Let $a \leq x_{n} \leq b$ for all n in N. If $x_{n} \rightarrow x$. Then prove that $a \leq x \leq b$ Let $a \leq x_{n} \leq b$ for all n in N. If $x_{n} \rightarrow x$. Then prove that $a \leq x \leq b$
Attempt - If I assume that $x$ is greater than both $a$ and $b$. Then since series is given convergent, so after certain stage its elements will lie between $(x-\epsilon , x + \epsilon )$. If I take epsilon to be such that $\epsilon = (b + x) /2$. Then sequence lies to right of $b$, which is contradiction. Same argument for if I take limit to be to left of a. Is this fine? Thanks
 A: I am showing one side only.
If $x<a$ then taking $\epsilon=a-x>0$
Now since $x_n \to x \exists m\in \mathbb N$ such that $x-\epsilon<x_n<x+\epsilon \forall n\geq m$
Then $x_n<x+\epsilon=x+a-x=a$ i.e $x_n<a$ for some $n$ contradiction as $x_n\geq a\forall n$
A: You are on the right track.  As requested in the comment, the formalism is as follows:
If $x>b$, take $\epsilon=x-b$.  Then if $n>N$ it follows that 
$$ |x_{n}-x| < x-b.$$
We know that this should mean that $x_n>b$ (which is a contradiction) so let's re-write the terms to make this appear very explicitly.
$$-(x-b) < x_{n}-x < (x-b)$$
Adding $x$ to all three parts of the inequality gives us:
$$ b < x_{n} < 2x-b$$
but of course, we only need that $b<x_{n}$ to get the contradiction.
A: Given $\varepsilon>0$, there is a positive integer $N$ such that
$$
|x-x_n|\le \varepsilon \quad \forall n\ge N.
$$ 
It follows that
$$
x_n-\varepsilon \le x\le x_n+\varepsilon \quad \forall n\ge N
$$
and therefore
$$
a-\varepsilon \le x\le b+\varepsilon.
$$
Letting $\varepsilon$ tend to $0$, we get $a\le x\le b$.
