How do I solve this limit?
$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$
This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$
The first one is a standard limit with value one, but I'n not sure about the other term.
Yes your idea is very good to put away the $\sin$ term, now we have that, for example by spherical coordinates
$$\frac{xyz}{x^2+y^2+z^2} =r \cos \theta\sin \theta\cos^2 \phi\sin \phi $$
or as an alternative by AM-GM we can use that
$$x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2y^2z^2} \implies \left|\frac{xyz}{x^2+y^2+z^2}\right| \le \frac{|xyz|^\frac13}{3}$$
concluding in both cases by squeeze theorem.
$0 \leq | \frac {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac 1 2 |z| \to 0$ since $|xy| \leq \frac 1 2{(x^{2}+y^{2})}$. Hence the given limit is $1$.
