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Continuity problem

  1. Sep 5, 2009 #1

    jgens

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    I've been reviewing my calculus text book and came accross this problem: Prove that the function [itex]f[/itex] defined by [itex]f(x) = \sqrt{x}[/itex] is continuous if [itex]x>0[/itex]. Would anyone mind verifying (or correcting) my proof? Suggestions are welcome. Thanks!

    Proof: Let [itex]\epsilon > 0[/itex] and choose [itex]\delta[/itex] such that [itex]\delta = \mathrm{min}(x_0, \epsilon\sqrt{x_0})[/itex]. Clearly if [itex]0 < |x - x_0| < \delta[/itex] then we have that [itex]0 < x < 2x_0[/itex] which proves that [itex]x > 0[/itex]. Since [itex]x > 0[/itex] we can apply a handy factoring technique and show that [itex]|x - x_0| = |\sqrt{x} - \sqrt{x_0}|(\sqrt{x} + \sqrt{x_0}) < \delta[/itex]. Using the fact that [itex]\sqrt{x} \geq 0[/itex] we know that [itex]\sqrt{x} + \sqrt{x_0} \geq \sqrt{x_0}[/itex] and consequently [itex]|\sqrt{x} - \sqrt{x_0}|\sqrt{x_0} < \epsilon\sqrt{x_0}[/itex] completing the proof.
     
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  3. Sep 5, 2009 #2
    Re: Continuity

    You seem to have chosen delta correctly and everything, but I think you could make it more clear by working with |f(x) - L| and end up with a chain of inequalities that shows that |f(x) - L| is less than epsilon. This makes it easier to follow the logic that you have set up.

    *EDIT* Of course, in this case I meant L = f(x_0).
     
  4. Sep 5, 2009 #3

    jgens

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    Re: Continuity

    Just to make sure I'm understanding your advice, I should show the chain of inequalities like this:

    [tex]\epsilon\sqrt{x_0} > |\sqrt{x} - \sqrt{x_0}|(\sqrt{x} + \sqrt{x_0}) \geq \sqrt{x_0}|\sqrt{x} - \sqrt{x_0}|[/tex]

    Thanks for the suggestion!
     
  5. Sep 5, 2009 #4
    Re: Continuity

    I was thinking of something more along the lines of starting with

    [tex]|\sqrt{x} - \sqrt{x_0}| = \frac{|x - x_0|}{\sqrt{x} + \sqrt{x_0}}.[/tex]

    The very first expression is |f(x) - f(x_0)|. Now if you start at your line in the original proof "Using the fact that..." and take reciprocals of the fact you demonstrated, then the right hand side of the line above is less than

    [tex]\varepsilon\sqrt{x_0}\cdot\frac{1}{\sqrt{x_0}},[/tex]

    which follows from the inequality you demonstrated and your choice of delta.
     
    Last edited: Sep 5, 2009
  6. Sep 5, 2009 #5

    jgens

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    Re: Continuity

    Ah! I understand now. Thanks for the suggestion!
     
  7. Sep 5, 2009 #6
    Re: Continuity

    Just a quick note. You don't have to prove that x>0 because you are only dealing with those x that are in that interval in the first place. Also, sqrt(x) is uniformly continuous on x>0, which means that the delta doesn't depend on the x0 chosen. This might be good practice to do as well.
     
  8. Sep 6, 2009 #7
    Re: Continuity

    It is not, because [itex]sqrt^\prime(x) \to +\infty[/itex] as [itex]x \to 0[/itex]. It is, however, uniformly continuous on every interval [itex](\varepsilon ; +\infty), \varepsilon > 0[/itex].
     
  9. Sep 6, 2009 #8
    Re: Continuity

    [itex]f(x) = \sqrt{x}[/itex] is uniformly continuous on [0, infinity). It is uniformly continuous on [0, N] for any N > 0 by continuity on compact metric space (or more easily, note that sqrt(x) is holder continuous for say, [0,1], and this is really all we need), and for any [itex]\delta > 0,[/itex] [itex]f'(x) = \frac{1}{2\sqrt{x}} \leq \frac{1}{2\sqrt{\delta}}[/itex] for x in [itex][\delta, \infty).[/itex]

    Then uniform continuity on [0, infinity) comes from uniform continuity on say [0,1] and [1, infinity). You can prove this using some basic analysis. It's even easier if you overlap the intervals say by working with [0,3] and [1, infinity).
     
  10. Sep 6, 2009 #9
    Re: Continuity

    Argh, yeah, you're right, of course. I must have been thinking about Lipschitz continuity rather than uniform continuity.
     
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