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Let $f$ be differentiable on $(a,b),$ then $f'(x)= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

I used that $\frac{f(x+h)-f(x-h)}{h}=\frac{f(x+h)-f(x)}{h} + \frac{f(x-h) + f(x)}{h}$ So

$$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2} \left( \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} + \lim_{h \rightarrow 0}\frac{-f(x-h) + f(x)}{h} \right)$$

The first limit is $f'(x)$, and for the second: because $f$ is differentiable the left sided limit equals the right sided limit, so i can change $h$ to $-a$ and get

\begin{align} & \lim_{h \rightarrow 0}\frac{f(x-h)+f(x)}{h}=\lim_{-a \rightarrow 0} \frac{-f(x+a)+f(x)}{-a} \\ = {} & \lim_{-a \rightarrow 0}\frac{-f(x+a)+f(x)}{-a} = \lim_{-a \rightarrow 0}\frac{f(x+a)-f(x)}{a}=f'(x) \end{align}

Now I wanted to know if the opposite direction also holds. So if $f'(x)= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$, then $f$ is differentiable. I tried to write $f'(x)= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$ as $$\frac{1}{2}(\lim_{\rightarrow 0}\frac{f(x+h)-f(x)}{h}+ \lim_{-a \rightarrow 0}\frac{f(x+a)-f(x)}{a})$$ but i cant conclude that $$\lim_{\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{-a \rightarrow 0} \frac{f(x+a)-f(x)}{a}$$

Edit: The anwser is that this does not hold (for counterexample see comments). To mention is that from the existance of $lim(x+y)$ the existance of $lim (x)$ and $lim (y)$ does not follow

John.W
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    Your title statement is confusing. If you can write $f'(x)=\textrm{something}$, then $f'(x)$ exists which is the same statement as $f$ is differentiable at $x$... – FShrike Aug 20 '21 at 18:15
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    Consider $f(x)=|x|$, $a=0$ – user429040 Aug 20 '21 at 18:16
  • There are some related question already: https://approach0.xyz/search/?q=OR%20content%3A%24%20%5Clim_%7Bh%20%5Crightarrow%200%7D%20%5Cfrac%7Bf(x%2Bh)-f(x-h)%7D%7B2h%7D%24 – Martin R Aug 20 '21 at 18:19
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    Now a much more interesting observation is that, if $$L=\lim_{h\to 0}\frac{f(x+h)-f(x-ah)}{(1+a)h}$$ exists and is finite for $|a|\neq1$, then $f$ is differentiable at $x$ and $f'(x)=L$. So $a=1$ is the only case where the existence of $L$ does not necessarily imply differentiability. – Sangchul Lee Aug 20 '21 at 21:32

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