$\sqrt x$ is uniformly continuous 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.  

In $[0,1]$ we want 
$$
\begin{split}
|f(x)-f(y)|=&\frac {|x-y|}{\sqrt{x}+\sqrt{y}} <\varepsilon\\
\Updownarrow\\
|x-y|<&\varepsilon(\sqrt{x}+\sqrt{y})<2\varepsilon\:\:\text{ so  }\:\:\delta=2\varepsilon  
\end{split}
$$
In $[1, \infty]$, 
$$
\frac {|x-y|}{\sqrt{x}+\sqrt{y}} < |x-y|\:\:\text{ so  }\:\:\delta=2\varepsilon
$$
