Explanation

(b)
$\lim_{x \rightarrow \frac{\pi}{6}}\frac{3\sin{x - \sqrt{3}\cos x}}{6x - \pi} = \lim_{x \rightarrow \frac{\pi}{6}}\frac{3\cos{x + \sqrt{3}\cos x}}{6}$
$= \frac{3\cos{\frac{\pi}{6} + \sqrt{3}}\sin\frac{\pi}{6}}{6}$
$= \frac{3.\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{6}$
$= \frac{1}{\sqrt{3}}$