Question

The value of $\underset{x\to {\tan }^{-1}3}{lim}\frac{{\tan }^{2}x-2\tan x-3}{{\tan }^{2}x-4\tan x+3}$ is-

• A. $0$
• B. $2$
• C. $\infty$
• D. $Noneofthese$

Answer 1

Answer: (B)

$2$

$\underset{x\to {\tan }^{-1}3}{lim}\frac{{\tan }^{2}x-2\tan x-3}{{\tan }^{2}x-4\tan x+3}$

${\tan }^{2}x-2\tan x-3={\tan }^{2}x+\tan x-3\tan x-3=\tan x(1+\tan x)-3(\tan x+1)=(\tan x+1)(\tan x-3)$

${\tan }^{2}x-4\tan x+3={\tan }^{2}x-\tan x-3\tan x+3=\tan x(\tan x-1)-3(\tan x-1)=(\tan x-1)(\tan x-3)$

$\frac{{\tan }^{2}x-2\tan x-3}{{\tan }^{2}x-4\tan x+3}=\frac{(\tan x+1)(\tan x-3)}{(\tan x-1)(\tan x-3)}=\frac{(\tan x+1)}{(\tan x-1)}$

$\underset{x\to {\tan }^{-1}3}{lim}\frac{{\tan }^{2}x-2\tan x-3}{{\tan }^{2}x-4\tan x+3}=\underset{x\to {\tan }^{-1}3}{lim}\frac{(\tan x+1)}{(\tan x-1)}=\frac{(3+1)}{(3-1)}=2$