Answer
$a-\lim_{x\rightarrow 3^+}\frac{2}{(x-3)^3}=\infty \\
b-\lim_{x\rightarrow 3^-}\frac{2}{(x-3)^3}=-\infty \\
c-\lim_{x\rightarrow 3}\frac{2}{(x-3)^3} \,does\,not\,exist \\
as
\lim_{x\rightarrow 3^+}\frac{2}{(x-3)^3}\neq \lim_{x\rightarrow 3^-}\frac{2}{(x-3)^3}$
Work Step by Step
$\lim_{x\rightarrow 3^+}\frac{2}{(x-3)^3}=\infty \\
(as\,x\,approach\,3\,from\,right\,\,\,((x-3)^3)\,is\,positive\,and\,approach\,0)\\
\lim_{x\rightarrow 3^-}\frac{2}{(x-3)^3}=-\infty \\
(as\,x\,approach\,3\,from\,left\,\,\,((x-3)^3)\,is\,negative\,and\,approach\,\,0)\\
\lim_{x\rightarrow 3}\frac{2}{(x-3)^3} \,does\,not\,exist \\
as
\lim_{x\rightarrow 3^+}\frac{2}{(x-3)^3}\neq \lim_{x\rightarrow 3^-}\frac{2}{(x-3)^3}$
