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In this case, $k,l,p,q\geq0$ and are integers. I have attempted the substitution $u=x^p$ and $v=y^q$ $$\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}=\lim_{(u,v)\to(0,0)} \frac{u^{k/p}v^{l/q}}{u^2+v^2}$$

Then, note that $|u|<\sqrt{|u|^2+|v|^2}$ and $|v|<\sqrt{|u|^2+|v|^2}$. So we have $$\bigg|\frac{u^{k/p}v^{l/q}}{u^2+v^2}\bigg|=\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}<\frac{(|u|^2+|v|^2)^{k/(2p)+l/(2q)}}{|u|^2+|v|^2}$$ Thus, the limit equals $0$ when $\frac{k}{p}+\frac{l}{q}>2$. We need to show that otherwise, the limit does not exist.

If $\frac{k}{p}+\frac{l}{q}=2$, taking the limit along the axes yields $0$ but letting $x^p=y^q$ gives $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}=\frac{|u|^{k/p+l/q}}{2|u|^2}\to\frac{1}{2}$$ so the limit does not exist.

Similarly, if $\frac{k}{p}+\frac{l}{q}<2$, taking the limit along the axes still yields $0$, but letting $x^p=y^q$ instead gives $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}>\frac{|u|^{2}}{2|u|^2}\to\frac{1}{2}$$ showing that the limit does not exist.

I'm not sure if all of the steps I've taken are correct - especially some of the inequalities as $(u,v)\to(0,0)$. Also, for the last two cases, I'm not exactly sure if the absolute values around $u$ and $v$ should be there or whether I should have let $|x|^p=|y|^q$.

Vasting
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    Everything you've done seems OK. – Ishan Deo Jun 10 '19 at 23:47
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    To me, it also seems you've done basically everything correctly, except for one minor point. For the final case of $\frac{k}{p}+\frac{l}{q}<2$, the inequality you use is only true when $\left|u\right| \lt 1$. Although you are taking the limit as $u \to 0$, so it's only important what happens when $u$ is close to $0$, I suggest you should still either explicitly state your assumption or also account for where $\left|u\right| \ge 1$. As for your question of using absolute values around $u$ & $v$, I believe you should to avoid the complications of handling the values if they're negative. – John Omielan Jun 11 '19 at 02:03
  • @JohnOmielan How would I be able to account for where $|u|\geq1$? It seems like what's happening when that's true is irrelevant to the limit. – Vasting Jun 11 '19 at 04:34
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    @Vasting I agree it's basically irrelevant to the limit, so you don't really need to account for it (I just mentioned it as a possibility, not necessarily the optimal one). Note I just stated it because your equation is not quite correct as currently stated. My suggestion is to just ensure it's clear you're using this restriction, e.g., with what you wrote of "$x^p = y^q$ instead", insert something like " and with $\left|u\right| \lt 1$". However, note this is a very minor point and it may be considered "nitpicking", but as you're asking here if your solution is correct, I stated this issue. – John Omielan Jun 11 '19 at 04:44
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    @Vasting I hope I'm not annoying you by pointing out relatively minor issues, but one more is the question allows $p, q$ or both to be $0$. However, in any such case, then $\frac{k}{p}+\frac{l}{q}$ isn't defined. I suggest initially handling these cases. The denominator goes to $1$ or $2$, but the numerator goes to $1$ if $k = l = 0$, else it goes to $0$. In all the cases, the limit exists. In a way, this makes some logical sense since if $p$, $q$ or both are $0$, plus if $k \neq 0$, $l \neq 0$ or both, then $\frac{k}{p}+\frac{l}{q}$ basically goes to "infinity", so it's definitely $\gt 2$. – John Omielan Jun 11 '19 at 23:45
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    @JohnOmielan Absolutely not, thanks for recognizing that - it definitely helps to see where all of the minor problems in a proof can arise. – Vasting Jun 12 '19 at 00:42
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    @Vasting With your own question, you can answer & accept it (note I did this with my one & only question so far, with text (& more context) of an interesting deleted question), you may wish to do that here. It'll help the site by removing this post from the unanswered queue, plus anybody reading it won't have to go through the comments to determine what the issues are but, instead, can just read a complete, accepted answer. I believe you'll also benefit as it'll give you more experience writing a rigorous & thorough proof. Also, although I don't see how, you may find better ways to write it. – John Omielan Jun 12 '19 at 20:04
  • Related : https://math.stackexchange.com/questions/66226/multivariable-limit-proof-lim-x-y-rightarrow-0-0-frac-leftx-righta?noredirect=1&lq=1 – Arnaud D. Aug 24 '20 at 08:27

1 Answers1

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Let $k,l,p,q\geq0$ be integers. First, note that if $p$ or $q$ (or both) are $0$, then the denominator approaches either $1$ or $2$. If $k=l=0$, the numerator goes to $1$, otherwise it goes to $0$. In all of these cases, the limit clearly exists.

Otherwise, let $u=x^p$ and $v=y^q$, which gives $$\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}=\lim_{(u,v)\to(0,0)} \frac{u^{k/p}v^{l/q}}{u^2+v^2}$$

Then, note that $|u|<\sqrt{|u|^2+|v|^2}$ and $|v|<\sqrt{|u|^2+|v|^2}$. Thus we have $$\bigg|\frac{u^{k/p}v^{l/q}}{u^2+v^2}\bigg|=\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}<\frac{(|u|^2+|v|^2)^{k/(2p)+l/(2q)}}{|u|^2+|v|^2}$$ We can see that if when $\frac{k}{p}+\frac{l}{q}>2$, the limit equals $0$. Now we will prove that if $\frac{k}{p}+\frac{l}{q}\leq2$ otherwise, the limit does not exist.

In the case of $\frac{k}{p}+\frac{l}{q}=2$, taking the limit along the axes yields $0$ but letting $x^p=y^q$ gives $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}=\frac{|u|^{k/p+l/q}}{2|u|^2}\to\frac{1}{2}$$ so the limit does not exist in this case.

Similarly, if $\frac{k}{p}+\frac{l}{q}<2$, taking the limit along the axes yields $\infty$ if $k=l=0$, and $0$ otherwise. However, letting $x^p=y^q$ instead with $|u|<1$ (since $u\to0$) gives $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}>\frac{|u|^{2}}{2|u|^2}\to\frac{1}{2}$$ showing that the limit does not exist in this case either.

Thus, the limit exists only when $p$ or $q$ (or both) are $0$ or when $\frac{k}{p}+\frac{l}{q}>2$.

Vasting
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