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?


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 $$ \begin{align} \lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h} &=\lim_{h\to0}h\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\ &=\lim_{h\to0}h\lim_{h\to0}\frac{1-\frac{f(x+h)+f(x-h)}{2f(x)}}{h^2}\\[4pt] &=0\cdot-\frac{f''(x)}{2f(x)}\\[12pt] &=0\tag{4} \end{align} $$

  • $\begingroup$ I think you forget $1/f(x)$ while deriving $(2)$. All the rest is nice, point $(3)$ is exactly what I was looking for. This have been also pointed out by Claude Leibovici. $\endgroup$ – DarioP Oct 25 '13 at 9:41
  • $\begingroup$ @DarioP: thanks for noting the omission. I have put back the missing $f(x)$. $\endgroup$ – robjohn Oct 25 '13 at 11:21

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 $0$.


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).

  • $\begingroup$ This is interesting, did you forget a factor $1/2$ or am I missing something important? That factor is in this framework math.stackexchange.com/questions/539094/… and being able to simplify in this way may lead me to a solution! $\endgroup$ – DarioP Oct 25 '13 at 8:59
  • $\begingroup$ You are right. I forgot the 1/2. I re-edited my answer $\endgroup$ – Claude Leibovici Oct 26 '13 at 12:14

Use $1 - \frac{f(x + h) + f(x - h)}{2 f(x)} = \frac{f(x) - f(x + h) + f(x) - f(x - h)}{2f(x)}$.


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