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 Oct 15 '16 at 12:45
  • $\begingroup$ I do. The limit is 0. $\endgroup$ – S. Yusj Oct 15 '16 at 13:09

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.