Prove that limit = 0. So the question is as in the title.
$$\lim_{(x,y,z)\to (0,0,0)}\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}$$
I decided to prove it from the definition.
so $L = 0$
$$\left\lvert\frac{\sin(xyz)}{\sqrt{(x^2+y^2+z^2)}}-0 \right \rvert \leq \frac {1}{\sqrt{(x^2+y^2+z^2)}} $$  
and then trnasform it to this
$${\frac{\sqrt{(x^2+y^2+z^2)}}{x^2+y^2+z^2}}$$
and now i don't know what to do.

Use $|\sin t|\le|t|$. Then
$$\left|\frac{\sin(xyz)}{\sqrt{x^2+y^2+z^2}}\right|
\le\left|\frac{xyz}{\sqrt{x^2+y^2+z^2}}\right|.$$
If we let $r=\sqrt{x^2+y^2+z^2}$ then $|xyz|\le r^3$ etc.
