MCQ
$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $
- A$\sqrt 3 $
- ✓$1/\sqrt 3 $
- C$ - \sqrt 3 $
- D$ - 1/\sqrt 3 $
$\mathop {\lim }\limits_{x \to \pi /6} \frac{{3\cos x + \sqrt 3 \sin x}}{6} = \frac{{3.\frac{{\sqrt 3 }}{2} + \sqrt 3 .\frac{1}{2}}}{6} = \frac{1}{{\sqrt 3 }}$.
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