# L'Hospital Rule in Multivariable Calculus

I was just wondering. Given functions $f(x,y), g(x,y)$, and the corresponding limit $\lim_{(x,y) \to (0,0) } \frac{f(x,y)}{g(x,y)}$ , where $\lim_{(x,y) \to (0,0) } f = \lim_{(x,y) \to (0,0) }g = 0$.

Assume that when moving to polar coordinates, we get that:

$\lim_{ r \to 0^+ } f(rcos\theta, rsin\theta) = \lim_{ r \to 0^+ } g(rcos\theta, rsin\theta)=0$. Can someone please tell me whether it is possible or not to differentiate $f,g$ with respect to $r$ and use l'Hospital rule?

I think that this is not legal, since $\theta$ can also be a function of $r$ , but I can't find any good example for the following claim: There exist functions $f,g$ such that $\lim_{ r \to 0^+ } f(rcos\theta, rsin\theta) = \lim_{ r \to 0^+ } g(rcos\theta, rsin\theta)=0$ but l'Hospital rule doesn't apply ...

Hope I made myself clear enough .

Thanks a lot

• This question is discussed in at least one other post: math.stackexchange.com/questions/177067/… [The short answer is: the single-variable LHR won't help you.] May 28 '13 at 16:31
• Additionally, see my question: math.stackexchange.com/questions/77723/… May 28 '13 at 16:45
• I still can't understand the answer... I am looking for a function $f(x,y)$ for which, when moving to polar coordinates and applying l'Hospital's rule gives a number result $C_1$ , while the limit does not exist at all, or the actual limit gives $C_2 \neq C_1$ ... Can you help me? May 28 '13 at 20:11
• or an example of a function where using l'Hopsital rule in polar coordinates gives some constant $C$ , but the limit does not exist at all... May 28 '13 at 20:39

I am looking for a function $f(x,y)$ for which, when moving to polar coordinates and applying l'Hospital's rule gives a number result $C_1$ , while the limit does not exist at all
A standard example is $$f(x,y)=\frac{x^2y}{x^4+y^2} = \frac{r \cos^2\theta\sin\theta}{r^2\cos^2\theta+\sin^2\theta} \tag1$$ For any fixed $\theta$, the limit as $r\to 0$ is $0$. Yet, $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist, as one can see by considering that $f(x,x^2) =1/2$.
... or the actual limit gives $C_2\ne C_1$
This can't happen. If $\lim_{(x,y)\to (0,0)}f(x,y)=C_2$, then the radial limits, or any other directional limit, will also be equal to $C_2$.