For what values of $n$ does the limit $\lim\frac{x^2y}{x^{2n}+y^2}$ exists? For what values of n does the following limit exist?
$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}$$
I've tried using the squeeze theorem but couldn't come up with anything good. Also converting it to cylindrical coordinates gave me the answer that the limit exists for all n (which I don't know if it's correct).
 A: Hint:
$$x^{2n} + y^2 = |x|^{2n} + |y|^2 \geqslant 2|x^n||y|$$
Thus
$$0 \leqslant \left|\frac{x^2y}{x^{2n} + y^2} \right| = \frac{|x|^2|y|}{x^{2n} + y^2} \leqslant \frac{|x|^{2-n}}{2}.$$
Consider the squeeze theorem to find a sufficient condition on $n$ for convergence to $0$.  Then consider what happens if this condition is not satisfied.
A: We know with $x=0$ as $y\to0$ then $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{y\to0} \frac{0}{0+y^2}=0$$
For $n>2$ this limit doesn't exist. Let $y=x^n$ so
$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{(x,y) \to (0,0)} \frac{x^{n+2}}{2x^{2n}}=\infty$$
This substituation shows for $n=2$ this limit doesn't exist. Let $y=x^2$ so
$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^{2n}+y^2}=\lim_{(x,y) \to (0,0)} \frac{x^4}{2x^4}=\frac12$$
For $n=1$ the limit exists. By polar coordinates
$$\Big|\frac{x^2y}{x^2+y^2}\Big|=\Big|\frac{r^3\cos^2\theta\sin\theta}{r^2}\Big|=|r\cos^2\theta\sin\theta|<r$$
A: Let's eliminate first the cases $n<0$ and $n\ge2$.


*

*$n<0$ converges


$$f_n(x,y)=\frac{x^2y\ .\  x^{2|n|}}{1+y^2x^{2|n|}}\sim x^{2|n|+2}y\to 0$$


*

*$n>2$ diverges


$$f_n(x,x^n)=\frac{x^2x^n}{x^{2n}+x^{2n}}=\frac{1}{2}x^{2+n-2n}=\frac{1}{2}x^{2-n}\to \infty$$


*

*$n=2$ has multiple limits, so doesn't converges


$$f_2(0,y)=\frac{0}{0+y^2}=0\to 0$$
$$f_2(x,x)=\frac{x^3}{x^4+x^2}\sim\frac{x^3}{x^2}\sim x\to 0$$
$$f_2(x,x^2)=\frac{x^2x^2}{x^4+x^4}\to\frac{1}{2}$$
So let's have now $n\in[0,2[$
We loose no generality in writting $y=x^ng(x,y)$, the function $g$ being arbitrary.
$$f(x,y)=\frac{x^2x^ng(x,y)}{x^{2n}+x^{2n}g(x,y)^2}=x^{2-n}\frac{g(x,y)}{1+g(x,y)^2}$$
According to this study : range of $\frac{x}{x^2+1}$ 
We can say that $|\frac{g}{1+g^2}|<1$
Thus $|f(x,y)|\le|x^{2-n}||\frac{g}{1+g^2}|<|x^{2-n}|\to 0$
$\color{teal}{Conclusion :}$ 
$\lim\limits_{x,y\to 0}f(x,y)$ exists for $n\in]-\infty,2[$ and this limit is $0$.

Of course we can also do this directly by setting $u=\frac{y}{x^n}$.
$$f_n(x,y)=\frac{x^2y}{x^{2n}+y^2}=\frac{x^2x^n\frac{y}{x^n}}{x^{2n}(1+\frac{y^2}{x^{2n}})}=x^{2-n}\frac{u}{1+u^2}$$
And since $|\frac{u}{1+u^2}|\le\frac 12$ is bounded, $f_n(x,y)\to 0$ for $n<2$.
Also since we can choose $y$ so that this expression take any desired value in $[-\frac12,\frac12]$ then $f_n(x,y)$ cannot converge for $n\ge 2$. 
Note: This is because it has multiple limits. And for $n>2$ the divergence to $\infty$ is not even required, just suffice to say that $-\infty$, $0$, $+\infty$ are all possible limits.
