Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{3}-\tan x}{\pi -3x}$

Answer 1

We have,

$\underset{x\to \frac{\pi }{3}}{\text{lim}}\frac{\sqrt{3}-\tan x}{\pi -3x}$

Put $x=\frac{\pi }{3}+h$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{3}-\tan (\frac{\pi }{3}+h)}{\pi -3(\frac{\pi }{3}+h)}$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{3}-\left(\frac{\tan \frac{\pi }{3}+\tan h}{1-\tan \frac{\pi }{3}\tan h}\right)}{\pi -3(\frac{\pi }{3}+h)}$

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

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{3}-\left(\frac{\sqrt{3}+\tan h}{1-\sqrt{3}\tan h}\right)}{\pi -\pi -3h}$

$=\underset{h\to 0}{\text{lim}}\frac{\sqrt{3}-3\tan h-\sqrt{3}-\tan h}{\pi -\pi -3h}$

$=\underset{h\to 0}{\text{lim}}\frac{-4\tan h}{-3h}$

$=\frac{4}{3}\underset{h\to 0}{\text{lim}}\frac{\tan h}{h}$ $[\because \underset{x\to 0}{\text{lim}}\frac{\tan x}{x}=1]$

$=\frac{4}{3}\times 1=\frac{4}{3}$