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!
 A: Hint: You can use spherical coordinates:

$$ x=r \, \sin\theta \, \cos\varphi \\
   y=r \, \sin\theta \, \sin\varphi \\
    z=r \, \cos\theta , $$

where $r=\sqrt{x^2+y^2+z^2}$, $0\leq \theta \leq \pi$, and $0\leq \phi \leq 2\pi$.
Note: consider taking the limit as $r\to 0$.
A: Hint: Note that $f(x,y,z)\to0$ if and only if $\lvert f(x,y,z)\rvert\to0$.  Now, 
$$
\left\lvert\frac{3xyz}{x^2+y^2+z^2}\right\rvert=\frac{3\lvert x\rvert\,\lvert y\rvert\,\lvert z\rvert}{x^2+y^2+z^2}.
$$
Now, note that
$$
\lvert x\rvert=\sqrt{x^2}\leq\sqrt{x^2+y^2+z^2},
$$
and similarly $\lvert y\rvert,\lvert z\rvert\leq\sqrt{x^2+y^2+z^2}$. 
A: 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,
$$
\frac{3|xyz|}{x^2+y^2+z^2}\le3\left(x^2+y^2+z^2\right)^{1/2}
$$
A: Hint:
$$\frac{x^2+y^2+z^2}{3} \geq |xyz|^{2/3}$$
A: $$
\left|\frac{3xyz}{x^2+y^2+z^2}\right|=\frac{|3xyz|}{\|(x,y,z)\|^2}\le
3\frac{\|(x,y,z)\|^3}{\|(x,y,z)\|^2}=3\|(x,y,z)\|
$$
