Suppose $a\le x_n\le b$ for all $n$ and suppose further that $x_n\rightarrow L$. Prove: $L \in [a, b]$.
The book provides a hint: if $L\lt a$ or if $L \gt b$, obtain a contradiction.
My reasoning is if $x_n\rightarrow L$, then $\left|x_n - L \right|\lt\epsilon$. (Definition of a limit). The question says $x_n\in [a, b]$.
So I put the sequence between the bounds $a$ and $b$.
$a\le\left|x_n - L \right|\le b$. When I break apart the absolute value brackets, this is the result:
$a- L \lt x_n\lt b+L$.
But this can't be true considering $x_n\in[a, b]$. Here, the left side is $a-L$ (where $L \lt a$) is too small to be between $[a, b]$. Same for the right side, $b+L$ (where $L \gt b$) is too large to be between $[a,b]$.
I'm not sure if I'm on the right track or not. Can someone help lead me to a better answer.