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Suppose $\lim\limits_{n \rightarrow \infty} a_n =\lim\limits_{n \rightarrow \infty} b_n = c$ and $a_n \le c_n \le b_n$ for all $n$. Prove that $\lim\limits_{n \rightarrow \infty} c_n = c$.

How would I do this?

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    $\begingroup$ Well, you could read any decent Calculus book or one of the first thousands of sites you'll find when googling "Sandwich or squeeze theorem"... $\endgroup$ – DonAntonio Feb 26 '14 at 19:23
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    $\begingroup$ @DonAntonio Are you suggesting it's unreasonable to ask a math question on a math question-and-answer site? $\endgroup$ – preferred_anon Feb 26 '14 at 19:43
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    $\begingroup$ Not at all, @DanielLittlewood: I am suggesting that some questions at a level of university/college can be first checked/read by other means...just like this precise question, say. $\endgroup$ – DonAntonio Feb 26 '14 at 19:49
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Concise answer

By the hypothesis we have $$|c_n-c|\le \max(|a_n-c|,|b_n-c|)$$ now let $\epsilon>0$ so by the limit's definition of $(a_n)$ and $(b_n)$ there's $N=\max(N_a,N_b)$ and if $n\ge N$ then $$|c_n-c|\le \max(|a_n-c|,|b_n-c|)\le \epsilon.$$

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Hints:

For all $\;\epsilon > 0\;\exists\,N\in\Bbb N\;s.t.\;\;n>N\implies\begin{cases}c-\epsilon<a_n<c+\epsilon\\{}\\c-\epsilon<b_n<c+\epsilon\end{cases}\;$

and from here that for $\;n>N\;$:

$$c-\epsilon<a_n<c_n<b_n<c+\epsilon\;\implies\ldots$$

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Given $\varepsilon>0$. There is $N\in \mathbb{N}$ such that $d(a_n,c)\le \varepsilon$ and $d(b_n,c)\le \varepsilon$ at the same time for all $n\ge N$ (why?). Then $c-\varepsilon\le a_n\le c_n\le b_n\le c+ \varepsilon$, so $|c_n-c|\le \varepsilon$. Since $\varepsilon$ was arbitrary it holds for each $\varepsilon>0$. Thus $\lim_{n\rightarrow \infty}c_n = c$ as desired.

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