Let $f:(a\,..b) \rightarrow \mathbb R$ or, for some $\xi \in (a\,..b),$ $f:(a\,..b)\setminus{\{{\xi}}\} \rightarrow \mathbb R$.
Let $x_0 \in (a\,..b)$ or $x_0 \in (a\,..b)\setminus{\{\xi}\}$.
Let $\displaystyle \lim_{\,x\to x_0}(f(x)-f(x_0))$ exist.
Is it possible to prove that for all such $f$, for all $x_0$,$\displaystyle \lim_{\,x\to x_0}(f(x)-f(x_0))=0$?
This happens if and only if $f$ is continuous at $x_0$. For example, if $$f(x)=\begin{cases}1\text{ if }x\neq 0\\0\text{ if }x=0\end{cases}$$ then $$\lim_{x\to 0}(f(x)-f(0))=1$$
As soon as $f(x)$ is not continous in $x_0$, the limit does not have to be zero.
Example:
$f = \begin{cases} |x|, ~\text{for}~|x|>0,\\ 1,~\text{otherwise} \end{cases} $
We do not have zero limit for $x_0 = 0$.
Nope! There are plenty of functions that can't full fill this requisite. All of them are discontinuous at a point within the bounded region. Let me give you an example:
Lets define $$f:= \left\{ \begin{array}{ccc} f(x)=1 &\mathrm{if}& x\in \mathbb Q\\ f(x)=0 & & o.w. \end{array} \right.$$
Now lets play the $\varepsilon$, $\delta$ game for say $x_0\in[0,1]$. This is where I pick an $\varepsilon$ and you try to find a $\delta$ that will meet the conditions. I will pick mine to be, $0<\varepsilon < 1$. Now can you find a $\delta$ that will meet those conditions?