Question

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

Answer 1

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

On Rationalising the Numerator , we get
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{\cos x}-\sqrt{\sin x}}{x-\frac{\pi }{4}}\times \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{\left(\right(\sqrt{\cos x}{)}^2-\left(\sqrt{\sin x}{)}^2\right)}{(x-\frac{\pi }{4})(\sqrt{\cos x}+\sqrt{\sin x})}$
$[\because (a+b\left)\right(a-b)=a^2-b^2]$

Multiply and divide the numerator by $\sqrt{2}$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x)}{(x-\frac{\pi }{4})(\sqrt{\cos x}+\sqrt{\sin x})}$

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

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

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

$=-\sqrt{2}\underset{x\to \frac{\pi }{4}}{lim}\frac{\sin (x-\frac{\pi }{4})}{(x-\frac{\pi }{4})}\times \frac{1}{(\sqrt{\cos x}+\sqrt{\sin x}}$
$[\because \sin (-\theta )=-\sin \theta ]$

$=-\sqrt{2}\times 1\times \frac{1}{\sqrt{\cos \frac{\pi }{4}}+\sqrt{\sin \frac{\pi }{4}}}$

$=-\sqrt{2}\times 1\times \frac{1}{{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}+{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}}$

$=\frac{-\sqrt{2}}{\frac{1}{2^{\frac{1}{4}}}+\frac{1}{2^{\frac{1}{4}}}}$

$==\frac{-\sqrt{2}}{\frac{2}{2^{\frac{1}{4}}}}$

$=\frac{-\sqrt{2}}{2^{1-\frac{1}{4}}}=\frac{-2^{\frac{1}{2}}}{2^{\frac{3}{4}}}$

$=\frac{-1}{2^{\frac{3}{4}-\frac{1}{2}}}=\frac{-1}{2^{\frac{1}{4}}}[\because \frac{a^m}{a^n}=a^{m-n}]$

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