If $\;\lim\limits_{x\rightarrow 10}g(x)=a\;$ and if$\;\lim\limits_{x \rightarrow10}\dfrac{f(x)}{g(x)}=b$, find $\lim\limits_{x\rightarrow10}f(x).$ If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$
The answer is insufficient information to determine. But I thought there was a "limit multiplication rule" I could use to determine it's $ab$? Can someone give an example when this doesn't hold?
[EDIT] based on below discussion, I guess the question boils down to, can we conclude $\lim\limits_{x\rightarrow10}f(x)$ exists based on above conditions?
I am also curious if the answer changes if $b=0$ or $a=0$?
 A: $f(x)= \frac{f(x)}{g(x)} \cdot g(x)$ for $x $ in a neighborhood of $10$, since $a \ne 0.$
Hence
$$f(x)= \frac{f(x)}{g(x)} \cdot g(x) \to b \cdot a$$
as $x \to 10.$
A: 
If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

Since $\;a\ne0\;,\;$ there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;.$
Consequently,
$f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$
and, by applying the limit multiplication rule, we get that
$\begin{align}
\lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\
&=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;.
\end{align}$

Now we are going to prove the following property

If there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;,\;$ if $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ then $\;\lim\limits_{x\rightarrow10}f(x)=a\cdot b\;.$

Since $\;f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$
by applying the limit multiplication rule, we get that
$\begin{align}
\lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\
&=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;.
\end{align}$
