Multivariable limit: $\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$ I need to solve the limit
$$\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$$
and I can't think on a possible upper bound for it.
Any ideas?
Thanks


$$\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$$

Let $$
f(x,y)=\dfrac{(y^2-x)^3}{x^2+y^4}
$$
Then $$
\lim_{(0,0)} f(x,y) = \lim_{(0,0)} f(x^2,y)\\
|f(x^2,y)|= \dfrac{|y^2-x^2|^3}{x^4+y^4}\le
\dfrac{(y^2+x^2)^3}{x^4+y^4} \\
=  \dfrac{r^6}{r^4 (\cos^4\theta + \sin^4\theta)}  = O(r^2)
$$using polar coordinates. Then the limit is 0.
