Question

Find :
$\underset{x\to \frac{\pi }{6}}{lim}\left(\frac{\sqrt{3}\sin x-\cos x}{x-\frac{\pi }{6}}\right)$

Answer 1

We have,

$\underset{x\to \frac{\pi }{6}}{lim}\left(\frac{\sqrt{3}\sin x-\cos x}{x-\frac{\pi }{6}}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

$\underset{x\to \frac{\pi }{6}}{lim}\left(\frac{\sqrt{3}\cos x-(-\sin x)}{1-0}\right)$

$\underset{x\to \frac{\pi }{6}}{lim}(\sqrt{3}\cos x+\sin x)$

$=(\sqrt{3}\cos \frac{\pi }{6}+\sin \frac{\pi }{6})$

$=(\sqrt{3}\times \frac{\sqrt{3}}{2}+\frac{1}{2})$

$=\frac{3}{2}+\frac{1}{2}$

$=2$

Hence, this is the answer.