Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

Answer 1

We have,
$\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

On rationalising the denominator, we get
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\tan x}{1-\sqrt{2}\sin x}\times \frac{1+\sqrt{2}\sin x}{1+\sqrt{2}\sin x}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{(1-\tan x)(1+\sqrt{2}\sin x)}{1^2-(\sqrt{2}\sin x{)}^2}$
$[\because (a+b\left)\right(a-b)=a^2-b^2]$

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

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

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

$[\because \cos 2x={\cos }^{2}x-{\sin }^{2}x]$

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

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

$=\frac{(1+\sqrt{2}\sin \frac{\pi }{4})}{\cos \frac{\pi }{4}(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})}$

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

$=\frac{2}{\frac{1}{\sqrt{2}}\left(\frac{2}{\sqrt{2}}\right)}=2$

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