calculus - Prove that f(x) can have any value between A and B

I need to prove the following statement:
Let f be a bounded and continuous function in the interval (a,b). if $\lim_{x\to a+}f(x)=A$ and $\lim_{x\to b-}f(x)=B$ and $A\neq B$ then f can get any value between and A and B in the interval (a,b).

What I did:
Let $ g(x) =
\begin{cases}
f(x), & \text{$aA, & \text{x=a} \\

B, & \text{x=b} \\
\end{cases}$
. We can see that $g(x)$ is a continuous function in the the interval [a,b]. Therefore, I can use the Intermediate value theorem to prove that for any value between A and B, there exists $a

My Questions:

1)Can i define a function that is divided to 3 cases?

2)I didn't use the fact that f(x) is bounded, and that seems weird. what is wrong with my proof?

3) I am sure there must be a better way to prove it because the next statement i need to prove the same just for $A=\infty$ and $B=-\infty$(f(x) isn't bounded) and my trick won't work there, that's why I have no idea how to prove it.

Answer

1) Yes, you can, if $A$ and $B$ are finite. You have just extended $f$ to a continuous function on $[a,b]$

2) You have implicitly used it in assuming that $A$ and $B$ are finite. In fact, $|A|, |B|<+\infty$ if and only if $f$ is bounded (why?)

3) If $A=+\infty, B = -\infty$, then for any real $\alpha\in \mathbb{R}$, there is $a'>a, b'<$ such that $f(a') < \alpha < f(b')$. So you apply intermediate value theorem to $[a',b']$