Answer
$a-\\\lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}=+\infty\\
b-\\
\lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}=-\infty \\
c-\\
\because \lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}\neq \lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}\\
\therefore \lim\limits_{x \to -2}\frac{x^3-5x^2+6x}{x^4-4x^2}\,does\,not\,exist.
$
$
d-\\\lim\limits_{x \to 2}\frac{x^3-5x^2+6x}{x^4-4x^2}=\frac{-1}{8}$
Work Step by Step
$a-\\\lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to -2^+}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)}
=\lim\limits_{x \to -2^+}\frac{x-3}{x(x+2)}\\
as\,x\,approach\,-2\,from\,right\,\,\,the\,numerator\,(x-3)\,approach\,\,-5\,\\and\,,x\,approach\,-2\,\,,(x+2),is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
so\,\lim\limits_{x \to -2^+}\frac{x-3}{x(x+2)}=+\infty \\
b-\\\lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to -2^-}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)}=\lim\limits_{x \to -2^-}\frac{x-3}{x(x+2)}\\
as\,x\,approach\,-2\,from\,left\,\,\,the\,numerator\,(x-3)\,approach\,\,-5\,\\and\,,x\,approach\,-2\,\,,(x+2),is\,negative\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
so\,\lim\limits_{x \to -2^-}\frac{x-3}{x(x+2)}=-\infty \\
c-\\\because \lim\limits_{x \to -2^+}\frac{x^3-5x^2+6x}{x^4-4x^2}\neq \lim\limits_{x \to -2^-}\frac{x^3-5x^2+6x}{x^4-4x^2}\\
\therefore \lim\limits_{x \to -2}\frac{x^3-5x^2+6x}{x^4-4x^2}\,does\,not\,exist.
$
$d-\\\lim\limits_{x \to 2}\frac{x^3-5x^2+6x}{x^4-4x^2}=\lim\limits_{x \to 2}\frac{x(x-3)(x-2)}{x^2(x-2)(x+2)}\\
=\lim\limits_{x \to 2}\frac{x-3}{x(x+2)}=\frac{2-3}{2(2+2)}=\frac{-1}{8}
$
