If limit of $f(x)$ exists and the limit of $f(x)g(x)$ exists, then does the limit of $g(x)$ exist? 
I was doing some exam preparation and I got stuck on this one. Whats the idea behind this question:
If $\lim_{x\to a}f(x)$ and $\lim_{x\to a} [f(x)g(x)]$ exist then does $\lim_{x\to a}g(x)$ exist?
Thanks..
 A: Consider $f(x)=x^2$ and $g(x)=\frac1x$ and let $x\to0$.
A: There is a theorem, it's called the arithmetic limit theorem I think, and it goes like this:
If $\lim_{x \to a}f(x) = L$ and $\lim_{x \to a}g(x) = K$ then $\lim_{x \to a}f(x)g(x) = KL$.
The statement in your question is sort of a converse to this and the point is probably to make you work out whether it actually holds or not. 
Although the OP is not asking for a hint or a solution of the problem:
Hint: Try to find a counter example. If you can't come up with one let me know and I'll add one to this answer.
A: In general the answer is NO. 

But there is a trivial case in which this is true i.e when $\lim_{n\rightarrow a}{f(x)}$ exists and is non-zero.
A sketch of the proof is as follows.
We know that
if $\lim\limits_{x\rightarrow a}{y(x)}=a$ and $\lim\limits_{x\rightarrow a}{w(x)}=b$ then $\lim\limits_{x\rightarrow a}{\left(y(x)\times w(x)\right)}=ab$
Here replace $y(x)=\frac{1}{f(x)}$ and $w(x)=f(x)g(x)$ to arrive at such a proof.

Our previous proof helps the task of finding a counter example easier. We can skip all cases $f(x)$ has a limit that is non-zero, rephrasing are limited to the case where $f(x)$ has a limit zero!. One such function is $f(x)=x^2$ and $g(x)=\frac{1}{x}$ at  $a=0$, see for yourself if $g(x)$ has a limit.
But this does not mean that if $f(x)$ has a limit $0$ then, $g(x)$ simply cannot have a limit. 
One interesting case of this is when $\lim\limits_{x\rightarrow a}{f(x)}=0$, $\lim\limits_{x\rightarrow a}{f(x)g(x)}$ exists and $\lim\limits_{x\rightarrow a}{\frac1{f(x)}}=\infty$. In this case we can prove that $\lim\limits_{x\rightarrow a}{g(x)}$ exist. This question has an answer here.
