Find the limit as $(x,y,z) \to (0,0,0)$ of a specific function $f(x,y,z)$ 
find $$\lim_{(x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}} $$ if exist

$\sin$ is a continuous function so $\lim_{(x,y,z) \to (0,0,0)} \sin(x^2+y^2+z^2)=0$.
$\sqrt{x}$ is not defined for $x<0$ over $\mathbb{R}$ I need to find $x,y,z$ function such that $\sqrt{x^2+y^2+z^2}<0$ to prove there is no limit?

HINT:
$$\frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\cdot\sqrt{x^2+y^2+z^2}\;.$$
What do you know about $\lim\limits_{x\to 0}\dfrac{\sin x}x$?
