Can this expression be reduced to a difference quotient? Setting up an equation I've come into this factor:
$\displaystyle \lim_{h\rightarrow0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h}; \quad f\in \mathcal{C}^\infty$
To me this looks more or less like a derivative, but I've not been able to reduce it to a common difference quotient.
Is that actually $df/dx$ or there's more?
 A: Using L'Hospital's rule, your limit is equivalent to
$$\lim_{h \to 0} \frac{- \frac{f'(x + h) - f'(x - h)}{2f(x)}}{1} = \lim_{h \to 0} \frac{f'(x - h) - f'(x + h)}{2f(x)}$$
Assuming that $f(x) \ne 0$, the fact that $f'$ is continuous implies that this limit is $0$.
A: Since $f\in C^2$, integration by parts gives Taylor's Theorem with remainder:
$$
\begin{align}
f(x)
&=f(a)+(x-a)f'(a)+\int_a^x(x-t)f''(t)\,\mathrm{d}t\\
&=f(a)+(x-a)f'(a)+\frac12(x-a)^2f''(\xi)\tag{1}
\end{align}
$$
for some $\xi$ between $a$ and $x$.
Using $(1)$ to represent $f(x+h)$ and $f(x-h)$, we get
$$
\begin{align}
\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}
&=-\frac{f(x+h)-2f(x)+f(x-h)}{2f(x)h^2}\\
&=-\frac{\frac12h^2f''(\xi_+)+\frac12h^2f''(\xi_-)}{2f(x)h^2}\\[4pt]
&=-\frac{f''(\xi)}{2f(x)}\tag{2}
\end{align}
$$
where $\xi_+\in(x,x+h)$ and $\xi_-\in(x-h,x)$ and $\xi\in(\xi_-,\xi_+)\subset(x-h,x+h)$.
Therefore,
$$
\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}=-\frac{f''(x)}{2f(x)}\tag{3}
$$
Of course, $(3)$ implies
$$
\begin{align}
\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h}
&=\lim_{h\to0}h\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\
&=\lim_{h\to0}h\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\[4pt]
&=0\cdot-\frac{f''(x)}{2f(x)}\\[12pt]
&=0\tag{4}
\end{align}
$$
A: If you reduce to the same denominator, you could notice that your expression is just
-h f"(x)/ [2 f(x)]. This assumes that f(x) is not zero, that f'(x) is continuous (then f"(x) exist).
A: Use $1 - \frac{f(x + h) + f(x - h)}{2 f(x)} = \frac{f(x) - f(x + h) + f(x) - f(x - h)}{2f(x)}$.
