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

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- Nov 10th 2009, 05:06 AMconcernShowing 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, 06:54 AMconcern
i'm thinking we may use squeeze theorem here. but again, not getting it.

- Nov 10th 2009, 07:00 AMDrexel28
Use that $\displaystyle \liminf_{n\to\infty}x_n\le\lim_{n\to\infty}x_n\le\ limsup_{n\to\infty}x_n$ to prove the one part. Next consider instead of $\displaystyle limsup,\liminf$ think about it as $\displaystyle \sup\text{ }X,\inf\text{ }X$ where $\displaystyle X$ is the set of all subsequential limits of $\displaystyle \left\{x_n\right\}$. 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, 07:08 AMconcern
- Nov 10th 2009, 07:10 AMDrexel28