<s>[INST] How to solve the limit:$ \lim_{x \to 0,y \to 0} \frac{x^2y^2}{x^3+y^3} $ I confronted a problem, that is
$$
\lim_{x \to 0,y \to 0} \frac{x^2y^2}{x^3+y^3}
$$
I tried to use $x=r\cos(\theta) , y=r\sin(\theta)$ to solve the problem, and I got 
$$\lim_{r \to 0}r(\frac{\cos^2(\theta)\sin^2(\theta)}{\cos^3(\theta)+\sin^3(\theta)})$$
And no matter what value $\theta$ takes, we all have the limit is zero.
My question is:
(1) Am I right? Or
(2) If we can change the limit into such "$rf(\theta)$" form, when r goes to zero, will the limit go to zero?
 [/INST] As an alternative note that


*

*$x=0 \,,y\to 0\implies \frac{x^2y^2}{x^3+y^3}=0$

*$x=t \,, y=t^2-t\,,t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^2(t^2-t)^2}{t^3+(t^2-t)^3}=\frac{t^6-2t^5+t^4}{t^3+t^6-3t^5+3t^4-t^3}=\frac{t^2-2t+1}{t^2-3t+3}\to \frac13$
</s>