Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{1-\sin 2x}{1+\cos 4x}$

Answer 1

We have,

$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{1-\sin 2x}{1+\cos 4x}$

Put $x=\frac{\pi }{4}+h$,

If $x\to \frac{\pi }{4}$, then $h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{1-\sin 2(\frac{\pi }{4}+h)}{1+\cos 4(\frac{\pi }{4}+h)}$

$=\underset{h\to 0}{\text{lim}}\frac{1-\sin (\frac{\pi }{2}+2h)}{1+\cos (\pi +4h)}$

$=\underset{h\to 0}{\text{lim}}\frac{1-\cos 2h}{1-\cos 4h}$

$[\because \sin (\frac{\pi }{2}+\theta )=\cos \theta ,\cos (\pi +\theta )=-\cos \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^{2}h}{2{\sin }^{2}2h}$ $[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^{2}h}{2(2\sin h\cos h{)}^2}$

$[\because \sin 2\theta =2\sin \theta \cos \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^{2}h}{2\times 4{\sin }^{2}h{\cos }^{2}h}$

$=\underset{h\to 0}{\text{lim}}\frac{1}{4{\cos }^{2}h}$

$=\frac{1}{4{\cos }^{2}0}=\frac{1}{4}$

Therefore,

$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{1-\sin 2x}{1+\cos 4x}=\frac{1}{4}$