# Showing Upper Limt= Lower Limit

• Nov 10th 2009, 06:06 AM
concern
Showing Upper Limt= Lower Limit
Hey, can anyone help in this theorem.

If {Xn } is abounded sequence,then its limit exist as (n goes to infinity) if and only iff

Lim supXn = Lim infXn
n-->infinity n-->infinity

• Nov 10th 2009, 07:54 AM
concern
i'm thinking we may use squeeze theorem here. but again, not getting it.
• Nov 10th 2009, 08:00 AM
Drexel28
Quote:

Originally Posted by concern
Hey, can anyone help in this theorem.

If {Xn } is abounded sequence,then its limit exist as (n goes to infinity) if and only iff

Lim supXn = Lim infXn
n-->infinity n-->infinity

Use that $\liminf_{n\to\infty}x_n\le\lim_{n\to\infty}x_n\le\ limsup_{n\to\infty}x_n$ to prove the one part. Next consider instead of $limsup,\liminf$ think about it as $\sup\text{ }X,\inf\text{ }X$ where $X$ is the set of all subsequential limits of $\left\{x_n\right\}$. And use the fact that if every subsequential limit converges to the same value that the limit exists and equals that value.
• Nov 10th 2009, 08:08 AM
concern
you are great Drexel28. Your help is truly awesome. In my 1st post also you were the only one who could come up with genuine solution. Hats off to you.

Quote:

Originally Posted by Drexel28
Use that $\liminf_{n\to\infty}x_n\le\lim_{n\to\infty}x_n\le\ limsup_{n\to\infty}x_n$ to prove the one part. Next consider instead of $limsup,\liminf$ think about it as $\sup\text{ }X,\inf\text{ }X$ where $X$ is the set of all subsequential limits of $\left\{x_n\right\}$. And use the fact that if every subsequential limit converges to the same value that the limit exists and equals that value.

• Nov 10th 2009, 08:10 AM
Drexel28
Quote:

Originally Posted by concern
you are great Drexel28. Your help is truly awesome. In my 1st post also you were the only one who could come up with genuine solution. Hats off to you.

There are a lot of great members on here. I just think that I have conicidentally answered a large amount of yours.