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Math Help - Show that sqrt(x) is uniformly continuous on [0,infinity)

  1. #1
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    Show that sqrt(x) is uniformly continuous on [0,infinity)

    Yikes! I feel like were all having trouble with uniform continuity...

    Use the definition of uniform continuity to show that \sqrt{x} is uniformly continuous on [0, \infty) (relative to [0, \infty))

    For this, do I have to show that \sqrt{x} is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1, \infty) before I can show its uniformly continuous on [0, \infty)?
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  2. #2
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    Quote Originally Posted by tn11631 View Post
    Yikes! I feel like were all having trouble with uniform continuity...

    Use the definition of uniform continuity to show that \sqrt{x} is uniformly continuous on [0, \infty) (relative to [0, \infty))

    For this, do I have to show that \sqrt{x} is uniformly continuous on [0,2] and then show that it is uniformly continuous on [1, \infty) before I can show its uniformly continuous on [0, \infty)?
    That is certainly the most convenient way to do it. The usual conditions ensuring that a function is uniformly continuous are either (a) it is continuous on a bounded closed interval, or (b) it is differentiable with a bounded derivative.

    The function \sqrt{x} on the interval [0,\infty) does not satisfy either of those conditions. But it satisfies (a) and (b) on the overlapping subintervals [0,2] and [1,\infty) respectively. From that, you can deduce that it is uniformly continuous on the whole interval.
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  3. #3
    Senior Member Pinkk's Avatar
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    Lemma: |\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|} for any x,y \ge 0.

    Proof:

    Let \epsilon > 0 and choose \delta = \epsilon^{2}. Then, for x,y \in [0,\infty ), |x-y|<\delta implies that |\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|} < \sqrt{\epsilon^{2}} = \epsilon. Thus, f(x) = \sqrt{x} is uniformly continuous on [0, \infty ). Q.E.D.
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