proving limits of monotone increasing function If $f(x) $ is a monotone increasing function on an open interval $(a,b)$
Then, how can I show $\lim \limits_{x \to x_0+}f(x)$ and $\lim \limits_{x \to x_0-}f(x)$ are finite and $\lim \limits_{x \to x_0-}f(x)\le f(x_0)\le\lim \limits_{x \to x_0+}f(x)$ if  $x_0\in (a,b)$?
They must be finite because $x_0 \in(a,b)$ and $f(x)$ approaches $x_0$ from both sides but how can I prove that?
The inequality makes sense because if $f(x_0) $ is a single point and both limits approach this single point from different sides, the inequalities must be true. 
I'm having trouble connecting these to proofs
 A: I'm not sure what level you are working at (Calc I vs Real Analysis I), but if the latter, then you should know that $$\lim_{x\to x_0^-} f(x) = \sup \{f(x) \mid x < x_0\},$$ and $f(x_0)$ is an upper bound for that set.
A: It's sufficient to show that the limits exist. By contradiction, given an arbitrary number $x_0\in(a,b)$, suppose the left limit of $f(x)$ at $x_0$ does not exist. (The left limit of $f(x)$ at $x_0$ exists means that for any $\epsilon>0$, there is a $\delta>0$ such that if $x_0-\delta<x_1<x_2<x_0$ then $|f(x_1)-f(x_2)|<\epsilon$.) Therefore, $\lim_{x\rightarrow{x_0^-}}f(x)$ does not exist implies that $\exists\epsilon_0>0$ such that for $\delta_1=1$, there is $x_0-1<x_1<x_1'<x_0$ so that
$$f(x_1')-f(x_1)>\epsilon_0,~i.e.,~f(x_1')>f(x_1)+\epsilon_0.$$
And for $\delta_2=x_0-x_1'$, there is $x_0-\delta_2=x_1'<x_2<x_2'<x_0$ so that
$$f(x_2')-f(x_2)>\epsilon_0,~i.e.,~f(x_2')>f(x_2)+\epsilon_0>f(x_1')+\epsilon>f(x_1)+2\epsilon_0.$$
Repeating this progress, we can find $x_0-1<x_1<x_1'<x_2<x_2'<\cdots<x_0$ such that
$$f(x_n')>f(x_1)+n\epsilon_0, n\geq2,$$
Let $N=\frac{f(x_0)-f(x_1)}{\epsilon_0}$, then for any $n>N$, we have $x_0-1<x_n'<x_0$ and
$$f(x_n')>f(x_1)+n\epsilon_0>f(x_0).$$
which is a contradiction since $f(x)$ is an increasing function. Thus the limit from left exists at each point.
We can show that the limit from right exists at each point in the same way.
