Question

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

Answer 1

We have,

$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{1-\tan x}{x-\frac{\pi }{4}}$

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

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

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

$[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}]$

$=\underset{h\to 0}{\text{lim}}\frac{1-\frac{(1+\tan h)}{1-1\times \tan h}}{h}$

$=\underset{h\to 0}{\text{lim}}\frac{1-\tan h-(1+\tan h)}{h(1-\tan h)}$

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

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

$=\frac{-2\times 1}{(1-\tan 0)}$

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

$=-2$

Therefore,

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