Answer
(a) $2$
(b) Does not exist.
(c) $0$
(d) Does not exist.
(e) $16$
(f) $2$
Work Step by Step
(a) Based on the graph, we have $$\lim_{x\to 2}[f(x)+g(x)]=\lim_{x\to 2}f(x)+\lim_{x\to 2}g(x)=2+0=2$$
(b) $$\lim_{x\to 1}[f(x)+g(x)]=\lim_{x\to 1}f(x)+\lim_{x\to 1}g(x)$$
As $$\lim_{x\to 1^-}g(x)\ne \lim_{x\to 1^+}g(x)$$ $$\lim_{x\to 1}g(x)$$ does not exist, and thus $$\lim_{x\to 1}[f(x)+g(x)]$$ does not exist.
(c) $$\lim_{x\to 0}[f(x)g(x)]=\lim_{x\to 0}f(x)\cdot\lim_{x\to 0}g(x)=0\times1.5=0$$
(d) $$\lim_{x\to -1}\frac{f(x)}{g(x)}=\frac{\lim_{x\to -1}f(x)}{\lim_{x\to -1}g(x)} $$ However, as $$\lim_{x\to -1}g(x)=0$$ the limit for this part of the question does not exist.
(e) $$\lim_{x\to 2}x^3f(x)=\lim_{x\to 2}x^3\cdot\lim_{x\to 2}f(x)=2^3\times2=16$$
(f) $$\lim_{x\to 1}\sqrt {3+f(x)}=\sqrt {3+\lim_{x\to 1}f(x)}=\sqrt {3+1}=2$$
