Answer
$\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})=-\infty.$
Work Step by Step
$\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})\to$
$\lim\limits_{x\to-4^-}x^2=(-4^-)^2=16.$
$\lim\limits_{x\to-4^-}\dfrac{2}{x+4}=\dfrac{2}{-4^-+4}=\dfrac{2}{0^-}=-\infty.$
By Theorem $1.15:$
$\lim\limits_{x\to-4^-}(x^2+\dfrac{2}{x+4})=-\infty.$
