# A simple question about convergence

1. Nov 3, 2012

### bedi

1. The problem statement, all variables and given/known data

Let x_n be a convergent sequence with a ≤ x_n for every n, where a is any number. Prove that a ≤ lim x_n when n→∞.

2. Relevant equations

Definition of limit. The usual ε, N stuff.

3. The attempt at a solution

Let lim x_n = x and choose ε=x_n-a. Hence we have |x_n - x| < x_n - a which shows that -x<-a and thus x>a.

Is this valid?

2. Nov 3, 2012

### micromass

Staff Emeritus
You do realize that $\varepsilon$ depends on n?? For a different n, you'll get a different $\epsilon$. Do you really want that??
Also, why is $\varepsilon>0$? Specifically, why is it nonzero?

3. Nov 3, 2012

### tiny-tim

hi bedi!

(try using the X2 button just above the Reply box )
but ε has to be independent of n

(ooh, micromass beat me to it! )

4. Nov 3, 2012

### bedi

Yes, you are right. However I still can't see the solution :(

5. Nov 3, 2012

### tiny-tim

try assuming the contrary

6. Nov 3, 2012

### bedi

Alright, so this will imply that x_n converges to both a and x which is a contradiction. Am I right?

7. Nov 3, 2012

### micromass

Staff Emeritus
Take $x_n=2$ for all n and take $a=0$. Then certainly $x_n$ does not converge to a. So no, you're not right.

8. Nov 3, 2012

### bedi

But I assumed that a>x. So -a<-x and x_n-a<x_n-x<ε. Hence x_n-a<ε ?

9. Nov 3, 2012

### micromass

Staff Emeritus
Although the idea is there, the proof is still not very nice. For example, what is $\varepsilon$?? You got to say things like this, not just introduce them without telling anybody what it is.

10. Nov 3, 2012

### tiny-tim

bedi, you know this is supposed to be a delta,epsilon proof …

so you must define delta, and you must define epsilon​