# Evaluating $\lim\limits_{(x,y) \rightarrow (0,0)} \frac{x^3 - y^3}{x^2 + y^2}$

I have tried approaching the limit from different functions like $x = 0$, $y = 0$, $y = x$, $y = x^2$, etc. But they all go to $0$, so my guess would be that the limit goes to $0$, but how do I show that the limit is definitely $0$?

$$\left|\frac{x^3 - y^3}{x^2 + y^2}\right|\leqslant \frac{|x|^3}{x^2 + y^2}+ \frac{|y|^3}{x^2 + y^2}\leqslant|x|+|y|$$

• How do you know $\left|\frac{x^3 - y^3}{x^2 + y^2}\right|\leqslant \frac{|x|^3}{x^2 + y^2}+ \frac{|y|^3}{x^2 + y^2}$? Is there a name for this/is there a proof of it I can read online? – mr eyeglasses May 17 '14 at 22:40
• The triangle inequality says that $|x^3 - y^3|\leqslant|x^3|+|y^3|=|x|^3+|y|^3$. – Did May 17 '14 at 22:42
• Can you please explain why $\frac{|x|^3}{x^2 + y^2}+ \frac{|y|^3}{x^2 + y^2}\leqslant|x|+|y|$. – Sandeep Silwal May 20 '14 at 14:52
• @SandeepSilwal Because $\frac{|x|^3}{x^2 + y^2}\leqslant\frac{|x|^3}{x^2}=|x|$ for instance. – Did May 20 '14 at 17:17

There is the ever-popular method of converting the expression into polar form, which will point out quickly if the effort to find the limit suffers from "directionality":

$$\ \frac{x^3 \ - \ y^3}{x^2 \ + \ y^2} \ \ \rightarrow \ \ \frac{r^3 \ (\cos^3 \theta \ - \ \sin^3 \theta)}{r^2} \ = \ r \ (\cos^3 \theta \ - \ \sin^3 \theta) \ .$$

This will approach zero from any direction $\ \theta \$ as $\ r \ \rightarrow \ 0 \$ .

• Like I say, it's "ever-popular" (which is not the same as saying that it works for everything)... – colormegone May 17 '14 at 22:46