Multivariable limit $\lim\limits_{(x,y,z)\to(0,0,0)}\frac{3xyz}{x^2+y^2+z^2}$ Can anybody give me a hint on how to bound 
$$\frac{3xyz}{x^2+y^2+z^2}$$
I'm trying to prove that it converges to zero. Thanks!

Hint: The AM-GM gives
$$
\left(x^2y^2z^2\right)^{1/3}\le\frac13\left(x^2+y^2+z^2\right)
$$
then
$$
\left|\frac{3xyz}{x^2+y^2+z^2}\right|\le\left|\frac{xyz}{\left(x^2y^2z^2\right)^{1/3}}\right|=|xyz|^{1/3}
$$

Alternate Approach
Easily, we have
$$
\begin{align}
x^2&\le x^2+y^2+z^2\\
y^2&\le x^2+y^2+z^2\\
z^2&\le x^2+y^2+z^2
\end{align}
$$
Multiplying and taking the square root yields
$$
|xyz|\le\left(x^2+y^2+z^2\right)^{3/2}
$$
Therefore,
$$
\frac{3|xyz|}{x^2+y^2+z^2}\le3\left(x^2+y^2+z^2\right)^{1/2}
$$
