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Prove continuity of sqrt(x) on (0,infinity)

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    This is a problem from my Analysis exam review sheet.

    Let L(x) = [itex]\sqrt{x}[/itex]. Prove L is continuous on E = (0,[itex]\infty[/itex])

    3. The attempt at a solution

    The way we've been doing these proofs all semester is to let [itex]\epsilon > 0[/itex] be given, then assume [itex]\left| x -x_{0} \right| < \delta[/itex] (which we figure out later) and [itex] x_{0} \in E [/itex]

    Then look at.
    [itex]\left| L(x) - L(x_{0}) \right| = \left| \sqrt{x} - \sqrt{x_{0}}\right|[/itex]
    and try to get a [itex]\left( x -x_{0} \right)[/itex] term which we can control, so that we can figure out a [itex]\delta[/itex] which will allow us to get the entire thing less than [itex]\epsilon[/itex]. So essentially I understand how to do the proof. My algebra skills are just really rusty and I can't figure out the long division or whatever I need to do to get a [itex]\left( x -x_{0} \right)[/itex] term out of that.
     
  2. jcsd
  3. Apr 8, 2012 #2

    Office_Shredder

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    One trick is to abuse inverse functions whose continuity you can deal with

    If you let [itex] y^2=x[/itex] and [itex] y_0^2 = x_0[/itex] then
    [tex] |x-x_0| = |y^2-y_0^2| = |y_0-y||y_0+y|<\delta[/tex]

    Now you want to show that [itex] |y-y_0|<\epsilon[/itex] if delta is small enough
     
  4. Apr 8, 2012 #3
    Thanks for the quick response, I think I see where to go from here, but just to make sure.
    Let [itex]\delta = \epsilon^{2}[/itex]
    We know [itex]\left| y + y_{0} \right| > \left| y - y_{0} \right|[/itex] since [itex]y[/itex] and [itex]y_{0}[/itex] are positive.
    and since [itex]\left| y + y_{0} \right| \left| y - y_{0} \right| < \delta = \epsilon^{2}[/itex]
    We can see that [itex]\left| y - y_{0} \right| < \sqrt{\delta} = \epsilon[/itex]
     
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