# Proving Convergence of Sequence Greater Than X | Help on Homework

• chocok

## Homework Statement

The question asks to prove that if (t_n) is a convergent sequence and suppose that its limit is great than a number x. The prove that it exists a number N such that n>N => t_n>x

## The Attempt at a Solution

I tried to say that as (t_n) converge, (t_n - x) converges too.
And let lim(t_n - x) = k, for k in R (real).
Then for some e>0, there's a number N s.t. n>N => |(t_n-x)-k|< e
Take e=0, then |(t_n-x)-k|=0 => t_n-k = x,
since lim(t_n)>x then k must be positive => lim(t_n)>x.

can anyone tell me if i am going the wrong way??
I am afraid that my concept somewhere is wrong..
Thanks

Generally speaking, when you write epsilon proofs, it's easier to follow the argument of the proof if you start by specifying what you want $\epsilon$ to be. (Even you worked your way back to it.)

You can't use $\epsilon = 0$ since that contradicts $\epsilon > 0$.

You need to be much more careful in the way you write things in general.

Then can i rephrase it as follows:
As (t_n) converge, (t_n - x) converges too.
And let lim(t_n - x) = k, for k in R, k>0.
Then for some e>0, there's a number N s.t. n>N => 0<|(t_n-x)-k|< e
|(t_n-x)-k|>0 implies t_n - k>x, since k is positive, it follows that t_n>x.

does this sound right? (it seems like i am ignoring the use of e?)
if not, can you give me some idea on how to make the proof works?
Thanks.

If L is the limit and L> x, then L-x> 0. What if you make $\epsilon= (L-x)/2$?

Thanks for replying!
When I have 2 situations, how can i show that when L-tn>0 (L>tn) and L>x actually implies tn>x?? Can you give me some idea?

All I can do is repeat: if the limit L> x, then L-x> 0. What happens if you take $\epsilon= (L-x)/2$?

Well, it might also be hand to review the definition of convergent sequence...