Answer
$\frac{3}{7}$
Work Step by Step
$\lim\limits_{x \to -3}\frac{x^2+3x}{x^2-x-12}=\frac{(-3)^2+3(-3)}{(-3)^2-(-3)-12}=\frac{0}{0}$
First step is to plug in the x value but since it comes out to 0/0 we have to do algebra.
$\lim\limits_{x \to -3}\frac{(x)(x+3)}{(x-4)(x+3)}=\lim\limits_{x \to -3}\frac{x}{x-4}=\frac{-3}{-3-4}=\frac{3}{7}$
