Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$

Question

Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$

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Let $\epsilon>0$, then there exists a $\delta=\epsilon^2$ such that $|x_1-x_2|<\delta$ for all $x_1,x_2\in[0,\infty)$.

$\begin{align}|f(x_1)-f(x_2)|&=|\sqrt{x_1}-\sqrt{x_2}|\\&<\sqrt{|x_1-x_2|}\\&<\sqrt{\delta}\\&=\epsilon\end{align}$

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Can someone tell me where I did wrong? I don't see where the error is. Thanks

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EDIT

From comment, people said the argument I have is right, I want to know how to show $|\sqrt{x_1}-\sqrt {x_2} |\leq\sqrt {|x_1-x_2|} $ is true.

Answer 1

The only thing I see wrong is using $<$ instead of $\leq$ for $|\sqrt{x_1}-\sqrt{x_2}|\leq \sqrt{|x_1-x_2|}$.

Answer 2

For $x,y\geq 0$, $$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y|.$$