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