Prove Sequence: x>0,<n>→x,N s.t. n>N implies xn>0

[silvermane]

Homework Statement

Prove that if x>0 and <_n> is a sequence with lim_{\rightarrow \infty} = x, then there is a real number N such that whenever n>N we have x_n>0.

The Attempt at a Solution

I was trying to prove this by contradiction, but got stuck. Any hints or tips would be greatly appreciated! No answers please :)

 

[micromass]

Do you mean this?
Prove that if x>0 and (x_n)_n is a sequence with \lim_{n\rightarrow +\infty}{x_n}=x, then there is a real number N such that whenever n>N we have xn>0.



Try to work with the definition of limits, i.e.

\lim_{n\rightarrow +\infty}{x_n}=x ~\Leftrightarrow~\forall\epsilon:\exists N:\forall n&gt;N: |x_n-x|&lt;\epsilon

Can you find an epsilon small enough such that the above definition garantuees xn>0 for n>N?

If you don't see it immediately, try finding this epsilon for some examples...

 

[silvermane]

Oh dear. I don't know what happened after I previewed it, but I meant this:
Prove that if x>0 and (x_n) is a sequence with \lim_{n\rightarrow \infty}{x_n}=x, then there is a real number N such that whenever n>N we have (x_n)>0.

That atrocious type-o changed the whole meaning of my problem; I'm terribly sorry :(

 

[╔(σ_σ)╝]

Hey, it been a while :-).

x-e < x_n < x+e

What happens if he pick e = x/2 ?

 

[silvermane]

lol I think I got it!

We have that x>0, and for every E>0, there exists an N such that for every n>N, we have
|x_n-x| < E,
and x-E <x_n<x+E

Since this is true for any E>0, then it must be true for E =1/2x.
For n>N, we have that x_n > x-1/2x = 1/2x >0.
Thus, x_n>0, and we are done.

haha, I think just wasn't thinking clearly when I initially looked at the question. It seems soooo easy after doing this :)

Thanks for your help