$\sqrt x$ 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.  

The explanation is from Jonathan Kane's textbook "Writing Proofs in Analysis". Which asks the reader to observe the behavior of the function $f(x,y)=\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}=\vert\sqrt{x}-\sqrt{y}\vert$. The natural step is to restrict the "size" of $\vert x-y\vert$ so that as $x,y\to\infty$ then so does $\sqrt{x}+\sqrt{y}$, which leads to $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}\to 0$. But a seeming roadblock arises as $x,y\to 0$ since that would make the denominator approach $0$ as well. However, the problem disappears when we realize that if $\sqrt{x}+\sqrt{y}\to 0$ then $\vert \sqrt{x}-\sqrt{y}\vert\to 0$ or if $\sqrt{x}+\sqrt{y}\to \infty$ then $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}=\vert\sqrt{x}-\sqrt{y}\vert\to 0$. Hence, if the given $\epsilon>0$ is such that $\sqrt{x}+\sqrt{y}<\epsilon$, then $\vert\sqrt{x}-\sqrt{y}\vert<\epsilon$ and we are done. On the other hand, if $\sqrt{x}+\sqrt{y}\geq\epsilon$, then $\frac{\vert x-y\vert}{\sqrt{x}+\sqrt{y}}<\frac{\vert x-y\vert}{\epsilon}$ and we only need to compute for $ \frac{\vert x-y\vert}{\epsilon}<\epsilon$ to get $\vert x-y\vert<\epsilon^2$.
