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Given a sequence $\{x_{n}\}_{n=1}^{\infty}$, then $\lim_{n\to\infty}\{x_n\}=-\infty$ if, for every K, there is an N such that, for every $n \geq N, x_n < K$.

Do we still have the same definition if the sequence indexed by $\mathbb Z$, i.e, $\{x_{n}\}_{n=-\infty}^{\infty}$?

More precisely, I'm looking for a definition of $\lim_{n\to-\infty}\{x_n\}=-\infty$, and $\lim_{n\to+\infty}\{x_n\}=+\infty$.

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2 Answers

Your first definition has an inequality backwards: you want $x_n<K$ for every $n\ge N$.

That said, the definitions that you want are indeed pretty much the same as for ordinary sequences:

$\lim\limits_{n\to -\infty}x_n=-\infty$ iff for each $K\in\mathbb{R}$ there is an $N_K\in\mathbb{Z}$ such that $x_n<K$ whenever $n\le N_K$,

and

$\lim\limits_{n\to \infty}x_n=\infty$ iff for each $K\in\mathbb{R}$ there is an $N_K\in\mathbb{Z}$ such that $x_n>K$ whenever $n\ge N_K$.

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Lol, blinded by the standards. I didn't see the typo in OP's question. –  Patrick Da Silva Feb 28 '12 at 1:02
    
@Brian, Thank you, I fixed that inequality. –  MathSt Feb 28 '12 at 2:00
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It is not usually done in a first analysis course because then we don't start worrying about such sequences, but sure. (Just so you know the standard for the term "sequence" is that they're indexed by $\mathbb N$, but that's possible for sure to index them by $\mathbb Z$, no problem with that.)

You could define analoguously to the $n \to \infty$ case that

$$ \lim_{n \to -\infty} x_n = + \infty \quad \Longleftrightarrow \quad \forall M > 0, \exists N \text{ s.t.} \quad \forall n < N, \quad x_n > M. $$

Note though that the $N$ existing here can be chosen negative.

For the other cases ($\pm \infty$ under the limit and $\pm \infty$ as a result), just "switch things" accordingly. There's no problem in doing that, but it's not commonly used though.

Hope that helps,

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