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I need to prove that $f(x)=\sqrt x$ is uniformly continuous on $[0, \infty)$.

I wrote $\displaystyle |\sqrt{x}-\sqrt{c}|=|\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{\sqrt{x}+\sqrt{c}}| \leq| \frac{x-c}{\sqrt{x}+\sqrt{c}}|$, but I am stuck after this. I found a solution that takes $\delta = \epsilon^2$, but I don't understand where this comes from. If we take this $\delta$, we will get the necessary result, but it seems like we are assuming $\sqrt{x}-\sqrt{c}=\epsilon$ rather than showing it.

Can someone explain the confusion or suggest another epsilon-delta solution?

M.H
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kiwifruit
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  • Note: $\left|\sqrt{x}-\sqrt{c}\right| = \left|\dfrac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{\sqrt{x}+\sqrt{c}}\right| = \left|\dfrac{x-c}{\sqrt{x}+\sqrt{c}}\right|$, rather than inequality. – Clarinetist Feb 17 '14 at 04:12
  • Hmm, it appears that someone edited my post incorrectly, and removed the expression I intended. I had that $\frac{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}{\sqrt{x}+\sqrt{c}} = \frac{x-c}{\sqrt{x}+\sqrt{c}} \leq \frac{x-c}{\sqrt{x} MINUS \sqrt{c}}$ – kiwifruit Feb 17 '14 at 04:16
  • Duplicate – NasuSama Feb 17 '14 at 04:25
  • I saw that question, but I did not understand the answers. – kiwifruit Feb 17 '14 at 04:26
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    Alternatively: Since $x^{1/2}$ is clearly continuous then is uniform continuous on $[0,1]$. Now we will show that also is uniform continuous on $[1, \infty)$.

    Let $(a_n){n=0}^\infty,,(b_n){n=0}^\infty$ be two sequences such that $a_n-b_n \rightarrow 0$ and $a_n,b_n \in [1,\infty)$. Then we'd like to show that $(a_n)^{1/2}-(b_n)^{1/2} \rightarrow 0$.

    $$|a^{1/2}-b^{1/2}|= \bigg|\frac{a_n-b_n}{a^{1/2}+b^{1/2}}\bigg|\le |a_n-b_n|/2$$

    Since $a_n,b_n\ge 1 $ for all $n \in \mathbb{N}$. Then since $(a_n-b_n)\rightarrow 0$, we're done.

     – Jose Antonio Feb 17 '14 at 04:54

2 Answers2

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Hint. Prove the inequality $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$.

Hanul Jeon
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Just divide it into two parts. On $[0,1]$ there is Heine–Cantor theorem. And on $[1, \infty)$ the denominator is greater (or equal) than 2.

sas
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  • Thank you, but this theorem was not covered in class.Could you elaborate why the denominator greater than two is important? – kiwifruit Feb 17 '14 at 04:27
  • What denominator? I think you were thinking of the derivative... – DonAntonio Feb 17 '14 at 04:28
  • I'm referring to "denominator is greater (or equal) than 2" – kiwifruit Feb 17 '14 at 04:29
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    $|\sqrt x - \sqrt y| = \left|\frac{x-y}{\sqrt x +\sqrt y}\right|\leq\frac 12 |x - y|<\frac 12 \delta$ – sas Feb 17 '14 at 04:30
  • Oh, ok, that makes sense for $[1, \infty]$, but is there another more basic theorem that can be used for [0,1]? – kiwifruit Feb 17 '14 at 04:36
  • More basic (at this stage) than the H-C theorem, @kiwifruit? I highly doubt it, and any try in this direction would probably be equivalent to a pretty close form of this theorem and/or its proof. – DonAntonio Feb 17 '14 at 10:58