$\lim_{(x,y)\to (0,0)} \frac{x^m y^n}{x^2 + y^2}$ exists iff $m+ n > 2$ I would like to prove, given $m,n \in \mathbb{Z}^+$, $$\lim_{(x,y)\to (0,0)}\frac{x^ny^m}{x^2 + y^2} \iff m+n>2.$$
(My gut tells me this should hold for $m,n \in \mathbb{R}^{>0}$ as well.) The ($\Rightarrow$) direction is pretty easy to show by contrapositive by familiar limit tricks. The ($\Leftarrow$) direction is giving me more trouble. 
So far my strategy has been to use the arithmetic mean: 
$$\left| \frac {x^n y^m} {x^2 + y^2}  \right| \leq \frac {x ^ {2n} + y ^ {2m}} {2 ( x^2 + y^2 ) } \leq \frac {x^ {2 (n-1)}} {2} + \frac { y^ {2 (m-1)} } {2},$$ 
but that's not helping if $m=1$, say.
Any ideas?
Apologies if this is a repeat...I couldn't find this on the site.

If $m+n>2$, you can divide into two cases, by observing that you can't have $m<2$ and $n<2$.
First case: $m\ge2$.
$$
\lim_{(x,y)\to (0,0)}\frac{x^ny^m}{x^2 + y^2}
=
\lim_{(x,y)\to (0,0)}\frac{y^2}{x^2 + y^2}x^ny^{m-2}
$$
where $n\ge1$ or $m-2\ge1$. The fraction is bounded, while the other factor tends to zero.
Similarly for $n\ge2$.
