Answer
(a) $6$
(b) $-2$
(c) Does not exist.
(d) Does not exist.
(e) $\frac{1}{4}$
(f) $2$
Work Step by Step
(a) $\lim_{x\to2}\frac{x^2+2x-8}{x-2}=\lim_{x\to2}\frac{(x+4)(x-2)}{x-2}=\lim_{x\to2}(x+4)=6$
(b) $\lim_{x\to2}\frac{x^2-2x-8}{x+2}=\lim_{x\to2}\frac{(x-4)(x+2)}{x+2}=\lim_{x\to2}(x-4)=-2$
(c) $\lim_{x\to2}\frac{1}{x-2}=\pm\infty$ Does not exist
(d) $\lim_{x\to2}\frac{x-2}{|x-2|}$ Does not exist, because $\lim_{x\to2^-}\frac{x-2}{|x-2|}\ne \lim_{x\to2^+}\frac{x-2}{|x-2|}$
(e) $\lim_{x\to4}\frac{\sqrt x-2}{x-4}=\lim_{x\to4}\frac{\sqrt x-2}{(\sqrt x+2)(\sqrt x-2)}=\lim_{x\to4}\frac{1}{\sqrt x+2}=\frac{1}{4}$
(f) $\lim_{x\to\infty}\frac{2x^2-4}{x^2+x}=\lim_{x\to\infty}\frac{2-4/x^2}{1+1/x}=\frac{2-0}{1+0}=2$
