Prove L $\in$ [a, b]

Question

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.

Answer 1

Your line $a \leq x_n - L \leq b$ should be $a-L \leq x_n - L \leq b-L$

Since $a \leq x_n \leq b$, it follows that $a - L \leq x_n - L \leq b- L $.

See what you can do with this now, and try to show that when $x$ is arbitrarily close to $L$, there are $x_n$ that leave [a,b], which is a contradiction.

Answer 2

Suppose $L<a$. Set $\varepsilon=(a-L)/2$. By definition of limit, there exists $N$ such that, for $n>N$, $$ |x_n-L|<\varepsilon $$ which implies $$ x_n-L<\frac{a-L}{2} $$ so $$ x_n<L+\frac{a-L}{2}=\frac{a+L}{2}<\frac{a+a}{2}=a $$ This is a contradiction.

Can you do the case $L>b$?