Q) $lim_{(x, y, z) \to (0,0,0)} \frac{xyz}{x^2 + y^2 + z^2} = 0$
Hint: $|x| \leq \sqrt{x^2 + y^2 + z^2}$
proof attempt:
Given $\epsilon > 0$, choose $\delta = \epsilon$. Then $||(x, y, z) - (0,0,0)|| = \sqrt{x^2 + y^2 + z^2}$, and so $||(x, y, z) - (0,0,0)|| < \delta$ implies that
$\left| \frac{xyz}{x^2+y^2+z^2} - 0 \right| = \frac{|xyz|}{x^2+y^2+z^2} \leq \frac{\sqrt{x^2+y^2+z^2} \cdot \sqrt{x^2+y^2+z^2} \cdot \sqrt{x^2+y^2+z^2}}{x^2 + y^2 + z^2} = \sqrt{x^2+y^2+z^2} = ||(x, y, z) - (0,0,0)|| < \delta = \epsilon$
By QM-GM or Cauchy-Schwarz $$ x^2+y^2+z^2 \geq 3\,\text{GM}^2(|x|,|y|,|z|) = 3 |xyz|^{2/3}$$ hence in a pointed neighbourhood of the origin $$ \left|\frac{xyz}{x^2+y^2+z^2}\right| \leq \tfrac{1}{3}\sqrt[3]{|xyz|}$$ and the wanted limit is zero by squeezing (which makes the use of $\varepsilon/\delta$ straightforward).