Find $\lim \limits_{(x,y) \rightarrow (0,0)}\frac{x^3 y^3 }{x^2+y^2}$ (epsilon delta proof)

Question

Suppose we have the following function: $f(x,y)=\dfrac{x^3 y^3 }{x^2+y^2}$.

Determine $\lim \limits_{(x,y) \rightarrow (0,0)}f(x,y)$.

I did the following, but I cannot continue.

Suppose $0 < \sqrt{x^2+y^2} < \delta$.

$$\left|\frac{x^3 y^3}{x^2 + y^2}-0\right| = \left|\frac{xyx^2y^2}{x^2+y^2}\right| \le \left|\frac{(x^2+y^2)^2(xy)}{x^2+y^2}\right|=|xy|\left|x^2+y^2\right|<|xy|\delta^2=\epsilon.$$ The $|xy|$ makes it impossible for me to relate $\delta$ to $\epsilon$. How can I solve it?

Answer 1

Hint: $\forall (x,y)\in \mathbb R^2\left(2|xy|\leq x^2+y^2\right)$.

This inequality helps not only for finding $\delta$, but also to get $$\forall (x,y)\in \mathbb R^2\left((x,y)\neq (0,0)\implies \left|\dfrac{xyx^2y^2}{x^2+y^2}\right|= \dfrac{x^2y^2}2\left|\dfrac{2xy}{x^2+y^2}\right|\leq \dfrac{x^2y^2}{2}\right),$$ which deviates slightly from your work.