Originally Posted by

**Scott H** Do you mean

$\displaystyle \lim_{\small (x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y^2}?$

If so, then using the Pythagorean identity $\displaystyle \cos^2\theta+\sin^2\theta=1$, we obtain

$\displaystyle \frac{x^3+y^3}{x^2+y^2}=\frac{r^3(\cos^3\theta+\si n^3\theta)}{r^2(\cos^2\theta+\sin^2\theta)}=\frac{ r^3(\cos^3\theta+\sin^3\theta)}{r^2\cdot 1}=r(\cos^3\theta+\sin^3\theta).$

Now notice that $\displaystyle r$ is the variable that goes to $\displaystyle 0$ (as we're approaching the origin), and

$\displaystyle |\cos^3\theta+\sin^3\theta|\le|\cos^3\theta|+|\sin ^3\theta|\le\cos^2\theta+\sin^2\theta=1.$

Edit: Added more useful inequality.