$\underset{x\to \frac{\pi }{6}}{lim} \frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$

By L–Hospital rule

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

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

$\Rightarrow \underset{\left(x\to \frac{\pi }{6}\right)}{lim} \left[\frac{\left(2\sqrt{3}\right)\left[\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x\right]}{6\left(x-\frac{\pi }{6}\right)}\right]$

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

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

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

$=\frac{1}{\sqrt{3}}$