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,
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
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
Now, let \(N \geq \max{(N_1, N_2)}\), then for all \(n \geq N\). Re-write the inequalities as follows
But we know that \(x_n \leq y_n \leq z_n\), so
Combining both inequalities, we get
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. \(\blacksquare\)