Answer
$a-\\
\lim\limits_{x \to 2^+ }\frac{(x^2-4x+3)}{(x-2)^2}=-\infty \\
b-\\
\lim\limits_{x \to 2^- }\frac{(x^2-4x+3)}{(x-2)^2}=-\infty \\
c-\\
\because \lim\limits_{x \to 2^+ }\frac{(x^2-4x+3)}{(x-2)^2}=\lim\limits_{x \to 2^- }\frac{(x^2-4x+3)}{(x-2)^2}\\
\therefore \lim\limits_{x \to 2 }\frac{(x^2-4x+3)}{(x-2)^2}=-\infty
$
Work Step by Step
$\lim\limits_{x \to 2^+ }\frac{(x^2-4x+3)}{(x-2)^2}=\lim\limits_{x \to 2^+ }\frac{(x-3)(x-1)}{(x-2)^2}=-\infty \\
(as\,x\,approach\,2\,from\,right\,\,\,the\,numerator\,(x-3)(x-1)approach\,\,-1\,\\and\,(x-2)^2\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
\lim\limits_{x \to 2^- }\frac{(x^2-4x+3)}{(x-2)^2}=\lim\limits_{x \to 2^- }\frac{(x-3)(x-1)}{(x-2)^2}=-\infty \\
(as\,x\,approach\,2\,from\,left\,\,\,the\,numerator\,(x-3)(x-1)approach\,\,-1\,\\and\,(x-2)^2\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
\because \lim\limits_{x \to 2^+ }\frac{(x^2-4x+3)}{(x-2)^2}=\lim\limits_{x \to 2^- }\frac{(x^2-4x+3)}{(x-2)^2}\\
\therefore \lim\limits_{x \to 2 }\frac{(x^2-4x+3)}{(x-2)^2}=-\infty
$
