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Thread: Sandwich principle

  1. #1
    Junior Member simplysparklers's Avatar
    May 2008
    Galway, Ireland

    Sandwich principle

    Hi guys!
    Ok, last problem today! I've already read any other posts on the sandwich principle, but I've never seen it before today, and my chapter doesn't explain it, just gives a question on it! If anyone could show me how this is done I would really appreciate it!

    Let $\displaystyle (a_{n}),(x_{n})$ and $\displaystyle (b_{n})$ be sequences with limits $\displaystyle \alpha,\xi,\beta$ respectively, and suppose that, for all $\displaystyle n\geq{1}$,
    $\displaystyle a_{n}\leq{x_{n}}\leq{b_{n}}$

    Show that $\displaystyle \alpha\leq\xi\leq\beta$ [This is sometimes called the sandwich principle. It can be useful if $\displaystyle (a_{n})$ and $\displaystyle (b_{n})$ are "known" sequences and $\displaystyle (x_{n})$ is unknown. It is especially useful when $\displaystyle \alpha=\beta$, for in this case we conclude that $\displaystyle \xi=\alpha=\beta$.]

    Nice explaination, if I did actually know what $\displaystyle (a_{n})$ and $\displaystyle (b_{n})$ were!In this case I don't know what to do, can anyone help please?
    Last edited by simplysparklers; May 12th 2008 at 10:17 AM. Reason: still don't know latex
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  2. #2
    MHF Contributor

    Aug 2006
    Letís do it for one half, you can expand.
    Suppose $\displaystyle \xi < \alpha $ then let $\displaystyle \varepsilon = \frac{{\alpha - \xi }}{2} > 0$.
    Use the sequence convergence definition twice to get $\displaystyle \left( {\exists N} \right)\left[ {k \geqslant N \Rightarrow \quad \left| {a_k - \alpha } \right| < \varepsilon \,\& \,\left| {x_k - \xi } \right| < \varepsilon } \right]$.
    Removing the absolute values and using the given we get: $\displaystyle \frac{{\xi + \alpha }}
    {2} < a_N \leqslant x_N < \frac{{\xi + \alpha }}{2}$.

    You can see the contradiction.
    Now do it again for the other half.
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  3. #3
    MHF Contributor arbolis's Avatar
    Apr 2008
    As you said, suppose that $\displaystyle {a_n}\leq{b_n}$ for all $\displaystyle n\geq1$.
    I give you an example. Usually the exercise will make you calcul the limit of the sequences $\displaystyle {a_n}$ and $\displaystyle b_n$. Suppose now that you found that $\displaystyle a_n$ tends to 10 (from below) when $\displaystyle n$ tends to positive infinite and $\displaystyle b_n$ tends to 10 (from bottom) when $\displaystyle n$ tends to positive infinite. If you can see that $\displaystyle a_n\leq x_n$ and $\displaystyle x_n\leq b_n$ for a given sequence $\displaystyle x_n$, then by the sandwich's theorem (or principle), $\displaystyle x_n$ tends to 10, which is very intuitive!
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