Answer
$a-\\\lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}=+\infty \\
b-\\\lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}=-\infty \\
c-\\\because \lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}\neq \lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}\\
\therefore \lim\limits_{x \to -2 }\frac{(x-4)}{x(x+2)}\,does\,not\,exist.
$
Work Step by Step
$\lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}=+\infty \\
(as\,x\,approach\,-2\,from\,right\,\,\,the\,numerator\,(x-4)approach\,\,-6\,\\and\,(x)approach\,-2\,,(x+2)\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
\lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}=-\infty \\
(as\,x\,approach\,-2\,from\,left\,\,\,the\,numerator\,(x-4)approach\,\,-6\,\\and\,(x)approach\,-2\,,(x+2)\,is\,negative\,and\,approach\,0 \,\\so\,the\,denominator\,is\,postive\,and\,approach\,\,\,0)\\
\because \lim\limits_{x \to -2^+ }\frac{(x-4)}{x(x+2)}\neq \lim\limits_{x \to -2^- }\frac{(x-4)}{x(x+2)}\\
\therefore \lim\limits_{x \to -2 }\frac{(x-4)}{x(x+2)}\,does\,not\,exist.
$
