Q: How to find the limit of $ y^{2}(x^{3}+2xy^{2})/(x^2+y^2)^{3/2}$ How to find the limit of $ y^{2}(x^{3}+2xy^{2})/(x^2+y^2)^{3/2}$ at $0$ if it exist?
May I ask for a epsilon-delta proof is possible? May I avoid the usage of the polar coordinate?

A: The numerator factors as $(xy^2)(x^2+2y^2)$ and $|xy^2|\leq(x^2+y^2)^{3/2}$.  Now it's easy.

A: Notice that
$$0 \le \left| \frac {y^2 (x^3 + 2xy^2)} {(x^2 + y^2)^{\frac 3 2}} \right| \le \left| \frac {y^2 x^3} {(x^2 + y^2)^{\frac 3 2}} \right| + \left| \frac {2xy^4} {(x^2 + y^2)^{\frac 3 2}} \right| \le \left| \frac {y^2 x^3} {(x^2 + 0)^{\frac 3 2}} \right| + \left| \frac {2xy^4} {(0 + y^2)^{\frac 3 2}} \right| = \\
\frac {y^2 |x|^3} {|x|^3} + 2 \frac {|x| y^4} {|y|^3} = y^2 + 2 |xy|$$
and the last expression tends to $0$ when $(x,y) \to (0,0)$, so by the squeeze theorem we get that $\left| \dfrac {y^2 (x^3 + 2xy^2)} {(x^2 + y^2)^{\frac 3 2}} \right| \to 0$, whence it follows that $\dfrac {y^2 (x^3 + 2xy^2)} {(x^2 + y^2)^{\frac 3 2}} \to 0$ too.
