Answer
$a-\lim_{x\rightarrow 2^+}\frac{1}{x-2}=\infty \\
b-\lim_{x\rightarrow 2^-}\frac{1}{x-2}=-\infty \\
c-\lim_{x\rightarrow 2}\frac{1}{x-2} \,does\,not\,exist \\
as\,\\
\lim_{x\rightarrow 2^+}\frac{1}{x-2}\neq \lim_{x\rightarrow 2^-}\frac{1}{x-2}$
Work Step by Step
$\lim_{x\rightarrow 2^+}\frac{1}{x-2}=\infty \\
(as\,x\,approach\,2\,from\,right\,\,\,(x-2)\,is\,positive\,and\,approach\,0)\\
\lim_{x\rightarrow 2^-}\frac{1}{x-2}=-\infty \\
(as\,x\,approach\,2\,from\,left\,\,\,(x-2)\,is\,negative\,and\,approach\,\,0)\\
\lim_{x\rightarrow 2}\frac{1}{x-2} \,does\,not\,exist \\
as
\lim_{x\rightarrow 2^+}\frac{1}{x-2}\neq \lim_{x\rightarrow 2^-}\frac{1}{x-2}$
