## Calculus (3rd Edition)

$-\frac{1}{5}$
Evaluating the function at -3, we get it of the form $\frac{0}{0}$. We try to rewrite the function cancelling the factors which are causing the limit to be of the form 0/0. Hence $\lim\limits_{x \to -3}\frac{x+3}{x^{2}+x-6}=\lim\limits_{x \to -3}\frac{x+3}{(x+3)(x-2)}=\lim\limits_{x \to -3}\frac{1}{x-2}= \frac{1}{-3-2}= -\frac{1}{5}$