Solutions

a)

Both

$$\lim_{x \to a} [f(x)+g(x)] \qquad\text{and}\qquad \lim_{x \to a} [f(x)g(x)]$$

can exist.

For arbitrary $a$, let

$$f(x)= \begin{cases} 0, & x \le a, \\ 1, & \text{otherwise}, \end{cases} \qquad g(x)= \begin{cases} 1, & x \le a, \\ 0, & \text{otherwise}. \end{cases}$$

Hence, we have that $f(x)+g(x)=1$ and $f(x)g(x)=0$. Both resulting functions are constant and thus have limits defined at $x=a$.

b)

Yes, $\lim_{x \to a}g(x)$ must exist.

Let

$$l_f = \lim_{x \to a} f(x), \qquad L = \lim_{x \to a} [f(x)+g(x)], \qquad l_g = L-l_f.$$

Choose any $\epsilon > 0$. By definition, there exist $\delta_1$ and $\delta_2$ such that

$$\begin{align*} 0<|x-a|<\delta_1 &\implies |f(x)-l_f| < \frac{\epsilon}{2}, \\ 0<|x-a|<\delta_2 &\implies |f(x)+g(x)-(l_f+l_g)| < \frac{\epsilon}{2}. \end{align*}$$

Let $\delta = \min(\delta_1,\delta_2)$. We thus have, for $0<|x-a|<\delta$,

$$\begin{align*} |g(x)-l_g| &= |g(x)-l_g+f(x)-f(x)+l_f-l_f| \\ &= |f(x)+g(x)-(l_f+l_g)-(f(x)-l_f)| \\ &\le |f(x)+g(x)-(l_f+l_g)| + |f(x)-l_f| \\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon. \end{align*}$$

c)

$$\lim_{x \to a} [f(x)+g(x)]$$

cannot exist if $\lim_{x \to a} f(x)$ exists and $\lim_{x \to a} g(x)$ does not exist. This can be proven with a proof by contradiction along the lines of the proof in b). We show an alternative proof that follows from the statement proven in b). We define the following propositions:

$$\begin{align*} P &: \lim_{x \to a} f(x) \text{ exists}, \\ Q &: \lim_{x \to a} g(x) \text{ exists}, \\ R &: \lim_{x \to a} [f(x)+g(x)] \text{ exists}. \end{align*}$$

We want to prove that

$$P \land \neg Q \implies \neg R.$$

This is equivalent to

$$\begin{align*} \neg(P \land \neg Q) \lor \neg R &\equiv \neg P \lor Q \lor \neg R \\ &\equiv \neg P \lor \neg R \lor Q \\ &\equiv \neg(P \land R) \lor Q \\ &\equiv P \land R \implies Q. \end{align*}$$

and because the last statement was proven in b), so the original implication is also true.

d)

$$\lim_{x \to a} g(x)$$

does not necessarily exist.

As a counterexample, let $f(x)=0$ and $g(x)=\frac{1}{x}$. Notice that

$$\lim_{x \to 0} f(x)g(x)=0,$$

but $\lim_{x \to 0} g(x)$ does not exist.