Let $(a\,;b)$ be an open interval of real numbers. Let $\xi$ be a limit point of $(a\,;b)$. Let functions $f:(a\,;b)\setminus\{\xi\}\rightarrow\mathbb R$ or $f:(a\,;b)\rightarrow\mathbb R$ be called differentiable on point $\xi$ iff $\displaystyle \lim_{x\rightarrow \xi}\dfrac {f(x)-f(\xi)}{x-\xi}$ exists.
Let $f$ be called differentiable on open interval $(a\,;b)$ iff $f$ is differantiable on every $\xi \in (a\,;b)$.
If I understand the above definitions correctly, then $f$ being differentiable on $(a\,;b)$ means that $\forall \xi \in (a\,;b):\exists \displaystyle \lim_{x\rightarrow \xi} \dfrac{f(x)-f(\xi)}{x-\xi}$. We can in such a way create a function $f':(a\,;b)\rightarrow \mathbb R$ such that $f'(x)=\displaystyle \lim_{x\rightarrow \xi} \dfrac{f(x)-f(\xi)}{x-\xi}$.
However, I encountered the definition of differentiability on open intervals with $f'$ as $\displaystyle \lim_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}$. Here is an example of such definition.
How is the definition of a function $f$ differentiable on open interval $(a\,;b)$ with $\displaystyle \lim_{h\rightarrow 0} \dfrac {f(x+h)-f(x)}{h}$ equivalent to the one saying "for all elements $\xi$ of $(a\,;b)$ there exists $\displaystyle \lim_{x\rightarrow \xi}\dfrac{f(x)-f(\xi)}{x-\xi}$".
Make the substitution $\xi = x+h$. Then $\lim_{h\rightarrow 0}=\lim_{x\rightarrow\xi}$. Furthermore: $$ \frac{f(x+h)-f(x)}{h} = \frac{f(\xi)-f(x)}{\xi - x} $$