We are given,

$\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sqrt 3 \sin x – \cos x}}{{x – \frac{\pi }{6}}}$

Simplifying the numerator,

$\sqrt 3 {\sin ^^\circ } – \cos x = 2\left( {\frac{{\sqrt 3 \sin x}}{2} – \frac{{\operatorname{cosx} }}{2}} \right)$

$ = 2\left( {\sin x\cos \left( {\frac{\pi }{6}} \right) – \cos x\sin \left( {\frac{\pi }{6}} \right)} \right)$

$ = 2\sin \left( {x – \frac{\pi }{6}} \right)$

$ = 2\sin \left( {x – \frac{\pi }{6}} \right)$

Evaluating the limit we get,

$\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sqrt 3 \sin x – \cos x}}{{x – \frac{\pi }{6}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{2\sin \left( {x – \frac{\pi }{6}} \right)}}{{x – \frac{\pi }{6}}}$

$ = 2\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin \left( {x – \frac{\pi }{6}} \right)}}{{x – \frac{\pi }{6}}}$

Applying, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ and substituting the limit value to get,

$2\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin \left( {x – \frac{\pi }{6}} \right)}}{{x – \frac{\pi }{6}}} = 2.1$

$ = 2$

Therefore, $\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sqrt 3 \sin x – \cos x}}{{x – \frac{\pi }{6}}} = 2$.