$\lim\limits_{x\to \large\frac{\pi}{6}}\large\frac{3\sin x-\sqrt 3\cos x}{6x-\pi}$
Using L Hospital rule
$\Rightarrow \lim\limits_{x\to \large\frac{\pi}{6}}\large\frac{3\cos x+\sqrt 3\sin x}{6}$
$\Rightarrow \large\frac{3(\Large\frac{\sqrt 3}{2})\normalsize +\sqrt 3(\Large\frac{1}{2})}{6}$
$\Rightarrow \large\frac{4\sqrt 3}{12}$
$\Rightarrow \large\frac{1}{\sqrt 3}$
Hence (b) is the correct answer.