I'm sure I could think my way through these, but I'm sick and on a tight schedule, so I was hoping someone here could help me out. I would appreciate a verification, with or without proof, of the following assertions:

$$(\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)$$

and

$$\sqrt{1 + \mathcal O(\epsilon^2)} = 1 + \mathcal O(\epsilon^2)$$

Thanks so much.

I think I've managed to show the first one. Suppose $f(\epsilon) = \mathcal O(\epsilon)$ (as $\epsilon \searrow 0$). Then there exists $C >0, \delta > 0$ such that $0 < \epsilon < \delta$ implies
$$\left| \frac{f(\epsilon)}{\epsilon} \right| \leq C.$$
To show that $(\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)$, one simply observes that
$$\left| \frac{f^2(\epsilon)}{\epsilon^2} \right| \leq C^2.$$