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

Question.

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

solution:

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

Consider $\sqrt{3} \sin x-\cos x=2\left(\frac{\sqrt{3} \sin x}{2}-\frac{\cos x}{2}\right)=2\left(\sin x \cos \left(\frac{\pi}{6}\right)-\cos x \sin \left(\frac{\pi}{6}\right)\right)$

On simplification the above equation can be written as

$\Rightarrow$$2\left(\sin x \cos \left(\frac{\pi}{6}\right)-\cos x \sin \left(\frac{\pi}{6}\right)\right)=2 \sin \left(x-\frac{\pi}{6}\right) \Rightarrow$$\sqrt{3} \sin x-\cos x=2 \sin \left(x-\frac{\pi}{6}\right)$

Now by substituting these values in given equation we get

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin \left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}} Taking constant term 2 outside \Rightarrow$$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin \left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}}=2 \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}}$

Now as $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

Now by applying the limit we get

$\Rightarrow$$2 \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}}=2.1=2 \Rightarrow$$\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{0}}=2$