Solutions

We have to show that for every $x_{n}, y_{n}, z_{n}$,

$$\forall l \in \mathbb{R}, \forall n \in \mathbb{N}, \quad x_{n} \leq y_{n} \leq z_{n} \wedge \lim x_{n} = \lim z_{n} = l \implies \lim y_{n} = l.$$

Suppose, then, that we have a three sequences $x_{n}, y_{n}, z_{n}$ satisfying the condition

$$\forall n \in \mathbb{N}, \quad x_{n} \leq y_{n} \leq z_{n}$$

and for which

$$\lim x_{n} = \lim z_{n} = l.$$

By definition, for any $\epsilon > 0$, there exists $N_1$ for which

$$x_{n} \in V_{\epsilon }(l)$$

when $n \geq N_1$. Furthermore, there also exists $N_2$ for which the inequality holds for the sequence $z_{n}$. That is,

$$\exists N_2 \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N_2 \implies z_{n} \in V_{\epsilon }(l)$$

When $N = \max \{N_1, N_2\}$, for every $n \geq N$, all terms in $x_{n}$ and $z_{n}$ are in the $\epsilon$ neighborhood centered at $l$. However, we ought to remind ourselves that the inequality also holds for this neighborhood. That is,

$$\forall n \geq N, \quad x_{n} \leq y_{n} \leq z_{n}.$$

Therefore, $y_{n}$ also belongs to this neighborhood $\forall n \geq N$. Thus, $y_{n}$ also converges to $l$.