Question

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

Answer 1

As we know,
$x^0=\frac{\pi }{180}x$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x^0}{x}=\underset{x\to 0}{lim}\frac{\sin \frac{\pi }{180}x}{x}$

$=\underset{x\to 0}{lim}\frac{\sin \frac{\pi }{180}x}{\frac{\pi }{180}x}\times \frac{\pi }{180}$

$=\frac{\pi }{180}\times 1[\because \underset{x\to 0}{lim}\frac{\sin \theta }{\theta }=1,\theta \text{is in radians}]=\frac{\pi }{180}$

Therefore,
$\underset{x\to 0}{lim}\frac{\sin x^0}{x}=\frac{\pi }{180}$