I'm resolving the limit of $\frac{(2x^2).y}{x^4+y^2}$ when $(x,y)\to(0,0)$
We didn't study any special theorem, we did only approach.
I tried first the changes $y=x^2$ and $y=0$.
In $f(x,x^2)$, operating I get that the $\lim =\frac{2x^4}{2x^4}=1$ So then I test the $f(x,0)$ operating I get $\frac{2x^2.0}{x^4+0}$, that is still $0/0$,but has a zero on the numerator, that is always enough to claim that the limit of $f(x,0)$ is $0$? Assuming that the limit $f(x,x^2)$=1 $f(x,0)$=0 so the limit don't exist.
Is right the procedure? Any way to do it easy? is right the assumption of the 0 on the numerator that the limit will be cero? Thanks.
