Question

If $\underset{x\to -3}{lim}\frac{x^2+x-6}{x+3}=k$, then the value of $|k|$ is

Answer 1

$\underset{x\to -3}{lim}\frac{x^2+x-6}{x+3}$
This is $\frac{0}{0}$ form
$=\underset{x\to -3}{lim}\frac{(x+3)(x-2)}{x+3}=\underset{x\to -3}{lim}(x-2)=-5\therefore |k|=5$