# Math Help - Sandwich principle

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

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

Nice explaination, if I did actually know what $(a_{n})$ and $(b_{n})$ were!In this case I don't know what to do, can anyone help please?

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

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