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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...

czash
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1 Answers1

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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.

Ted Shifrin
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  • I'm sorry, but I still can't figure it out ... Let's look at the second example you suggested. When moving to polar coordinates, and differentiate with respect to $\theta$ , we get the limit of $ \frac{1.5 \sqrt{r} cos(\theta) \sqrt{|sin \theta | } }{2rcos^2 \theta + |sin \theta| } $ , and this limit does not exist (we can take $ \theta =r $ and then the limit is $0/0$ ... Can you please re explain why did you say that this limit is a good example for my request? I'm not even sure that this limit does not exist... Can you please give some more details? Thanks ! – czash May 29 '13 at 18:10
  • No, you don't get to put $\theta$ in as a function of $r$ unless you intend to do so beforehand and use the chain rule. If you intend the limit to be $C$ no matter what function you put in for $\theta$, then this can only happen when $f$ and $g$ are independent of $\theta$ — at least to the leading order. If you allow the functional dependence, then you are effectively allowing us to approach on all curves to the origin, which is sufficient to guarantee a limit. – Ted Shifrin May 29 '13 at 18:25
  • In case you're not familiar with these sorts of examples, I've added an edit above. – Ted Shifrin May 29 '13 at 18:26
  • I understand everything you said, but still can't generlize this idea:let's look at the following example: $\frac{2^{xy}-1}{|x|+|y|} $ and we want to calculate the limit at (0,0). Is it legitimate to move to polar coordinates and then use l'Hospital rule ? If it is valid, then why ?

    Hope you'll be able to help

     – czash May 29 '13 at 19:19
  • Well, I guess now I'm confused. I'm basically saying that your L'Hôpital's rule method won't work to prove the limit unless, in fact, you basically do consider all possible paths given by varying $\theta$ as a function of $r$, which is, indeed cumbersome. But, obviously, if you have a universal bound $|f(x,y)|\le Ar^a$ and $|g(x,y)|\ge Br^b$, with $a>b$, then $\lim\frac{f(x,y)}{g(x,y)} = 0$ will work. I honestly don't see L'Hôpital's rule as relevant here. For your example, $|2^{xy}-1|\le \frac12\ln 2 r^2 + o(r)$ and $|x|+|y|\ge r$, so the quotient is bounded by $Cr+o(1)$ and does $\to 0$. – Ted Shifrin May 29 '13 at 19:49
  • This is way too long for comments. But if you try to use L'Hôpital's rule, $\frac{\partial}{\partial r}f(r,\theta)=(\ln 2)r2^{r^2\cos\theta\sin\theta}(\sin2\theta)$ and $\frac{\partial}{\partial r}g(r,\theta)=|\cos\theta|+|\sin\theta|$, so, thinking about bounds as above, you can see the limit is $0$ no matter how $\theta$ depends on $r$. But, yeah, you need to do that step. I'm not sure it's any easier. – Ted Shifrin May 29 '13 at 19:52
  • Great , Thanks a lot !! – czash May 30 '13 at 06:30