l'Hôpital's Rule and Multivariable Limits I decided to post another message regarding this problem because I still didn't understand it at all:
Can someone give me an example of function $f(x,y),g(x,y)$ for which:
$\lim\limits_{r\to 0^+} \dfrac {f(r\cos \theta , r\sin \theta ) }{ g(r\cos\theta , r\sin \theta)} =\dfrac{0}{0}$, and $\lim\limits_{r\to 0^+} \dfrac {\frac{\mathrm df(r\cos \theta , r\sin \theta )}{\mathrm dr} }{ \frac{\mathrm dg(r\cos\theta , r\sin \theta )}{\mathrm dr} } = C $ for some constant $C$ , but the actual limit $\lim\limits_{(x,y)\to (0,0)} \dfrac {f(x,y)}{g(x,y)}$ does not exist at all? 
All I need is an example of a case where l'Hôpital's rule for multivariable limits when appearing in polar coordinates is not helpful (the reason such an example must exist is because $\theta$ can also depend on $r$ , but when using l'Hôpital's rule wrt $r$, we consider $\theta$ to be a constant... ).
Hope someone will be able to help me this time.
Thanks in advance.
Just to clarify things-
This is not a homework question... Only something I thought about... 
 A: The classic example of a function that is continuous along lines but discontinuous will do it, if you allow repeated L'Hôpital:
$$f(x,y)=x^2y \quad\text{and}\quad g(x,y)=x^4+y^2\,.$$
So, in polar coordinates, it looks like $\dfrac{r^3(...)}{r^2(...)}$ and approaches $0$ as $r\to 0$ (with $\theta\ne 0,\pi$).
Recall that if we let $y=cx$, then $$\frac{f(x,cx)}{g(x,cx)} = \frac{cx^3}{x^4+c^2x^2} =\frac{cx}{c^2+x^2}\to 0 \quad\text{as }x\to 0\,,$$
unless $c=0$, but that case is easily handled directly. On the other hand, if we let $y=x^2$, then
$$\frac{f(x,x^2)}{g(x,x^2)} = \frac{x^4}{x^4+x^4} = \frac12 \quad\text{for all }x\ne 0\,,$$
If you want a single application of L'Hôpital to do it, let's try
$$f(x,y) = x|y|^{1/2} \quad\text{and}\quad g(x,y)=x^2+|y|\,.$$
The same sort of analysis works here: Along $y=cx$, $\dfrac{f(x,cx)}{g(x,cx)} \to 0$, and along $y=x^2$, $\dfrac{f(x,x^2)}{g(x,x^2)}=\dfrac{|x|}{x} = \pm 1$.
I believe that if your limit $C\ne 0$ and $f(r,\theta)$ and $g(r,\theta)$ are actually differentiable functions, then your result works, because we'll have
$$f(r,\theta) = arh(\theta) + o(r) \quad\text{and}\quad g(r,\theta)=brk(\theta) + o(r)\,,$$
from which we'll have
$$\frac{f(r,\theta)}{g(r,\theta)} = \frac{a r h(\theta) + o(r)}{b r k(\theta) + o(r)} \to \frac{ah(\theta)}{bk(\theta)}\,;$$
if this is a constant $C$, we must have $h(\theta) = k(\theta)$, and it works.
