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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


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 $L-\epsilon<a_n\leq b_n <\epsilon+M \implies L - \epsilon < \epsilon +M \implies L < M + 2\epsilon$

OKay I am stuck now, but I feel I am getting close

EDIT: alternate proof from text


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 $|b_n - a_n - (M - L)| <L -M$ 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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You haven’t proved anything: you’ve simply asserted the conclusion. You will need to use the definition of limit somewhere; I suggest that you assume that $L>M$ and use the definition of limit to derive a contradiction. – Brian M. Scott Sep 23 '12 at 0:32
@BrianM.Scott, yeah I felt that the proof seemed circular. Is contradiction the only way? – Hawk Sep 23 '12 at 0:46
It may not be the only way, but it’s by far the easiest. – Brian M. Scott Sep 23 '12 at 0:48
OKay I will edit what I did. I thnk I got it – Hawk Sep 23 '12 at 0:48
@jak: There's a gap between $M$ and $L$, so what should you choose $\epsilon$ to be in order to get a contradiction? Also, the flow of your argument could be improved slightly. – wj32 Sep 23 '12 at 0:56
up vote 5 down vote accepted

Suppose $L>M$. Let $\epsilon=\frac{L-M}{2}>0$. Then there are positive integers $A$ and $B$ such that $$L-\epsilon<a_n<L+\epsilon$$ for all $n>A$ and $$M-\epsilon<b_n<M+\epsilon$$ 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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How the heck did you come up with $\epsilon=\frac{L-M}{2}$? – Hawk Sep 23 '12 at 1:16
@jak: – wj32 Sep 23 '12 at 1:17
@jak It's a classic. – Pedro Tamaroff Sep 23 '12 at 1:17
@WillHunting, in your picture, does that mean I should omit the assumption that $a_n \leq b_n$? – Hawk Sep 23 '12 at 1:31
@jak: You are given certain facts, and they tell you that you can pick any $\epsilon > 0$ you want and you will receive a $N$ such that $|a_n - L| < \epsilon$ whenever $n>N$. The key thing here is that the choice of $\epsilon$ is up to you; you must use this to your advantage. It happens that choosing $\epsilon = (L-M)/2$ contradicts $M<L$, so $M<L$ cannot be true. (Have you taken a proof writing course? If not, you should!) – wj32 Sep 23 '12 at 4:19

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.

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