TITLE: Proof of 'sandwich theorem' for sequences
QUESTION [5 upvotes]: Please bear with me here and please try to read it all and spot any mistakes or errors as I'm trying to prove this result but I'm unsure of whether I have done it or not. THANK YOU.
Suppose we have the following statement $(a_n)\rightarrow \ell , (b_n)\rightarrow \ell $ and we have $$a_n\leq c_n\leq b_n$$ then $(c_n)\rightarrow \ell $.
I think I have a proof which goes as follow ;
$$a_n\leq c_n\leq b_n\leq  \ \Rightarrow 0\leq c_n-a_n\leq b_n-a_n $$, as the terms are all larger than 0, taking the absolute value will not change any of the signs of the inequalities. So we have $$0\leq |c_n-a_n|\leq|b_n-a_n|. $$
Now consider $$|b_n-a_n|=|(b_n-\ell )+(\ell - a_n)|\leq |b_n-\ell |+|a_n - \ell | \text{ (by triangle inequality)} .$$
Using the definition of a sequence tending to a value, if $(a_n)\rightarrow \ell $ then $\exists N_1\in \mathbb{N} \text{ s.t} \ \forall n>N, |a_n-\ell |<\epsilon \ ,\forall \epsilon >0 .$
We do the same for $(b_n) $ but replacing $N_1 $ with $N_2$ and using the same $\epsilon $ without loss of generality. 
So we can now say that $$|b_n-a_n|\leq |b_n-\ell |+|a_n - \ell |<2\epsilon . $$
So we have $$0\leq |c_n-a_n |\leq|b_n-a_n|<\epsilon _0, \text{ where } \epsilon _0=2\epsilon .$$
So we can conclude (using sandwich theorem for null sequences) that $(c_n-a_n)\rightarrow 0 \Rightarrow (c_n)\rightarrow (a_n)\Rightarrow (c_n)\rightarrow \ell $ since $(a_n)\rightarrow \ell . \ \ \ \square $

REPLY [4 votes]: For any $n$ such that $a_n,b_n \in (l-\epsilon,l+\epsilon),$ convexity shows $ [a_n,b_n]\subset (l-\epsilon,l+\epsilon).$ Since $c_n \in [a_n,b_n], $ we have $c_n \in (l-\epsilon,l+\epsilon)$ for the same $n.$