Answer
$\lim\limits_{x\to -3}\dfrac{x^2+x-6}{x^2-9}=\dfrac{5}{6}.$
Work Step by Step
$\lim\limits_{x\to -3}\dfrac{x^2+x-6}{x^2-9}=\lim\limits_{x\to -3}\dfrac{(x-2)(x+3)}{(x-3)(x+3)}=\lim\limits_{x\to -3}\dfrac{x-2}{x-3}$
$=\dfrac{-3-2}{-3-3}=\dfrac{5}{6}.$
