Spivak's Calculus - Chapter 5 Problem 8 (For making things simple - everywhere where I've written "$\lim f(x)$" I meant $\lim_{x\to a} f(x)$)
I'm trying to do this excercise:

And apparently the answers in the answer book are yes, yes, no, no. I have not read explanations yet but I came to a different conclusion and need info whether my logic is flawed. So let's assume $f(x)$ is defined everywhere "near" $a$ and $g(x)$ is not (numbers near $a$ are not in the domain). That means $\lim g(x)$ does not exist. Then, clearly $\lim [f(x)+g(x)]$ exists and it's the same as $\lim f(x)$, same for $\lim f(x)g(x)$. So following this logic the answers should be yes, no, yes, no. But I think I might have done something that is not allowed when trying to solve the problem this way. Am I right?
 A: As explained in the comment on your post, if you can't approach $a$ with $g$ then you can't approach it in neither $g(x)f(x)$ nor $g(x) + f(x)$.
With this said here are some hints:
(a) Consider $f(x) = \frac{1}{x-a}, g(x) = \frac{-1}{x-a}$ for addition, and two step functions where $f$ is $1$ for $x<a$ and $2$ for $x \ge a$ and $g$ is $2$ for $x < a$ and $1$ for $x \ge a$ for multiplication.
(b) $g(x) = f(x) + g(x) - f(x)$
(c) Assume for contradiction that $\lim_{x \to a} \left[ f(x) + g(x) \right]$ does exist and work like in (b)
(d) Consider $f(x) = 0, g(x) = \frac{1}{x-a}$
A: So it seems you and Mr. Spivak disagree on (b) and (c). Let's see.
Assume $\lim f(x)$ and $\lim (f(x)+g(x))$ exist. Since
$$
g(x) = (f(x)+g(x))-f(x),
$$
we deduce that $\lim g(x)$ exists by the standard rule for the limit of a sum: the answer to (b) is yes.
Assume, for the sake of contradiction, that $\lim (f(x)+g(x))$ exists. Then we apply (b) and discover that $\lim g(x)$ exists. This contradicts the assumption that $\lim g(x)$ does not exist: the answer to (c) is no.
