lim sup iff conditions - please help explain Deciphering the definition of the upper limit, we see that $\limsup x_n=L$ is and only if the following two conditions are fulfilled:


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*(a) $\forall\epsilon>0\;\;\exists N\in\mathbb{N}$ such that $\forall n\geq N\ x_n<L+\epsilon$;

*(b) $\forall\epsilon>0\;\;\forall N\in\mathbb{N} \exists n > N$ such that $x_n>L-\epsilon$.


Hi all, could you please explain this for me? What does condition (a) mean, and what does condition (b( mean?
thanks!
 A: Here's another way to write these conditions that may help:
(a) means that for every $\epsilon > 0$, every $x_n$ except for at most finitely many satisfy $x_n < L + \epsilon$.  That is, there are no (infinite) subsequences that stay above $L + \epsilon$
(b) means that for every $\epsilon > 0$, there are infinitely many $x_n$ that satisfy $x_n > L - \epsilon$.  That is, there exists a subsequence that stays above $L - \epsilon$.
A: Very briefly condition (a) says that the limsup is not larger than $L$ and condition (b) says it's not smaller than $L$.  
In a bit more detail, (a) says that if you take a number $M$ larger than $L$, then after maybe a finite number of exceptions the sequence is always less than $M$.  
(b) says that if you take a number $M$ less than $L$, there are infinitely many members of the sequence that are greater than $M$.  
Again, informally, I like to think of them as (a) kind of pushes the sequence down to $L$ as time goes on, but condition (b) always draws the sequence back up towards $L$, no matter how often they stray away. Note that (b) is kind of weaker.  It lets the sequence go far below $L$ infinitely often, as long as it always comes back, whereas (a) says the sequence must stay not too high above $L$ eventually.   
A: First fix an $\varepsilon>0$.
(a) means $L+\varepsilon$ is an upper bound for all $x$ of sufficiently high order.
(b) means $L-\varepsilon$ is not an upper bound for all such $x$.
It indeed implies $L=\limsup x_n$ since


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*$\sup\limits_{n\ge N}x_n\le L+\frac1N\quad $ by (a), hence, passing to the limit, $\,\,\limsup x_n \le L$.

*$\sup\limits_{n\ge N}x_n>L-\frac1N\quad $ by (b), hence, passing to the limit, $\,\,\limsup x_n \ge L$.

