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It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at that point exists or not. To see the above we will use polars. \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{5xy^2}{x^2+y^2}&=\lim_{r\rightarrow0}\frac{5(r\cos(\theta))(r\sin(\theta))^2}{(r\cos(\theta))^2+(r\sin(\theta))^2}\\ & =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))}\\ & =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2}\\ & =\lim_{r\rightarrow0}5r\cos(\theta)\sin^2(\theta)=0. \end{align*} Now let's test for the definition $\epsilon-\delta$, to see that the limit exists and is zero. \begin{align*} \left|\frac{5xy^2}{x^2+y^2}-0\right| & =\left|\frac{5xy^2}{x^2+y^2}\right|\\ & =\frac{5|x|y^2}{x^2+y^2}\\ & \leq 5|x| \text{ since $y^2\leq x^2+y^2$}\\ & \leq 5\sqrt{x^2+y^2} \text{ since $|x|\leq\sqrt{x^2+y^2}$}\\ & <5\delta = \epsilon. \end{align*} Therefore we have: \begin{align*} \forall\epsilon>0, \exists\delta>0 \text{ such that, if } 0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-0| &\leq5 \sqrt{x^2+y^2}\\ &<5 \delta = 5\left(\frac{\epsilon}{5}\right)=\epsilon. \end{align*} Thus the limit exists and is zero, therefore the discontinuity of $f$ is avoidable.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

BML
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2 Answers2

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Your first sentence is wrong; since $f$ is undefined at $(0,0)$, it is neither continuous nor discontinuous at that point.

Otherwise, you proved correctly that $\lim_{(x,y)\to(0,0)}f(x,y)=0$ and that therefore you can extend $f$ in such a way that it becomes continuous at $(0,0)$.

José Carlos Santos
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  • +1. More about the first point can be found here: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark Aug 18 '21 at 14:10
  • @HansLundmark Thank you. – José Carlos Santos Aug 18 '21 at 14:14
  • The first thing I say is exactly what you say, because I say it does not exist and it is obvious why – BML Aug 18 '21 at 14:47
  • And since $f$ is undefined at $(0,0)$ it makes no sense to talk about continuity at that point. – José Carlos Santos Aug 18 '21 at 14:48
  • The title explains well what you are looking for (with all due respect my comment) – BML Aug 18 '21 at 14:54
  • @BML: The first (or maybe second) thing you say is “therefore the function is discontinuous”, and it's this phrasing that is arguably wrong (as José is pointing out), since the function actually is continuous – after all, it's continuous at every point of its domain! See the question that I linked to. – Hans Lundmark Aug 18 '21 at 15:04
  • Sure the function is continuous in its domain, but we are talking about the point $(0,0)$. – BML Aug 18 '21 at 15:12
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    @BML: Which is not in the domain, which is why it's inappropriate to say that it's discontinuous there. It's would be weird for a continuous function to have discontinuities, don't you think? Again, see the question that I linked to for more discussion. – Hans Lundmark Aug 18 '21 at 17:46
  • I understand what they want to tell me, but I insist, I am talking about the point $(0,0)$ (not the domain of $f$) and I am not proving that it is continuous or not. – BML Aug 18 '21 at 17:58
  • But you wrote “and therefore the function is discontinuous”. – José Carlos Santos Aug 18 '21 at 17:59
  • I mean in the point, because in the writing you should avoid fillers ... – BML Aug 18 '21 at 18:00
  • Which point? The point $(0,0)$? – José Carlos Santos Aug 18 '21 at 18:01
  • Yes, the point $(0,0)$. – BML Aug 18 '21 at 18:02
  • Then, as I wrote in my answer, you are wrong. The function $f$ is not discontinuous at $(0,0)$ (and neither it is continuous at that point). – José Carlos Santos Aug 18 '21 at 18:03
  • Okay, I understand what you want to tell me and I think the problem is writing the ideas, in any case thank you for your comments. – BML Aug 18 '21 at 18:25
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Both your proofs (the analytic proof, and the proof using polar coordinates) that $\lim_{(x,y) f(x,y)\rightarrow (0,0)}=0$ are correct.

Medo
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  • It is that I wrote it wrong because it must go $\cos^2{\theta}+r^2 \sin^{2}{\theta}=1$ (I already corrected it) – BML Aug 18 '21 at 11:44
  • I see that. And to make it perfect you could also note that the limit is zero, as the factor $\cos{\theta}\sin^2{\theta}$ is bounded. – Medo Aug 18 '21 at 11:47
  • Thank you for your comments – BML Aug 18 '21 at 14:47