Answer
$\dfrac{3}{2}$
Work Step by Step
$\lim\limits_{x \to -3}\dfrac{x^2-9}{x^2+2x-3}$
Factor:
$\lim\limits_{x \to -3}\dfrac{(x+3)(x-3)}{(x+3)(x-1)}$
The $x+3$ cancels out:
$\lim\limits_{x \to -3}\dfrac{x-3}{x-1}$
Plug in -3:
$\lim\limits_{x \to -3}\dfrac{-3-3}{-3-1}=\dfrac{-6}{-4}=\dfrac{3}{2}$
