- #1

- 2,571

- 1

## Homework Statement

I already have the solutions, but I am not sure what the solutions are trying to say.

[PLAIN]http://img194.imageshack.us/img194/2595/unledlvc.jpg [Broken]

So in

If N is any integer greater than [itex]\frac{1}{\epsilon}[/itex], the implication n > N will hold

I don't understand this, we have

[tex]n > \frac{1}{\epsilon}[/tex]

and

**If**(and I am guessing we really want this to hold)

[tex]N > \frac{1}{\epsilon}[/tex]

How does that deduce n > N? Wouldn't it make more sense to say that [tex]N < \frac{1}{\epsilon}[/tex] and hence [tex]N < \frac{1}{\epsilon} < n[/tex]?

One final question from me is, how does N relate to epsilon? I understand that

[tex]|a_n - L | < \epsilon[/tex] is speaking how close a_n is to our limit L for a very very small epsilon and the goal of the proof-theorem is to

**find**such an epsilon, what role does n > N play with respect to [tex]|a_n - L | < \epsilon[/tex]

I have accompanied this with a picture

[PLAIN]http://img213.imageshack.us/img213/5001/unledsy.jpg [Broken]

I cna see that the epsilon is trying to "squeeze" our limit and how n > N is supporting this too. So is the intuition that they (both inequalities) are saying the same thing but they are both supporting each other?

I'll clarify if possible

Last edited by a moderator: