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So for instance we have this limit of a function $\displaystyle\lim_{(x,y)\to(0,0)}{xy\over \sqrt{2x^2+y^2}}$, and the function isn't continuous at the point $(0,0)$. Now we can try to find the limit, and there's a possibility that it doesn't exist in case we find that the function approaches different values from different paths. Is that correct?

So for instance I tried approaching the point from y-axis ($x=0$) and x-axis ($y=0$), as well as $y=x$. The limit for all these is $0$. So it seems like the limit of the function as $(x,y)$ approach $(0,0)$ is $0$, but how do I know I haven't missed some other path, from which we might have approached different value?

Max
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    In short, you don't. No finite class of paths will ever be enough to know if a limit exists. Edit: This question might interest you. – Git Gud Feb 01 '15 at 13:36
  • So we try to make the most safe assumption based on our experience, in a way? – Max Feb 01 '15 at 13:40
  • You don't have to stay ignorant about the limit. You gain some intuition and then try to prove you're correct. If you suspect the limit is $0$, you try to prove it. On you're way towards proving it, you should either actually prove it or find out why it should fail and construct an example of a path on where the limit isn't $0$. – Git Gud Feb 01 '15 at 13:43
  • @GitGud Actually, not even an infinite class of paths will be enough. – Braindead Feb 01 '15 at 15:12
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    @Max A very common misconception is that somehow polar coordinates gives the definite answer. But this is false. The only real way to know is either a counterexample or a proof. – Braindead Feb 01 '15 at 15:14
  • @Braindead Yes, my bad. No ammount of sublimits is enough, unless it's all sublimits. – Git Gud Feb 01 '15 at 15:15
  • So we are starting to feel that there seem to be no counterexamples and are beginning to think how would we prove that it actually exists, as I understood? And in this case we could prove it with epsilon-delta definition of a limit, maybe, as one of the options? (I am not actually asking for a proof here). – Max Feb 01 '15 at 15:31
  • @Max Unfortunately that's the way to go. A graphic device will also be helpful in deciding whether you want to look for a counterexample or try for a full-on proof. There is no real systematic "works-all-the-time" method. – Braindead Feb 01 '15 at 15:33
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    @Max $2|xy|\leq x^2+y^2\leq 2x^2+y^2$. – Git Gud Feb 01 '15 at 15:34

2 Answers2

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One can use paths, or particular sequences $(x_n,y_n)\to(0,0)$, in order to prove that such a limit does not exist. Producing a single path or sequence where the limit does not exist, or two different paths or sequences with different limits of the expression in question closes the case.

But in the case at hand you (correctly) conjecture that the limit does exist, and you want to prove this. Under these circumstances it is of no help considering paths: You would have to prove that the limit is the same for all paths, but there are just to many of them. In addition I'm sure you can't even imagine how complicated a path towards $(0,0)$ can look like.

Instead we have to use the actual definition of convergence in this case. This definition works with $\epsilon$, $\delta$, and neighborhoods. Fortunately you have a definite idea what the limit is, namely $0$. You therefore have to prove the following: Given an $\epsilon>0$ there is a $\delta=\delta(\epsilon)>0$ such that $$0\leq r:=\sqrt{x^2+y^2}<\delta\qquad\Longrightarrow\qquad \bigl|f(x,y)\bigr|<\epsilon\ .$$ Of course it is allowed to use standard limit theorems, like the squeeze theorem, instead of $\epsilon$ and $\delta$. In the case at hand one has $$|f(x,y)|={r^2\>|\cos\phi\sin\phi| \over r \sqrt{2\cos^2\phi+\sin^2\phi}}\leq r\ ,$$ and this implies $\lim_{(x,y)\to0} f(x,y)=0$.

Christian Blatter
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You can use polar coordinates to evaluate your limit which becomes

$$\lim_{r\to 0} \frac{r^2\cos \theta \sin \theta}{r\ \sqrt{2\cos^2 \theta +\sin^2 \theta}} = \lim_{r\to 0} \frac{r \cos \theta \sin \theta}{ \sqrt{2\cos^2 \theta +\sin^2 \theta}} =0 $$

