Answer
$a-\\\lim_{x \to 4^+}
\frac{x-5}{(x-4)^2}=-\infty \\
b-\\\lim_{x \to 4^-}
\frac{x-5}{(x-4)^2}=-\infty \\
c-\\
\because \lim_{x \to 4^+}
\frac{x-5}{(x-4)^2}=\lim_{x \to 4^-}
\frac{x-5}{(x-4)^2}=-\infty \\
\therefore \lim_{x \to 4}
\frac{x-5}{(x-4)^2}=-\infty \\
$
Work Step by Step
$\lim_{x \to 4^+}
\frac{x-5}{(x-4)^2}=-\infty \\
(as\,x\,approach\,4\,from\,right\,\,\,the\,numerator\,(x-5)approach\,\,-1\,\\and\,(x-4)^2\,is\,positive\,and\,approach\,0 )\\
\lim_{x \to 4^-}
\frac{x-5}{(x-4)^2}=-\infty \\
(as\,x\,approach\,4\,from\,left\,\,\,the\,numerator\,(x-5)approach\,\,-1\,\\and\,(x-4)^2\,is\,positive\,and\,approach\,0 )\\
\because \lim_{x \to 4^+}
\frac{x-5}{(x-4)^2}=\lim_{x \to 4^-}
\frac{x-5}{(x-4)^2}=-\infty \\
\therefore \lim_{x \to 4}
\frac{x-5}{(x-4)^2}=-\infty \\
$
