Question

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

Answer 1

We have,

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

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

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

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

$=\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{\sqrt{2}(1-\sin (x+\frac{\pi }{4}))}{{(\frac{\pi }{4}-x)}^2}$

$[\because \sin (A+B)=\sin A\cos B+\cos A\sin B]$

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

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

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

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{2}(1-\cos h)}{(-h{)}^2}$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{2}\left(2{\sin }^2\frac{h}{2}\right)}{(-h{)}^2}$

$[\because 1-\cos \theta =2{\sin }^2\frac{\theta }{2}]$

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

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

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

$=\frac{2\sqrt{2}}{4}\times (1{)}^2=\frac{1}{\sqrt{2}}$