Prove that $\lim_{(x,y)\to(0,0)}{\frac{x^ny^m}{x^2+y^2}}=0$ if $n+m>2$.

Question

I wrote a proof for this, but it doesn't require that $n+m>2$, only that $n+m\neq 2$. So my question is, what is wrong with my proof? Note that for the question, $n$ and $m$ are non-negative integers.

Proof. Let $\varepsilon>0$ and suppose that $\delta>\sqrt{x^2+y^2}>0$. Now observe that $$\left|\frac{x^ny^m}{x^2+y^2}-0\right|=\frac{\left|x^ny^m\right|}{x^2+y^2}\leq\frac{\left(|x|^2\right)^{n/2}\left(|y|^2\right)^{m/2}}{x^2+y^2}$$ Clearly, $|x|^2=x^2\leq x^2+y^2$ and $|y|^2=y^2\leq x^2+y^2$. Thus $$\frac{\left(|x|^2\right)^{n/2}\left(|y|^2\right)^{m/2}}{x^2+y^2}\leq\frac{\left(x^2+y^2\right)^{n/2}\left(x^2+y^2\right)^{m/2}}{x^2+y^2}=\left(x^2+y^2\right)^{(n+m-2)/2}<\delta^{n+m-2}$$ Choose $\delta=\varepsilon^{\frac{1}{n+m-2}}$. Then $$\left|\frac{x^ny^m}{x^2+y^2}-0\right|<\delta^{n+m-2}=\varepsilon,$$ as desired.

So, obviously $n+m\neq2$, but I don't see why I can't have $n+m<2$. Any help would be greatly appreciated.

Answer 1

Note that you use $$ \sqrt{x^2+y^2} < \delta\ \Longrightarrow \bigl(\sqrt{x^2+y^2}\bigr)^{n+m-2} < \delta^{n+m-2}, $$ which relies on the fact that the function $$ t \longmapsto t^{n+m-2} $$ is monotonically increasing. This is false if the exponent is negative, i.e. like picture the graph of $1/x = x^{-1}$

Answer 2

I would use polar coordinates representation for $x$ and $y$. $$x=r\cos\theta\\y=r\sin\theta$$ The limit of $(x,y)\to 0$ is equivalent to $r\to 0$, independent of $\theta$. Then you have $$\lim_{r\to 0}\frac{r^n\cos^n\theta r^m\sin^m\theta}{r^2}=\cos^m\theta\sin^n\theta\lim_{r\to0}r^{m+n-2}$$ If $m+n-2>0$ the limit is $0$. If $m+n<2$ then the last limit is infinite. If $m+n=2$, then the last limit is $1$, but the overall value depends on the path $(\theta)$, so the limit does not exist.