Squeeze Theorem/Sequences/Real Numbers

Theorem
Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.

Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:
 * $\ds \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$

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:
 * $\ds \lim_{n \mathop \to \infty} x_n = l$

Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ 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:
 * $\size {x - l} < \epsilon \iff l - \epsilon < x < l + \epsilon$

Let $\epsilon > 0$.

We need to prove that:
 * $\exists N: \forall n > N: \size {x_n - l} < \epsilon$

As $\ds \lim_{n \mathop \to \infty} y_n = l$ we know that:
 * $\exists N_1: \forall n > N_1: \size {y_n - l} < \epsilon$

As $\ds \lim_{n \mathop \to \infty} z_n = l$ we know that:
 * $\exists N_2: \forall n > N_2: \size {z_n - l} < \epsilon$

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

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:
 * $\forall n > N: \size {x_n - l} < \epsilon$

Hence the result.