Question

Evaluate:

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

Answer 1

${lim}_{x\to \frac{\pi }{6}}\frac{\sqrt{3}\sin x-\cos x}{x-\frac{\pi }{6}}$ $(\frac{0}{0}\to \text{indeterminate form})$

$={lim}_{x\to \frac{\pi }{6}}\frac{2[\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x]}{x-\frac{\pi }{6}}$

$={lim}_{x\to \frac{\pi }{6}}\frac{2[\cos \frac{\pi }{6}\sin x-\sin \frac{\pi }{6}.\cos x]}{x-\frac{\pi }{6}}$

$={lim}_{x\to \frac{\pi }{6}}\frac{2\sin (x-\frac{\pi }{6})}{(x-\frac{\pi }{6})}$

$(\because \sin (A-B)=\sin A\cos B–\cos A\sin B)$
Put $x-\frac{\pi }{6}=y$

So, when $x\to \frac{\pi }{6}\Rightarrow y\to 0$

$={lim}_{y\to 0}2\times \frac{\sin y}{y}$

$=2$ $[\because {lim}_{x\to 0}\frac{\sin x}{x}=1]$