Squeeze / Sandwich Theorem
Show that if \(x_n \leq y_n \leq z_n\) for all \(n \in \mathbb{N}\), and if \(\lim x_n = \lim z_n = l\), then \(\lim y_n = l\).

Proof

To show that \(\lim(y_n) = l\), let \(\epsilon > 0\) be arbitrary. We want to prove that there exists some \(n \geq N\) such that,

$$ \begin{align*} \lvert y_n - l \rvert < \epsilon. \end{align*} $$

To do so, we are given that \(\lim x_n = l\). This means that there exists some \(N_1 \in \mathbb{N}\) such that when \(n \geq N_1\), then

$$ \begin{align*} \lvert x_n - l \rvert < \epsilon. \end{align*} $$

Similarly, we are given that \(\lim z_n = l\). This means that there exists some \(N_2 \in \mathbb{N}\) such that when \(n \geq N_2\), then

$$ \begin{align*} \lvert z_n - l \rvert < \epsilon. \end{align*} $$

Now, let \(N \geq \max{(N_1, N_2)}\), then for all \(n \geq N\). Re-write the inequalities as follows

$$ \begin{align*} l - \epsilon &< x_n < l + \epsilon. \\ l - \epsilon &< z_n < l + \epsilon. \end{align*} $$

But we know that \(x_n \leq y_n \leq z_n\), so

$$ \begin{align*} l - \epsilon &< x_n \leq y_n \\ y_n \leq z_n &< l + \epsilon. \end{align*} $$

Combining both inequalities, we get

$$ \begin{align*} l - \epsilon < x_n \leq y_n \leq z_n &< l + \epsilon \\ l - \epsilon < y_n &< l + \epsilon \\ - \epsilon < y_n - l &< \epsilon \\ |y_n - l| &< \epsilon. \end{align*} $$

From this we can conclude that for any \(\epsilon > 0\) there exists an \(N \geq n\) such that \(|y_n - l| < \epsilon\). Therefore, \(\lim y_n = l\) as we wanted to show. \(\blacksquare\)

References