0}{lim}\frac{\sin 3x}{5x}=\frac{3}{5}$