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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 $\displaystyle \lim_{x\rightarrow \xi}\dfrac{f(x)-f(\xi)}{x-\xi}$".

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  • $\begingroup$ $h$ is just $x-\xi$ (here $\xi$ is fixed). Note that in this case "$x\to\xi$" is equivalent to "$x-\xi\to 0$" (i.e., "$h\to 0$"). $\endgroup$
    – MPW
    Commented Feb 20, 2017 at 16:06
  • $\begingroup$ You did not encounter the definition of differentiability on an open interval at that link; it was assumed there that the reader already knew the definition. They are simply emphasizing that $f'$ is then itself a function on $(a,b).$ $\endgroup$
    – zhw.
    Commented Feb 20, 2017 at 16:06
  • $\begingroup$ @zhw. What's the definition of differentiability on an open interval then? $\endgroup$
    – asdasdf
    Commented Feb 20, 2017 at 17:58
  • $\begingroup$ It's what you wrote aboe. $\endgroup$
    – zhw.
    Commented Feb 20, 2017 at 18:21

1 Answer 1

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Make the substitution $\xi = x+h$. Then $\lim_{h\rightarrow 0}=\lim_{x\rightarrow\xi}$. Furthermore: $$ \frac{f(x+h)-f(x)}{h} = \frac{f(\xi)-f(x)}{\xi - x} $$

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  • $\begingroup$ How to prove that $\displaystyle \lim_{h\rightarrow 0}=\displaystyle \lim_{x\rightarrow \xi}$? What's the proof that $x+h$ is in the interval? $\endgroup$
    – asdasdf
    Commented Feb 20, 2017 at 19:08
  • $\begingroup$ $\xi = x+h$ What happens when you let $h\rightarrow 0$? Furthermore if $\xi\in(a,b)$ then there will always exist an $h$ for which $x+h\in(a,b)$, namely any $h<b-x$. $\endgroup$
    – Damian Sowinski
    Commented Feb 21, 2017 at 16:33

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