## Calculus 10th Edition

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