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Find $\displaystyle \lim_{(x,y)\to(0,0)} x^2\sin(\frac{1}{xy}) $ if exists, and find $\displaystyle\lim_{x\to 0}(\lim_{y\to 0} x^2\sin(\frac{1}{xy}) ), \displaystyle\lim_{y\to 0}(\lim_{x\to 0} x^2\sin(\frac{1}{xy}) )$ if they exist.

Hey everyone. I've tried using the squeeze theorem and found $0 \le |x^2\sin(\frac{1}{xy})| \le |x^2|\cdot 1 \xrightarrow{x\to0} 0 $ and so the "double" limit exists and equals zero. Now, I know $\lim_{x\to 0}\sin(\frac{1}{ax})$ diverges, so both $lim_{y\to 0}(\lim_{x\to 0} x^2\sin(\frac{1}{xy}) )$ and $lim_{x\to 0}(\lim_{y\to 0} x^2\sin(\frac{1}{xy}) )$ do not exist(?)

I don't think I understand multi-variable limits, I would love your help on this basic one so I can understand better. Thanks in advance :)

Noy
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    Yes the solution by squeeze theorem is correct! What are your other doubts? – user May 02 '18 at 20:56
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    You are right that $\lim_{(x,y)\to(0,0)} \sin(\frac{1}{xy})$ doesn't exist (I wouldn't use the term diverges) but the term $x^2$ suffices to make the whole limit exists and it is equal to $0$. – user May 02 '18 at 20:59
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    Well done! You are welcome, Bye! – user May 02 '18 at 21:02
  • @gimusi That must mean one of us is wrong, because you claim that $\lim_{(x,y)\to(0,0)} x^2\sin(\frac{1}{xy})$ exists and I claim it doesn't. Shall we consult the gods? – The Phenotype May 02 '18 at 21:04
  • @ThePhenotype ops I didn't noticed that, let me check! :) – user May 02 '18 at 21:06
  • The question is, whether only if both the second and the third limit exist, and they are both different- then we can conclude the first one does not exist. – Noy May 02 '18 at 21:11
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    @NoyPerel What you are saying is almost correct. The first one might also not exist if we have that one of the two other limits doesn't exist. – The Phenotype May 02 '18 at 21:23
  • @ThePhenotype Thank you, that's a confusing topic- I should read about this more and see different examples – Noy May 02 '18 at 21:25
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    @NoyPerel You can find many example here on MSE, here for example https://math.stackexchange.com/q/2647468/505767, https://math.stackexchange.com/q/2624273/505767, etc – user May 02 '18 at 21:34
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    @NoyPerel Please give the accept gimusi's answer, without him I would have given you fake mathematics. Everything I wrote in my answer (even though it has more words) is based on what he said to me. – The Phenotype May 02 '18 at 21:35
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    Thank you both, I am much more enlightened now! :) I will check out those posts – Noy May 02 '18 at 21:39
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    @NoyPerel You are free to accept the answer you prefer, also after editing, it is a community and it doesn't mind wheter or not I gave some hint for other answer. Please feel free to choose the answer that you prefer! – user May 02 '18 at 21:40
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    @ThePhenotype Thank for your kindness, but I think that Noy must be free to choose the answer that him/her prefer! No problem for me, It's fine to help one each other and solve the OP in the best way for the asker! – user May 02 '18 at 21:43

3 Answers3

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You only need simply to note that

$$0\le \left|x^2\sin(\frac{1}{xy})\right|\le x^2\to 0$$

then conclude by squeeze theorem.

user
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  • I don't want to be that guy, but you only repeated what OP said and then added "conclude"... Please don't turn into Dr. Graubner (no offense to him, but I seem him notoriously often among the Low Quality Posts). – The Phenotype May 02 '18 at 21:14
  • @ThePhenotype I've just indicated the plane way to solve this kind of limit, then I've added some comment to the other doubts formulated by the asker. The argument used also by you seems very confusing to me and not correct, What is your thought about that and about the comment I've added under your answer? – user May 02 '18 at 21:16
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$0 \le |x^2\sin(\frac{1}{xy})| \le |x^2| \xrightarrow{x\to0} 0$ means indeed that for any trajectory of $(x,y)$ going to $(0,0)$ we have that $x^2\sin(\frac{1}{xy})$ approaches $0$.

Note that just by taking any $y\neq 0$ fixed we cannot just conclude that it would not approach $0$, since eventually you also want $y$ to approach $0$. Indeed we can see that $y\neq 0$ fixed is a trajectory that does not intersect or approach $(0,0)$.

The Phenotype
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  • I'm not sure about your argument, you fix $x\neq 0$ but recall that $x\to 0$. To show that limit doesn't exist we shold find two different trajectories with $(x,y)\to(0,0)$ with different limits. By squeeze theorem it seem really clear that the limit is equal to $0$. Am I wrong? – user May 02 '18 at 21:14
  • @gimusi I think you do have a point. I should actually try to find a trajectory with a non-convergent limit, but I can't seem to find one... – The Phenotype May 02 '18 at 21:17
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    $\sin \frac{1}{xy}$ is bounded between $[-1,1]$ therefore $\lim_{x \to 0} x^{2}\sin \frac{1}{xy} = 0$. – Alex May 02 '18 at 21:20
  • @ThePhenotype Indeed squeeze theorem seems indicate that such trajectories don't exist at all. – user May 02 '18 at 21:20
  • @gimusi Alright, I'm convinced! Deleting my post in 3... 2... 1... – The Phenotype May 02 '18 at 21:20
  • @NoyPerel gimusi is right about this one. The trajectory I gave is a line that doesn't actually go to $(0,0)$. – The Phenotype May 02 '18 at 21:21
  • @ThePhenotype That's ok, thank you still :) – Noy May 02 '18 at 21:23
  • @ThePhenotype You are welcome! I really appreciate your sense of humor and frankness! ;) – user May 02 '18 at 21:23
  • @ThePhenotype You can always revise the answer! – user May 02 '18 at 21:23
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    @gimusi Thanks haha. Well okay, I will revise it. Also thanks for enlightening me :) – The Phenotype May 02 '18 at 21:24
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    @ThePhenotype You are very welcome! I hope you will be hearty with me when the next time I will be not completely correct! Bye :) – user May 02 '18 at 21:31
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So the moral of this question is that, one might think that if the double limit exists, then it entails the existence of iterated limits, this question gives a counterexample, but what is true is that,

Given that $\lim_{(x,y)\rightarrow(a,b)}f(x,y)=L$ exists and that for each $x\in B_{\delta}(a)-\{a\}$, $\lim_{y\rightarrow b}f(x,y)=M_{x}$ exists, then $\lim_{x\rightarrow a}M_{x}=L$, in other words, $\lim_{x\rightarrow a}\lim_{y\rightarrow b}f(x,y)=\lim_{(x,y)\rightarrow(a,b)}f(x,y)$.

The assumption that for a deleted neigborhood of $a$ that the existence of $\lim_{y\rightarrow b}f(x,y)$ cannot be relaxed.

user284331
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