Answer
(a) $5$
(b) $9$
(c) $2$
(d) $-\frac{1}{3}$
(e) $-\frac{3}{8}$
(f) $0$
(g) Does not exist.
(h) $-\frac{6}{11}$
Work Step by Step
(a) Use the basic Limit Laws, we have: $$\lim_{x\to a}[f(x)+h(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}h(x)=-3+8=5$$
(b) $$\lim_{x\to a}[f(x)]^2=[\lim_{x\to a}f(x)]^2=(-3)^2=9$$
(c) $$\lim_{x\to a}\sqrt[3] {h(x)}=\sqrt[3] {\lim_{x\to a}h(x)}=\sqrt[3] 8=2$$
(d) $$\lim_{x\to a}\frac{1}{f(x)}=\frac{1}{\lim_{x\to a}f(x)}=\frac{1}{-3}=-\frac{1}{3}$$
(e) $$\lim_{x\to a}\frac{f(x)}{h(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}h(x)}=\frac{-3}{8}=-\frac{3}{8}$$
(f) $$\lim_{x\to a}\frac{g(x)}{f(x)}=\frac{\lim_{x\to a}g(x)}{\lim_{x\to a}f(x)}=\frac{0}{-3}=0$$
(g) $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}=\frac{-3}{0}$$ Does not exist.
(h) $$\lim_{x\to a}\frac{2f(x)}{h(x)-f(x)}=\frac{2\lim_{x\to a}f(x)}{\lim_{x\to a}h(x)-\lim_{x\to a}f(x)}=\frac{2(-3)}{8+3}=-\frac{6}{11}$$
