How to solve the limit:$ \lim_{x \to 0,y \to 0} \frac{x^2y^2}{x^3+y^3} $

Question

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?

Answer 1

If $x=0$, then the limit is $0$. However$$\lim_{n\to\infty}\frac{\left(\frac1n+\frac1{n^2}\right)^2\left(-\frac1n\right)^2}{\left(\frac1n+\frac1{n^2}\right)^3+\left(-\frac1n\right)^3}=\lim_{n\to\infty}\frac{(n+1)^2}{3n^2+3n+1}=\frac13.$$

Answer 2

As an alternative note that

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

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