I've been struggling to prove that the following limit exists and calculate it. Squeeze theorem must be used. $$\lim_{(x,y)\to(0,0)} \frac{x^3-2xy^2}{x^2+y^2}$$

Note: using polar coordinates is not allowed.

  • $\begingroup$ Do you know what the limit is? It could help. $\endgroup$
    – ajotatxe
    Commented Oct 15, 2016 at 12:45
  • $\begingroup$ I do. The limit is 0. $\endgroup$
    – S. Yusj
    Commented Oct 15, 2016 at 13:09

1 Answer 1


$$\left|\frac{x^3-2xy^2}{x^2+y^2}\right|=|x|\left|\frac{x^2-2y^2}{x^2+y^2}\right|\leq|x| \frac{x^2+2y^2}{x^2+y^2}\leq |x| \frac{2x^2+2y^2}{x^2+y^2}=2|x|\to0$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .