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Did I underestimate the limit proof?

Let $(a)_{n\in \Bbb N}$ and $(b)_{n\in \Bbb N}$ be sequences of real numbers such that $a_n$ $\le$ $b_n$ for all $n\in \Bbb N$. Prove that if $a_n \to a$ and $b_n \to b$; then a $\le$ b.

I have this question as homework. I have some sort of solution in mind but the professor wants us to hand in a formal proof. What I thought so far is to assume contrary, that a $\gt$ b, and to examine $n$ for large numbers to conclude $b_n \gt a_n$ and to have contradiction. But my problem is I don't even know what formal proof is. Could you please tell me what it is using this question.

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marked as duplicate by Qiaochu Yuan Oct 15 '12 at 5:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Consider the sequence $b_n - a_n$. What does this converge to? – Ink Oct 14 '12 at 18:24
b - a. Where does this lead me to? – Mert Toka Oct 14 '12 at 18:51
up vote 1 down vote accepted

Assume by contradiction that $a >b$.

Let $\epsilon >0$, we will pick it later.

Then there exists an $N$ so that for all $n> N$ we have

$$|a_n -a| < \epsilon \Rightarrow a_n > a- \epsilon$$

$$|b_n -b| < \epsilon \Rightarrow b_n < b+\epsilon $$

Now, if you pick some $\epsilon$ so that $b+ \epsilon \leq a-\epsilon$ you get

$$ b_n < b+ \epsilon \leq a-\epsilon < a_n \,,$$


You figure now the right $\epsilon$, and start your argument by "Let $\epsilon= ...$.

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The correct $\epsilon$ is $a-b$ to get: $b_n < a \leq b < a_n$ ? – CodeKingPlusPlus Oct 14 '12 at 19:23
I suppose $a - b$ can be used for the conradiction. Is it what you have in mind @n-s ? – Mert Toka Oct 14 '12 at 20:04
@MertToka $b+\epsilon \leq a- \epsilon$ means $2\epsilon \leq a-b$... So nope $\epsilon = a-b$ won't work, but figuring one so that $2 \epsilon \leq a-b$ should be really easy.... – N. S. Oct 14 '12 at 20:44

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