If two limits don't exist, are we sure the limit of sum doesn't exist? This is a true/false question.

If the limits $$\lim_ {x\to a}\ f(x)\quad \text{ and }\quad  \lim_ {x\to a}\ g(x)$$ don't exist, limits $$\lim_ {x\to a}\ (f(x)+g(x))\quad \text{ and }\quad \lim_ {x\to a}\ (f(x)g(x))$$ don't exist also.

My idea is, that, because $$\lim_ {x\to a}\ (f(x)+g(x))=\lim_ {x\to a}\ f(x)+\lim_ {x\to a} g(x)$$ then the limits doesn't exist, correct?
 A: Consider $f(x) = \frac{1}{x}$ and $g(x) = -\frac{1}{x}$. Take the two-sided limit as $x \to 0$.
A: $\mathbb{FALSE}$.
Let $f=1/(x-a)$ and $g=-1/(x-a)$
Your product hypothesis is false too;
Let $f=1/\sin(1/(x-a))$ and $g=\sin(1/(x-a))$
A: Generally. pick a function $f$ with
$$\lim_{x\to a} f(x) \neq C \in \mathbb K$$
and then chose $g(x) = -f(x)$ for a counter-example on addition and
$h(x) = \frac1{f(x)}$ (where defined and $\lim\limits_{x\to a} h(x) \neq D \in \mathbb K$) for a counter-example on multiplication.
A: It depends: if at least one among $f(x)$ and $g(x)$ is not well defined in any neighborhood of the limit point $a$ (except point $a$ itself), then of course nor is $f(x)+g(x)$ or $f(x)g(x)$ and therefore the limits of the sum and the product do not exist. E.g. take $f(x)=\sqrt{x}$ and $a=-1$, then
$$
\lim_{x\to -1}\sqrt{x}
$$
does not exist (beause $f(x)$ is defined only for $x\geq0$), nor does 
$$
\lim_{x\to -1}\big(\sqrt{x} + g(x)\big)
$$
for any function $g(x)$ (even if the limit of $g$ exists).
Otherwise, i.e. if you can find a suitable neighborhood of $a$ in which both $f(x)$ and $g(x)$ are well defined (except, at most, in point $a$), then you cannot tell. In fact if $\lim_{x\to a}f(x)$ does not exist, then take $g(x)=-f(x)$ and you have
$$
\lim_{x\to a} \big(f(x) + g(x)\big) = 0
$$
so the limit of the sum exists. As a counterexample, instead, take $f(x)=g(x)=\frac{1}{x}$ and $a=0$. Then
$$
\lim_{x\to 0} f(x)~,~\lim_{x\to 0}g(x)~,~\text{and}~\lim_{x\to 0}\big(f(x)+g(x)\big)
$$ 
do not exist.
The same argumentations hold for the product as well.
A: False. The gist of it is that the "non-existence" of the limits of $f$ and $g$ might "cancel" each other out as you get close to $a$. For instance, one might go to $+\infty$ while the other goes to $-\infty$, and if this occurs in the right way, they'll cancel out exactly and $f+g$ will tend to, say, $0$. Look at the examples in the other two answers.
You can't do:
$$\lim_ {x\to a}\ (f(x)+g(x))=\lim_ {x\to a}\ f(x)+\lim_ {x\to a} g(x)$$
Because that rule only works if the limits of $f$ and $g$ exist at $a$.
