Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.
So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.
$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.
 A: Your proof seems to be correct! Another more intuitive way would be to use polar coordinates in this situation (which always is a good thing to consider when you have a $x^2+y^2$ term somewhere): $$\lim \limits_{(x,y) \to 0} \frac{x^ny^m}{x^2+y^2}=\lim \limits_{r\to 0}\frac{(r\cos(\theta))^n(r\sin(\theta))^m}{(r\cos(\theta))^2+(r\sin(\theta))^2}=\lim \limits_{r\to 0} r^{n+m-2}\cos(\theta)^n\sin(\theta)^m \rightarrow 0 \quad \text{when} \; n+m>2$$
(since $cos(\theta)^n\sin(\theta)^m$ is bounded)
A: Problem like this is fun to solve for sure. Let's do the backward direction: If $m+n > 2$, then either $m \ge 2$ or $n \ge 2$. Let's say $m \ge 2$, then rewrite it as $\dfrac{x^2y}{x^2+y^2}\cdot x^{m-2}y^{n-1}$. Observe that $x^{m-2}y^{n-1} \to 0$ or $1$, and $0  \le \left|\dfrac{x^2y}{x^2+y^2}\right| \le |y|$, thus it goes to $0$, and the whole thing goes to $0$. Conversely, if the limit exists, say $L < \infty$ and if $m+n \le 2 \implies m = n = 1\implies \dfrac{x^my^n}{x^2+y^2} = \dfrac{xy}{x^2+y^2}$ does not have a limit as you can choose $2$ paths to $(0,0)$, contrary to the existence of $L$, thus it is the case that $m+n > 2$.
