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
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$\begingroup$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 $C_2$.
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