Advanced Calc. Continuity problem

  • #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.


Thanks!
 

Answers and Replies

  • #2
35,231
7,050
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]
 
  • #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.
 
  • #4
35,231
7,050
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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