Exercise 2.3.3

Show that if $x_n\leq y_n\leq z_n$ for all $n\in\mathbb N$ and if $\lim x_n = \lim z_n=\ell$, then $\lim y_n=\ell$ as well.

Comment: This is a pretty well known theorem, even in calculus.

Note that $|y_{n} - \ell | \leq |y_{n} - x_{n} | + | x_{n} - \ell| $ by the triangle inequality. This statement can be re-written as it is below since the distance between $x_{n}$ and $y_{n}$ can be no greater than the distance between $z_{n}$ and $x_{n}$.

$$|y_{n} - \ell | \leq |z_{n} - x_{n} | + |x_{n} -\ell |.$$

Now, from theorem 2.3.3 we know the limit of $(z_{n}-x_{n})$ exists (and is zero). Thus there exists some $N \in \mathbb{N}$ such that for all $ n>N$,

$$|z_{n} - x_{n} | \text{ and } | x_{n} - \ell | \text{ are both less than }\frac{\epsilon}{2}.$$

Therefore $\forall n>N$,

$$| y_{n} - \ell | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Which concludes the proof.