Show $\lim_{(x,y)\to(0,0)}\frac{x^3-xy^3}{x^2+y^2}=0$ I need to show if the following limit is exists (and exists $0$),
$$
f(x,y)=\frac{x^3-xy^3}{x^2+y^2},
$$
for $\vec x\to 0$. I tried out the following:
$$
\frac{x^3-xy^3}{x^2+y^2}\leq\frac{x^3-xy^3}{x^2}=1-\frac{y^3}{x^2},
$$
but obviously I'm stuk here. A similar approach by omitting $y^2$ gives the same problem. What could I do next?

With polar coordinates we have $$x=r\cos { \theta ,y=r\sin { \theta  }  } \\ \lim _{ (x,y)\to (0,0) } \frac { x^{ 3 }-xy^{ 3 } }{ x^{ 2 }+y^{ 2 } } =\lim _{ r\rightarrow 0 } \frac { { r }^{ 3 }\cos ^{ 3 }{ \theta  } -{ r }^{ 4 }\cos { \theta \sin ^{ 3 }{ \theta  }  }  }{ { r }^{ 2 } } =0$$
