No limit for $\frac{x^2 y}{x^2-y^2}$ when $(x,y) \rightarrow (0,0)$ I'm trying to calculate
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2}$$
I've tried a lot of paths: $y=x$, $y=x+1$, $y=x^2$... Always I end up with $0$.
The answer of my homework and the Wolfram Alpha say me it doesn't exist.
What I'm doing wrong?
I appreciate any help.
 A: This solution is more like what you were trying.
Start with $x=2y$, you will get
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2}
=\lim_{(x,y)\to(0,0)}\dfrac{4y^2 y}{4y^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{4y^3}{3y^2} = \lim_{(x,y)\to(0,0)}\dfrac{4y}{3}=0$$
But for $x=y^2-y$ you will get
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{(y^2-y)^2y}{(y^2-y)^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{y^5-2y^4+y^3}{y^4-2y^3} = \lim_{(x,y)\to(0,0)}\dfrac{y^2-2y+1}{y-2} = -\dfrac{1}{2}$$
A: For $x\ne 0$ you can rewrite the function as
$$\frac{y}{1-(y/x)^2}\;.$$
In order for this to approach something other than $0$, it’s going to have to be an indeterminate $\frac00$ form, so the ratio $\frac{y}x$ is going to have to approach $1$. Let’s try $y=xe^x$: $e^x$ approaches $1$ as $x$ approaches $0$. Then
$$\frac{x^2y}{x^2-y^2}=\frac{x^3e^x}{x^2-x^2e^x}=\frac{xe^x}{1-e^x}\;,$$
and you can evaluate the limit as $x\to 0$ using l’Hospital’s rule, for instance.
A: Given any $\alpha>0$ and any $x$, pick a $y$ so that $|x^2-y^2|<\alpha^{-1}|x|^3$ and $|y|<|x|$.
Then $\frac{|x|^3}{|x^2-y^2|}>\alpha$, and $|(x,y)|<2|x|$.
