$\lim\limits_{x \rightarrow \tan ^{-1}3} \frac{\tan ^2 x-2 \tan x-3}{\tan ^2 x-4 \tan x+3}$

1 2 0 3

2

$\lim\limits_{x \rightarrow \tan ^{-1} 3} \frac{(\tan x-3)(\tan x+1)}{(\tan x-3)(\tan x-1)}$ $\lim\limits_{x \rightarrow \tan ^{-1} 3} \frac{\tan x+1}{\tan x-1}=\frac{3+1}{3-1}=2$ Hence (2) is the correct answer.