Verified Answer :

\(\lim\limits_{x\to4}\frac{x-4}{x^2-8x+16}=\lim\limits_{x\to4}\frac{x-4}{(x-4)^2}=\lim\limits_{x\to4}\frac{1}{x-4}\\\\\lim\limits_{x\to4^-}\frac{1}{x-4}=\frac{1}{0^-}=-\infty\\\\\lim\limits_{x\to4^+}\frac{1}{x-4}=\frac{1}{0^+}=\infty\)

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