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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

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    @jak: That's absolutely correct - you need to do some experimentation to get the right "magic" values. But you don't need to put that working in the finished product (the proof), because it's irrelevant to the logic and it can usually be inferred.2012-09-23

2 Answers 2

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Suppose $L>M$. Let $\epsilon=\frac{L-M}{2}>0$. Then there are positive integers $A$ and $B$ such that $L-\epsilon for all $n>A$ and $M-\epsilon for all $n>B$. It follows that $a_n>L-\epsilon=M+\epsilon>b_n$ for all $n>A+B$, a contradiction.

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    I am reading my other book and they used a different method (kind of) and I don't understand it either. I am going to add it here and maybe you could give me some comments2012-09-23
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Use these facts. If $x_n\to L$ then for any open interval $I$ containing $L$, $x_n\in I$ for all but finitely many $n$. You can structure an argument from this fact. Give it a whirl.