Answer
$\lim\limits_{x\to-3^-}\dfrac{x+3}{x^2+x-6}=-\dfrac{1}{5}.$
Work Step by Step
$\lim\limits_{x\to-3^-}\dfrac{x+3}{x^2+x-6}=\lim\limits_{x\to-3^-}\dfrac{(x+3)}{(x+3)(x-2)}=\lim\limits_{x\to-3^-}\dfrac{1}{x-2}$
$=\dfrac{1}{-3^--2}=-\dfrac{1}{5}.$
