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I have to prove that if a function $f$ is differentiable on $(a,b)$, then \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*}

Using the fact that $f'(x) = \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$, I wrote my proof in the following manner:

\begin{align*} \lim\limits_{h \rightarrow 0}f(x+h) - 2f(x) = \lim\limits_{h \rightarrow 0} - f(x-h) \\ \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0}\dfrac{-f(x-h)}{2h} \\ \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)}{h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x-h)}{2h} \\ \lim\limits_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*}

However, I believe that it is actually incorrect, because when I divide by $2h$ I am potentially making the limit undefined. How would I go about correcting my proof?

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  • $\begingroup$ Hint: First show $$f'(x) = \lim_{h \to 0} \frac{f(x) - f(x-h)}{h}$$ $\endgroup$ – Simon S Oct 27 '15 at 22:43
  • $\begingroup$ And when you have finished the exercise ponder over what you have just done. It's not merely an exercise in handling meaningless limits. You have shown that a function that has a derivative at a point also has a symmetric derivative at that point and that the two derivatives have the same value. Think for a while about the mysteries of the "symmetric derivative." It has kept many of us quite busy and highly entertained. $\endgroup$ – B. S. Thomson Oct 27 '15 at 23:24
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\begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \end{align*} but also \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x)-f(x-h)}{h} \end{align*} sum them up and divide by 2 to get

\begin{align*} f'(x) = \lim\limits_{h \rightarrow 0}\frac{ \dfrac{f(x+h)-f(x)}{h} + \dfrac{f(x)-f(x-h)}{h}}2 =\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*}

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You cannot just divide by $h$ from nowhere, it is not correct. Consider splitting $$ \frac{f(x+h)-f(x-h)}{2h}=\frac{f(x+h)-f(x)+f(x)-f(x-h)}{2h}= \frac12\left(\frac{f(x+h)-f(x)}{h}+\frac{f(x-h)-f(x)}{-h}\right). $$ Both terms go to $f'(x)$ as $h\to 0$.

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$$ f(x+h) = f(x)+f'(x)h + o(|h|) $$ together with: $$ f(x-h) = f(x)-f'(x)h + o(|h|) $$ gives: $$ f(x+h)-f(x-h) = 2h\cdot f'(x) + o(|h|).$$

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Since $f$ is differentiable, we may use l'Hopitals rule:

$$\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}=\lim_{h\to 0}\frac{f'(x+h)-(-1)f'(x-h)}{2}=\lim_{h\to 0}\frac{f'(x+h)+f'(x-h)}{2}=\frac{2f'(x)}{2}=f'(x).$$

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  • 1
    $\begingroup$ The other proofs use only the existence of the derivative at the one point. But it is fair to use the stronger hypothesis since the poster included that (he didn't say $f$ was continuously differentiable though). (If it was a homework assignment, though, he can't use yours since it must surely have had the weaker hypothesis). $\endgroup$ – B. S. Thomson Oct 27 '15 at 23:47

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