Squeeze Theorem/Sequences/Real Numbers

Theorem
Let $\left \langle {x_n} \right \rangle, \left \langle {y_n} \right \rangle$ and $\left \langle {z_n} \right \rangle$ be sequences in $\R$.

Let $\left \langle {y_n} \right \rangle$ and $\left \langle {z_n} \right \rangle$ be convergent to the following limit:
 * $\displaystyle \lim_{n \to \infty} y_n = l, \lim_{n \to \infty} z_n = l$

Suppose that $\forall n \in \N: y_n \le x_n \le z_n$.

Then $x_n \to l$ as $n \to \infty$, that is, $\displaystyle \lim_{n \to \infty} x_n = l$.

That is, if $\left \langle {x_n} \right \rangle$ is always between two other sequences that both converge to the same limit, $\left \langle {x_n} \right \rangle$ is said to be sandwiched or squeezed between those two sequences and itself must therefore converge to that same limit.

Proof
From Negative of Absolute Value: Corollary 1:
 * $\left|{x - l}\right| < \epsilon \iff l - \epsilon < x < l + \epsilon$

Let $\epsilon > 0$.

We need to prove that $\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$.

As $\displaystyle \lim_{n \to \infty} y_n = l$ we know that:
 * $\exists N_1: \forall n > N_1: \left|{y_n - l}\right| < \epsilon$

As $\displaystyle \lim_{n \to \infty} z_n = l$ we know that:
 * $\exists N_2: \forall n > N_2: \left|{z_n - l}\right| < \epsilon$

Let $N = \max \left\{{N_1, N_2}\right\}$.

Then if $n > N$, it follows that $n > N_1$ and $n > N_2$.

So:
 * $\forall n > N: l - \epsilon < y_n < l + \epsilon$
 * $\forall n > N: l - \epsilon < z_n < l + \epsilon$

But:
 * $\forall n \in \N: y_n \le x_n \le z_n$

So:
 * $\forall n > N: l - \epsilon < y_n \le x_n \le z_n < l + \epsilon$

and so:
 * $\forall n > N: l - \epsilon < x_n < l + \epsilon$

So:
 * $\forall n > N: \left|{x_n - l}\right| < \epsilon$

Hence the result.

Also known as
This result is also known, in the UK in particular, as the sandwich theorem.