$\sqrt x$ 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.  
 A: Let's try a more general approach using the mean value theorem. Let $f=x^\alpha$ and suppose $x<y$. Since  $y^{\alpha}-x^\alpha=(y-x) \alpha c^{\alpha-1}$ for $x<c<y$. So for $0< \alpha<1$ we thus have 
$$y^\alpha-x^{\alpha}\le (y-x)\alpha y^{\alpha-1}\le \alpha(y-x)$$
for $y\ge 1$, which shows that $f$ is uniformly continuous on $[1, \infty)$ and clearly is uniformly continuous on $[0,1]$. Thus, if $0<\alpha<1$, $f$ is uniformly continuous on $[0,\infty)$.
A: We let $ \epsilon > 0 $ 
We have that $ | x - y | < \delta  $ .
We choose that $ \delta = \epsilon^2 $ 
We then note the following result.  
$ | f(x) - f(y) | = | \sqrt{x} - \sqrt{y} | = \sqrt{ | \sqrt{x} - \sqrt{y}|^2}  \leq \sqrt{ | \sqrt{x} - \sqrt{y}| | \sqrt{x} + \sqrt{y} |} = \sqrt{ | x - y | } < \sqrt{\delta} = \epsilon$ 
$ \square $ 
A: Let $\epsilon > 0.$ Pick $\delta = \epsilon^2.$ Then for $|x-y| < \delta$ we have
$$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y| = |x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon. $$
A: We'll prove that $f(x) = \sqrt{x}$ is uniformly continuous on $\mathbb{R}_+$. Indeed, $[0,1]$ being a compact set, $f$ is uniformly continuous on this interval. On the other hand, on $[1,\infty), f$ is Lipschitz, and hence is uniformly continuous. Hence we are now done.
A: 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 $[1, \infty]$, 
$$
\frac {|x-y|}{\sqrt{x}+\sqrt{y}} < |x-y|\:\:\text{ so  }\:\:\delta=2\varepsilon
$$
A: 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 $\vert x-y\vert<\epsilon^2$.
