Proving t_n > x when limit of convergent sequence exceeds x

[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

 

[NateTG]

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.

 

[chocok]

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.

 

[HallsofIvy]

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

 

[chocok]

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?

 

[HallsofIvy]

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

 

[NateTG]

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