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Prove that $f(x) = \sqrt x$ is uniformly continuous on $[0,∞)$.

take any $x,y \in [0, \infty)$. If $|x-y| < \sigma$ then $|f(x)-f(y)|=|\sqrt x - \sqrt y|= |(x-y)(x+y)/(\sqrt x + \sqrt y)| < |\sigma (x+y)/(\sqrt x + \sqrt y)|$

Now, I suppose, here must be shown that $|(x+y)/(\sqrt x + \sqrt y)|$ is bounded to show the continuity. Is it right? How do I proceed?


marked as duplicate by mathcounterexamples.net, José Carlos Santos, Nosrati, Henning Makholm, user1729 Aug 20 '18 at 13:33

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You are wrong where you have written $|\sqrt x - \sqrt y|= |(x-y)(x+y)/(\sqrt x + \sqrt y)|$. You should have written $$|\sqrt x - \sqrt y|= |\dfrac{x-y}{\sqrt x+\sqrt y}|<\dfrac{\sigma}{\sqrt x+\sqrt y}$$without lose of generality let's assume $x>y$. We prove that $\sigma =\epsilon^2$. We have $$|\sqrt x-\sqrt y|^2=x+y-2\sqrt{xy}<x-y=|x-y|$$therefore if $|x-y|<\epsilon ^2$ then $|\sqrt x-\sqrt y|<\epsilon$ and we complete our proof here


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