TITLE: Difference Quotient And Derivatives
QUESTION [5 upvotes]: Prove that the limit of the difference quotient $$\frac{f(c+h)-f(c-h)}{2h}$$ as $h \to 0$ is equal to $f'(c)$ if it exists

Now we can write this difference quotient as $$\frac{f(c+h)-f(c)+f(c)-f(c-h)}{2h}=\frac{f(c+h)-f(c)}{2h}+\frac{f(c)-f(c-h)}{2h}$$ Now taking the limits as $h \to 0^+$ we see that $$\lim_{h \to 0^+}\frac{f(c+h)-f(c)}{2h}=f'(c^+)/2=f'(c)/2$$ as $f'(c)$ exists and $$\lim_{h \to 0^+}\frac{f(c)-f(c-h)}{2h}=f'(c^-)/2=f'(c)/2$$ as $f'(c)$ exists So we get the limit of the difference quotient as $h \to 0^+$ equal to $f'(c)$ Similarly we can prove this as $h \to 0^-$ 
Is my proof correct?

REPLY [2 votes]: The proof looks essentially correct, but there is a small technicality hidden in here. In the second displayed equation you asserted that the limit
$$
\lim_{h\to 0^+} \frac{f(c) - f(c-h)}{2h}
$$
is equal to $f'(c^-)/2$, which is true. However, the reasoning here is slightly obscure. Explicitly, you could insert the step 
$$
\lim_{h\to 0^+} \frac{f(c) - f(c-h)}{2h} =\lim_{h \to 0^-} -\frac{f(c + h) - f(c)}{2(-h)} = f'(c)/2,
$$
where I have made a change of variables $h = -h$ in the limit, thus replacing $\lim_{h\to 0^+}$ by $\lim_{h\to 0^-}$ and necessitating the insertion of the $-$ in the denominator.