Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{1-\cos 6x}}{\sqrt{2}(\pi /3-x)}$

Answer 1

We have,

$\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{1-\cos 6x}}{\sqrt{2}(\pi /3-x)}$

$=\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{1-\cos 2\left(3x\right)}}{\sqrt{2}(\pi /3-x)}$

$=\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{2{\sin }^{2}3x}}{\sqrt{2}(\pi /3-x)}$ $[\because 1-\cos 2x=2{\sin }^{2}x]$

$=\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{2}\sin 3x}{\sqrt{2}(\pi /3-x)}$

$=\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sin 3x}{(\pi /3-x)}$

Put $x=\frac{\pi }{3}+h,h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 3(\frac{\pi }{3}+h)}{(\frac{\pi }{3}-(\frac{\pi }{3}+h))}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin (\pi +3h)}{-h}$ $[\because \sin (\pi +\theta )=-\sin \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{-\sin 3h}{-h}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 3h}{\frac{1}{3}\times 3h}$ $[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1]$

$=3\left(1\right)=3$

Therefore,

$\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{1-\cos 6x}}{\sqrt{2}(\pi /3-x)}=3$