Answer
The limit does not exist
Work Step by Step
$\lim\limits_{(x, y, z) \to (0, 0, 0)}\frac{\sin(\sqrt {x^2+y^2+z^2})}{x^2+y^2+z^2}$
Let $t = \sqrt {x^2+y^2+z^2}$
$\lim\limits_{(t) \to 0^+} = \frac{\sin t}{t^2} = \lim\limits_{(t) \to 0^+} \frac{\cos t}{2t} = +\infty$ [NB: We apply L'hopitals rule here in finding the limit]
$\lim\limits_{(t) \to 0^-} = \frac{\sin t}{t^2} = \lim\limits_{(t) \to 0^-} \frac{\cos t}{2t} = -\infty$
