Limit of a sequence: $a_n \le b_n \le c_n$ Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be sequences such that $a_n ≤ b_n ≤ c_n$ for all
$n ≥ N_0$, $N_0 ∈ \mathbb{N}$. Suppose that $\{a_n\}$ and $\{c_n\}$ both converge to $l ∈ \mathbb{R}$.
Prove that $\{b_n\}$ also converges to $l$.
I am in Grade 12 and doing some self-learning with sequences. Could someone please let me know how to prove this?
I was thinking to construct $l - e < a_n \le b_n \le c_n < l + e$.
Do I have to "choose" $|a_n - l | <$ some epsilon and $|c_n - l|  <$ some epsilon?
Thanks!!
 A: Right. That's pretty much how you would go about. In detail, first fix $\varepsilon>0$. Then find $N_{1}$ and $N_{2}$ such that $|a_{n}-l|<\varepsilon$ and $|c_{k}-l|<\varepsilon$ for all $n\geq N_{1}$ and $k\geq N_{2}$. Choose $N_{\varepsilon}=\max\{N_{0},N_1,N_2\}$. Then
\begin{align*}
b_{n}-l\leq c_{n}-l\leq |c_{n}-l|<\varepsilon
\end{align*}
and
\begin{align*}
l-b_{n}\leq l-a_{n}\leq |l-a_{n}|<\varepsilon
\end{align*}
for all $n\geq N_{\varepsilon}$, then
\begin{align*}
|b_{n}-l|<\varepsilon
\end{align*}
for all $n\geq N_{\varepsilon}$. So $b_n \to l$.
A: Hint:  I find it useful to think of $\epsilon - N$ proofs as challenge-response.  If you claim $b_n \to l$ I am allowed to challenge you with an $\epsilon$ and you have to give me an $N$ such that $|b_n-l| \lt \epsilon$ for all $n \gt N$.  Now you were told by somebody that $a_n \to l$ and $c_n \to l$, so you get to challenge them with the same type of question.  It doesn't have to be the same $\epsilon$, often you want to divide it by some number, but the same one works here.  So ask them for an $M$ so that $|a_n - l| \lt \epsilon$ whenever $n \gt M$ and an $M'$ so that $|c_n - l| \lt \epsilon$ whenever $n \gt M'$  Now you can come back to me with (what?)
