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Question

What condition must non-negative integers m, n and p satisfy so that

$$\lim_{(x,y)\to(0,0)}\frac{x^my^n}{(x^2+y^2)^p}$$

exist? Prove your answer.

[Note: if $m=n=p=0$, then the limit trivially exists and equals 1.]

Solution (my attempt)

$$\sqrt{x^2+y^2}<\delta \implies |\frac{x^my^n}{(x^2+y^2)^p}-L|<\epsilon$$

I proved a similar question where if $m=n=p=1$ the limit did not exist shown below:

Assume that it does have a limit, such that $\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)}=L$. This means $ \forall\epsilon>0$ $\exists \delta>0 $

$$ ||(x,y)||<\delta \implies |f(x,y) -L|<\epsilon $$

Note that, $\forall a\neq 0$, $a \in \mathbb{R} $, we have $f(a,a) = \frac{a^2}{2a^2}=\frac{1}{2}$

Therefore, $\forall \epsilon>0$, if we chose $a \in \mathbb{R}:0<a<\frac{\delta}{\sqrt{2}} \implies ||(a,a)||<\delta $ and thus $|\frac{1}{2}-L|<\epsilon$. Hence $L=\frac{1}{2}$

However, $\forall b\neq 0$, $b \in \mathbb{R} $, we have $f(b,-b) = \frac{-b^2}{2b^2}=-\frac{1}{2}$. So we can conclude $L=-\frac{1}{2}$ in a similar way to the one above, which is a contradiction.

Can I use a similar proof to this one for my question or should I take a different approach? Any help is appreciated.

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    $\begingroup$ Polar coordinates will help in this one. $\endgroup$ – zhw. Aug 30 '15 at 5:40
  • $\begingroup$ for m,n,p>o you can count 2 cases, either m+n>2p then the limit exists and equal to 0 or, $m+n\leqq 2p$ then the limit does not exists $\endgroup$ – Rupsa Aug 30 '15 at 6:37
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Hint:

$x=r\cos(a)$, $y=r\sin(a)$, $r>0$ and the limit becomes:

$$\lim_{r\to0}\frac{r^{m+n}\cos^m(a)\sin^n(a)}{r^{2p}}$$

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