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