Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi {)}^2}$

Answer 1

We have:
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi {)}^2}$

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

$=\underset{h\to 0}{lim}\frac{\sqrt{2}-(\cos [\frac{\pi }{4}+h]+\sin [\frac{\pi }{4}+h])}{{(4[\frac{\pi }{4}+h]-\pi )}^2}$
$\because \sin (A+B)=\sin A\cos B+\cos A\sin B\because \cos (A+B)=\cos A\cos B-\sin A\sin B$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}[\cos \frac{\pi }{4}\cos h-\sin \frac{\pi }{4}\sin h+\sin \frac{\pi }{4}\cos h+\cos \frac{\pi }{4}\sin h]}{(\pi +4h-\pi {)}^2}$

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

$=\underset{h\to 0}{lim}\frac{\sqrt{2}-\left[\frac{2}{\sqrt{2}}\cos h\right]}{(\pi +4h-\pi {)}^2}$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}-\sqrt{2}\cos h}{(\pi +4h-\pi {)}^2}$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}-\sqrt{2}\cos h}{(\pi +4h-\pi {)}^2}$

$=\underset{h\to 0}{lim}\frac{2\sqrt{2}{\sin }^2\frac{h}{2}}{(4h{)}^2}$

$=\underset{h\to 0}{lim}\frac{2\sqrt{2}}{16}\times \frac{{\sin }^2\frac{h}{2}}{4\times \frac{h^2}{4}}$

$=\frac{2\sqrt{2}}{16\times 4}\underset{h\to 0}{lim}{\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2$

$=\frac{2\sqrt{2}}{16\times 4}(1{)}^2$
$[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]$

$=\frac{1}{16\sqrt{2}}$

Therefore,
$\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi {)}^2}=\frac{1}{16\sqrt{2}}$