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Definition Let $c \in (a, b)$ and let $f:(a,b) \rightarrow \mathbb{R}$. Then $f$ is symmetrically differentiable at $c$ if $$\lim_{h \to 0} \frac{f(c + h) - f(c - h)}{2h}$$ exists and is finite.

Problem Show that if $f$ is differentiable at $c$, then $f$ is symmetrically differentiable at $c$, and the derivatives are the same.

I'm not entirely sure where to start. We want to show that, given $f'(c)$ exists, $$\lim_{h \to 0} \frac{f(c + h) - f(c - h)}{2h} = f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h}$$ but I'm not sure how to do that. Can anyone provide minimal assistance? This seems like such a simple problem, but I'm stuck nonetheless. Thanks in advance for any response.

Sprinkle
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Hint: $$\lim_{h\to0}\frac{f(c+h)-f(c-h)}{2h}=\lim_{h\to0}\frac{f(c+h)-f(c)+f(c)-f(c-h)}{2h}=\lim_{h\to0}\frac{\left[f(c+h)-f(c)\right]-\left[f(c-h)-f(c)\right]}{2h}=\frac{1}{2}\lim_{h\to0}\left[\frac{f(c+h)-f(c)}{h}-\frac{f(c-h)-f(c)}{h}\right]$$ and make a change of variable $(-h)\mapsto h$ in the second limit (I mean, in the second fraction).

zipirovich
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