Answer by WHLin for Why is $ \lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy}$ defined while $\lim_{(x,y,z) \to (0,0,0)} \frac{\sin(xyz)}{xyz}$ is not?

By definition, we have $sin(t)=t-\frac{t^3}{3!}+\frac{t^5}{5!}-...$ and hence $sin(xyz)=xyz-\frac{(xyz)^3}{3!}+\frac{(xyz)^5}{5!}-...$

Therefore, when $(x,y,z)$ approaches to zero, the behavior of $\frac{sin(xyz)}{xyz} \approx 1 - \frac{(xyz)^2}{3!} + [\text{higher order terms}]$. The ratio limit exists and the residue converges to zero quadratically with the product $xyz$.