Question

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

• A. 0
• B. $-\frac{1}{2}$
• C. $\infty$
• D. 12

Answer 1

The correct option is B $-\frac{1}{2}$

Given, $\underset{x\to -3}{lim}\frac{{\tan }^{-1}(x+3)}{(x+3)(x+1)}=\underset{x\to -3}{lim}\left(\frac{{\tan }^{-1}(x+3)}{(x+3)}\right).\frac{1}{(x+1)}$
$=1.\left(\frac{1}{-3+1}\right)=-\frac{1}{2}$