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This question already has an answer here:

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


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

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


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


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.

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marked as duplicate by Jack, user91500, Ian Miller, BLAZE, draks ... Dec 10 '15 at 6:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ See here. $\endgroup$ – mattos Dec 10 '15 at 2:13
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For $x,y\geq 0$, $$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y|.$$

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

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    $\begingroup$ I am wondering how to show $|\sqrt {x_1}-\sqrt {x_2}|\leq \sqrt {|x_1-x_2|}$ is true $\endgroup$ – Simple Dec 10 '15 at 1:49
  • $\begingroup$ Try squaring that. $\endgroup$ – Umang Dec 10 '15 at 2:55

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