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Advanced Calc. Continuity problem

  1. Jun 24, 2010 #1
    So I've been trying to figure this out. The question is:

    If the limit x->infinity of Xn=Xo

    Show that, by definition, limit x->infinity sqrt(Xn)=sqrt(Xo)

    I'm pretty sure I need to use the epsilon definition.
    I worked on it with someone else and we think that what we have to show is the this:

    Want to show:
    For all e>0 there is an N>0 s.t. for all n>N, |sqrt(Xn) - sqrt(Xo)|<e

    I just don't know how to show this.

  2. jcsd
  3. Jun 24, 2010 #2


    Staff: Mentor

    Does this help?
    [tex]\sqrt{x_n} - \sqrt{x_0} = \sqrt{x_n} - \sqrt{x_0} \frac{\sqrt{x_n} + \sqrt{x_0}}{\sqrt{x_n} + \sqrt{x_0}} = \frac{x_n - x_0}{\sqrt{x_n} + \sqrt{x_0}}[/tex]
  4. Jun 24, 2010 #3
    ^ If it does, I can't see it. I feel like I need to find an N in terms of e to show that this si continuous or something.
  5. Jun 24, 2010 #4


    Staff: Mentor

    You're given that
    [tex]\lim_{n \to \infty} x_n = x_0[/tex]

    What does that mean in terms of the epsilon-N definition of a limit?
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