Daniel Fischer
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Mhenni Benghorbal
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    The OP isn't asking how to prove that this particular limit is $0$. – Git Gud Feb 01 '15 at 13:45
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    @GitGud: Leave the OP take his time to read my answer! Thanks for your comment! – Mhenni Benghorbal Feb 01 '15 at 13:46
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    This does not prove that the limit is 0. There are many examples where the polar coordinates may give you 0 as the limit, but the actual limit does not exist. – Braindead Feb 01 '15 at 14:55
  • @Braindead: When it exists we can use it? – Mhenni Benghorbal Feb 01 '15 at 14:55
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    What are you talking about? Of course if the limit exists, you can use any path to compute the limit. Consider $x^2/(y+x)$. Taking $r\to0$ gives you $0$. But the limit clearly fails to exist. – Braindead Feb 01 '15 at 14:59
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    And even if the proof of the limit was correct, this doesn't answer the question. – Git Gud Feb 01 '15 at 15:02
  • The limit exists!!!!!! – Mhenni Benghorbal Feb 01 '15 at 15:04
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    The limit exists and it is $0$, but your method is incorrect. – Git Gud Feb 01 '15 at 15:04
  • @GitGud: Sorry I do not have much time! – Mhenni Benghorbal Feb 01 '15 at 15:05
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    You are missing the point. If the limit exist, then the OP could have used the path $(0,y)$ or $(x,0)$. – Braindead Feb 01 '15 at 15:11
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    The question was how to tell whether the limit exists. And the method of using polar coordinates does not work. – Braindead Feb 01 '15 at 15:11
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    @Braindead So using polar coordinate substitution is just like taking another path? (and thus it doesn't prove the existence of the limit?) – Max Feb 01 '15 at 15:23
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    @Max To be more exact, you are looking at a special family of paths. (Rays to $(0,0)$). If you look at $x^2/(y+x)$, if you take a parabolic path $(x,x^2-x)$ and make $x\to 0$, you'll see that you would get a limit of 1. This is not detected by polar coordinates because they only account for straight line segments. – Braindead Feb 01 '15 at 15:30
  • @Braindead: I fail to see why using polar coordinates would not work? Here the denominator $$\sqrt{2\cos^2\theta+\sin^2\theta}\ge\sqrt{\cos^2\theta+\sin^2\theta}\ge1$$ is bounded away from zero, and the numerator is of the form $r$ times something bounded to the interval $[-1,1]$. The point is that $|f(x,y)|\le Kr$, where the constant $K$ does not depend on $\theta$! Mhenni is apparently too hot-tempered to explain this to you, so you are welcome to vote his answer any which way you see fit. – Jyrki Lahtonen Feb 01 '15 at 16:47
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    @JyrkiLahtonen But that's not Mhenni's solution. Simply changing to polar coordinates and making $r\to0$ is not sufficient to determine whether a limit exists. – Braindead Feb 01 '15 at 18:34
  • I'm not ruling whether that is Mhenni's solution or not. But it is sufficient in this case. Think sandwich! $f(x,y)$ is between $r$ and $-r$. Surely it $\to0$. About the flag. Yeah. It seemed to me there were still two flags. Yours and Mhenni's. I thought I would decline both, but Daniel had already declined his earlier. Sorry about that. Protesting like that is not good behavior. The moderators are discussing the case. – Jyrki Lahtonen Feb 01 '15 at 19:10
  • @JyrkiLahtonen I didn't flag this problem? – Braindead Feb 02 '15 at 00:32
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    @JyrkiLahtonen I am not disputing your solution, and your tone suggests that I don't understand your argument -- I do, but that's not Mhenni's solution. Again, I feel like you are missing my point. The OP was asking for a general method for determining whether the limit exists. Converting to polar works SOMETIMES and it alone isn't enough -- you had to use other inqualities. Also note that your polar solution is not very different from the epsilon delta proof. – Braindead Feb 02 '15 at 00:35
  • @Braindead: I apologize for the non-constructive tone. Another user asked about the flag in a now deleted comment. I confused you two. And also I was somehow affected by the heat of the argument. My point was that with polar coordinates you can approach the origin by paths other than rays. But you never really contested that. – Jyrki Lahtonen Feb 02 '15 at 05:29
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    @JyrkiLahtonen You keep saying things that are obviously true. What I have been saying, is that if you convert to $(r,\theta)$ and only make $r\to 0$, like it was done by Mhenni's solution, you are essentially fixing $\theta$ while $r\to 0$. (Just like if you have $(x,y)$ and just make $x\to 0$.) I honestly don't see the point of continuing this discussion. If you wish to post an answer to OP and explain how you can use polar coordinates properly, then by all means do so. But I am done. – Braindead Feb 02 '15 at 05:59
  • @Braindead. I agree that we should stop this exchange. I apologize for derailing it from your original point. – Jyrki Lahtonen Feb 02 '15 at 06:50
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    (1) I agree that the answer is lacking a little detail and explanation, but (2) it does prove the limit exist in this particular case, as here we have the case of a function that converges to zero times a bounded function, and that's enough. Certainly using polar coordinates can be dangerous and many times misleading, yet in this, as in many other case, it works pretty fine. – DonAntonio May 16 '17 at 22:04