This is the problem:
Prove that if $a_n \le b_n$ for $n \ge 1, L = \lim_{n \to \infty} a_n$ and $M = \lim_{n \to \infty} b_n$, then $L \le M$
EDIT: Progress
Proof
Assume $L >M$ and $a_n \leq b_n$, then
(1) $|a_n -L| < \epsilon$ when n > $N_1$
(2) $|b_n-M| < \epsilon$ when n > $N_2$
Expanding (1) and (2) gives
$L - \epsilon < a_n < \epsilon + L$ and $M - \epsilon < b_n < \epsilon + M$
Since $a_n \leq b_n$, we have $L-\epsilon
OKay I am stuck now, but I feel I am getting close
EDIT: alternate proof from text
Proof
Let $\lim_{n\to\infty}b_n -a_n=M -L$. Therefore for any $\epsilon > 0$, $\exists N:$
$|b_n - a_n - (M - L)| <\epsilon$ whenever $n > N$
Take $\epsilon = L - M$ and we get $|b_n - a_n - (M - L)|
whenever $n > N$ and since $a \leq |a|$, we have $b_n - a_n - (M - L) < L -M \iff a_n >b_n$, but this contradicts the assumption and therefore $L > M$ must be false