I'm asked to prove using an $\epsilon - \delta$ - argument that $\displaystyle \lim_{(x,y) \to (0,0)} \frac { 4x^3 + 8xy^2}{3\sqrt[3]{(x^2 + y^2)^4}} = 0 $. Here is the proof I came up with:
Let $\epsilon > 0$ then:
$\left|\dfrac { 4x^3 + 8xy^2}{3\sqrt[3]{(x^2 + y^2)^4}}\right| = \dfrac {\left|x\right| \left| 4x^2 + 8y^2 \right|}{\left|3\sqrt[3]{(x^2 + y^2)^4}\right|} \le \dfrac {\left|x\right| \left| 8x^2 + 8y^2 \right|}{\left|3\sqrt[3]{(x^2 + y^2)^4}\right|} = \dfrac {8\left|x\right| (x^2 + y^2)}{3\sqrt[3]{(x^2 + y^2)^4}} = \dfrac {8 \sqrt{x^2}}{3\sqrt[3]{(x^2 + y^2)}} \le\dfrac {8 \sqrt{x^2}}{3\sqrt[3]{x^2}} =\dfrac{8}{3}\sqrt[3]{x}$.
Choose $\delta = \left(\dfrac{3}{8}\epsilon\right)^3$. Then $\sqrt{x^2+y^2} < \delta \implies x < \delta \implies \dfrac{8}{3}\sqrt[3]{x} < \dfrac{8}{3}\sqrt[3]{\delta} = \epsilon $. So:
$\left|\dfrac { 4x^3 + 8xy^2}{3\sqrt[3]{(x^2 + y^2)^4}}\right| < \epsilon$
Is this a valid proof? I am not very comfortable with the inequalities in these kind of proofs so I might have made a mistake there.