Multivariable limit which should be simple ! How to calculate the following limit WITHOUT using spherical coordinates? 
$$
\lim _{(x,y,z)\to (0,0,0) } \frac{x^3+y^3+z^3}{x^2+y^2+z^2} 
$$
?
Thanks in advance

Since $x^2 \le x^2+y^2+z^2$, we have $|x| \le \sqrt{x^2+y^2+z^2}$, and similarly for $|y|$ and $|z|$. This gives
$$
\left|\frac{x^3+y^3+z^3}{x^2+y^2+z^2}\right|
\le \frac{|x|^3+|y|^3+|z|^3}{x^2+y^2+z^2}
\le \frac{(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2}
.
$$
Can you take it from there?
