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 $\forall n \in \N: \left \langle {y_n} \right \rangle \le \left \langle {x_n} \right \rangle \le \left \langle {z_n} \right \rangle$.

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 sequence and itself must therefore converge to that same limit.

Proof
Note from the corollary to Negative of Absolute Value, we have:
 * $\left|{x - l}\right| < \epsilon \iff l - \epsilon < x < l + \epsilon$

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:
 * $\forall n \in \N: \left \langle {y_n} \right \rangle \le \left \langle {x_n} \right \rangle \le \left \langle {z_n} \right \rangle$

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.