Can you 'take $\limsup$' in both sides of an inequality? I'm reading the proof for $(1)$ of this paper, and I can't get the hang of how the author concludes the "hence we have that $L-\epsilon<\limsup b_n<L+\epsilon$", could anybody explain this?
I think he takes the $\limsup$ of each part, but even if that's the case, I don't get how to get that answer.
 A: I assume you've understood the proof up to 
$${C\over n}+{(n-N)(L-\epsilon)\over n}<b_n<{C\over n}+{(n-N)(L+\epsilon)\over n}$$
We have $\limsup x_n\le \limsup b_n$ whenever $x_n\le b_n$. This is in Baby Rudin 3.19 so I won't write a proof out here.  We also know that if $x_n$ converges to a limit, then that limit equals lim sup and lim inf.  
Now, take the limits on the left and right above.  The limit of $C/n$ is of course zero, $C$ being independent of $n$.  The limit of $(n-N)/n$ is of course $1$, $N$ being again independent of $n$.  $L-\epsilon$ and $L+\epsilon$ are independent of $n$. Thus, we get that the limit on the left is $L-\epsilon$ and the one on the right $L+\epsilon$.
Edited to add proof of Baby Rudin 3.19.  We are working in the extended reals, with $\pm\infty$ included, per 3.16.  We want show that if $s_n\le t_n$ for sufficiently large $n$, then $\limsup s_n\le \limsup t_n$ and the same for lim inf.  Let's do lim sup.  
Lim sup is by definition the sup of the set of subsequential limits. If $\limsup s_n=-\infty$ then the inequality we want is not constraining.  Otherwise, let $s_{n_i}$ converge to $\limsup s_n$ (3.17(a)). 
For large $i$, the sequence $t_{n_i}$ is bounded below by $\limsup s_n-\epsilon$ for any positive $\epsilon$.  Because of the lower bound, either a subsequence of $t_{n_i}$ converges to a limit $L$ or one goes to $+\infty$.  In the first case, the limit $L$ must be $\ge$ the limit of $s_{n_i}$ because of $s_n\le t_n$, and so $\limsup s_n\le L\le \limsup t_n$ as desired.  In the second case, $\limsup t_n=+\infty$ and so the desired inequality is not constraining.
