Originally Posted by

**Ulysses** What you think about this one?

$\displaystyle \displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}$

I think the limit is zero, but I'm having some trouble with its demonstration. I think I've already demonstrated it, but I'm not completely sure.

$\displaystyle \displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{6xy^2}{x^2+y^4}}=\d isplaystyle\lim_{(x,y) \to{(0,0)}}{6x}\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}$

I've tried to solve the part on the right using polar coordinates, and I get to

$\displaystyle \lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{y^2}{x^2+y^4}}=\lim _{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2\c os^2\theta+r^4\sin^4\theta}}=\lim_{r \to{0}+}{\displaystyle\frac{r^2\sin^2\theta}{r^2(\ cos^2\theta+r^2\sin^4\theta)}}=\lim_{r \to{0}+}{\displaystyle\frac{\sin^2\theta}{\cos^2\t heta+r^2\sin^4\theta}}=\displaystyle\frac{\sin^2\t heta}{cos^2\theta}$

Which means this limit doesn't exists, but as it is multiplied by something that tends to zero, the limit should be zero. Is this right?