<s>[INST] 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.
 [/INST] 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
$$\lim_{(x,y)\to(0,0)}\dfrac{x^2 y}{x^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{(y^2-y)^2y}{(y^2-y)^2-y^2} = \lim_{(x,y)\to(0,0)}\dfrac{y^5-2y^4+y^3}{y^4-2y^3} = \lim_{(x,y)\to(0,0)}\dfrac{y^2-2y+1}{y-2} = -\dfrac{1}{2}$$
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