Answer
See proof below.
Work Step by Step
Let $\{a_{n}\}, \{b_{n}\}$ and $\{c_{n}\}$ be such that there exists an N, for which
$(n\gt N)\Rightarrow a_{n}\leq b_{n}\leq c_{n}$.
Also, let $\displaystyle \lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}c_{n}=L.$
Select a positive number $\epsilon$.
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=L$ means that there is an $N_{1}$ such that
$(n\gt N_{1})\Rightarrow|L-a_{n}|\lt\epsilon$
That is, $ L-\epsilon\lt a_{n}\lt L+\epsilon$
$\displaystyle \lim_{n\rightarrow\infty}c_{n}=L$ means that there is an $N_{2}$ such that
$(n\gt N_{2})\Rightarrow|L-c_{n}|\lt\epsilon$
That is, $ L-\epsilon\lt c_{n}\lt L+\epsilon$
So, if we take $M=\displaystyle \max\{N_{1},N_{2}\}$, we can say
$L-\epsilon\lt a_{n}$ and $c_{n}\lt L+\epsilon\qquad(*).$
We have the hypothesis $a_{n}\leq b_{n}\leq c_{n}$
Applying (*), we can say that
$ L-\epsilon\lt b_{n}\lt L+\epsilon$
that is, for any chosen positive number $\epsilon$, there exists a M such that
$n\gt M\Rightarrow|L-b_{n}|\lt\epsilon.$
By definition, $\displaystyle \lim_{n\rightarrow\infty}b_{n}=L.$
