Prove that the function $\sqrt x$ is uniformly continuous on $\{x\in \mathbb{R} | x \ge 0\}$.
To show uniformly continuity I must show for a given $\epsilon > 0$ there exists a $\delta>0$ such that for all $x_1, x_2 \in \mathbb{R}$ we have $|x_1 – x_2| < \delta$ implies that $|f(x_1) – f(x_2)|< \epsilon.$
What I did was $\left|\sqrt x – \sqrt x_0\right| = \left|\frac{(\sqrt x – \sqrt x_0)(\sqrt x + \sqrt x_0)}{(\sqrt x + \sqrt x_0)}\right| = \left|\frac{x – x_0}{\sqrt x + \sqrt x_0}\right| < \frac{\delta}{\sqrt x + \sqrt x_0}$
but I found some proof online that made $\delta = \epsilon^2$ where I don't understand how they got? So, in order for $\delta =\epsilon^2$ then $\sqrt x + \sqrt x_0$ must $\le$ $\epsilon$ then $\frac{\delta}{\sqrt x + \sqrt x_0} \le \frac{\delta}{\epsilon} = \epsilon$. But then why would $\epsilon \le \sqrt x + \sqrt x_0? $ Ah, I think I understand it now just by typing this out and from an earlier hint by Michael Hardy here.
Best Answer
Let $\epsilon > 0.$ Pick $\delta = \epsilon^2.$ Then for $|x-y| < \delta$ we have
$$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y| = |x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon. $$