Compute the following limit if they exist: $\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$ Compute the following limit if they exist:
$\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$

Attempt:
Given $\epsilon > 0$. Choose $\delta = \text{Not sure yet}$. Then $||(x, y, z) - (0,0,0)|| = \sqrt{x^2 + y^2 + z^2}$, and so $||(x, y, z) - (0,0,0)|| < \delta$
$\left| \frac{2x^2y c oz(z)}{x^2 + y^2} \right| \leq \frac{2x^2|y|}{x^2+y^2} \leq \frac{2 (x^2 + y^2)(\sqrt{x^2+y^2})}{x^2 + y^2} = 2 \sqrt{x^2 + y^2}$
Okay, I am stuck here. As you can see I need $\sqrt{x^2 + y^2 + z^2}$ in order for $\delta = \frac{\epsilon}{2}$. 
Also what does it compute to?

Well, we have $$2\sqrt{x^2+y^2} \le 2\sqrt{x^2+y^2+z^2} $$
and you can pick  $\delta = \frac{\epsilon}2$.
