1. Jun 24, 2010

### BustedBreaks

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!

2. Jun 24, 2010

### Staff: Mentor

Does this help?
$$\sqrt{x_n} - \sqrt{x_0} = \sqrt{x_n} - \sqrt{x_0} \frac{\sqrt{x_n} + \sqrt{x_0}}{\sqrt{x_n} + \sqrt{x_0}} = \frac{x_n - x_0}{\sqrt{x_n} + \sqrt{x_0}}$$

3. Jun 24, 2010

### BustedBreaks

^ 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. Jun 24, 2010

### Staff: Mentor

You're given that
$$\lim_{n \to \infty} x_n = x_0$$

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