Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$ I have been wondering whether the following limit is being used somehow, as a variation of the derivative:
$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} .$$
Edit:
I know that this limit is defined in some places where the derivative is not defined, but it gives us some useful information.
The question is not whether this limit is similar to the derivative, but whether it is useful somehow.
Thanks.
 A: This cannot be used as a definition of the derivative. First the result is half the sum of the left and right derivatives at $x$, when these exist. Second the limit can be well defined even when the sided derivatives do not exist, consider for example $f(x)=|x|^a$ around $x=0$ for suitable values of $a$. More generally, the limit at $x$ exists and is $g'(x)$ as soon as $f=g+s$ with $g$ differentiable at $x$ and $s$ symmetric around $x$ in the sense that $s(x+z)=s(x-z)$ for every $|z|$ small enough hence this notion can be used to get rid of symmetric but badly behaved parts of $f$ around $x$.
A: If $f$ is allowed to be discontinuous we have this example:
$$ x \in \mathbb{Q} \implies \lim_{h \to 0} \frac{1_\mathbb{Q}(x+h) - 1_\mathbb{Q}(x-h)}{2h} = \lim_{h \to 0} \frac{0}{2h} = \lim_{h \to 0} 0 = 0.$$
That doesn't seem particularly useful to me.
A: The "symmetric difference" form of the derivative is quite convenient for the purposes of numerical computation; to wit, note that the symmetric difference can be expanded in this way:
$$D_h f(x)=\frac{f(x+h)-f(x-h)}{2h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}h^2+\frac{f^{(5)}(x)}{5!}h^4+\dots$$
and one thing that should be noted here is that in this series expansion, only even powers of $h$ show up.
Consider the corresponding expansion when $h$ is halved:
$$D_{h/2} f(x)=\frac{f(x+h/2)-f(x-h/2)}{h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}\left(\frac{h}{2}\right)^2+\frac{f^{(5)}(x)}{5!}\left(\frac{h}{2}\right)^4+\dots$$
One could take a particular linear combination of this half-$h$ expansion and the previous expansion in $h$ such that the term with $h^2$ zeroes out:
$$4D_{h/2} f(x)-D_h f(x)=3f^\prime(x)-\frac{f^{(5)}(x)}{160}h^4+\dots$$
and we have after a division by $3$:
$$\frac{4D_{h/2} f(x)-D_h f(x)}{3}=f^\prime(x)-\frac{f^{(5)}(x)}{480}h^4+\dots$$
Note that the surviving terms after $f^\prime(x)$ are (supposed to be) much smaller than either of the terms after $f^\prime(x)$ in the expansions for $D_h f(x)$ and $D_{h/2} f(x)$. Numerically speaking, one could obtain a slightly more accurate estimate of the derivative by evaluating the symmetric difference at a certain (well-chosen) step size $h$ and at half of the given $h$, and computing the linear combination $\dfrac{4D_{h/2} f(x)-D_h f(x)}{3}$. (This is akin to deriving Simpson's rule from the trapezoidal rule). The procedure generalizes, as one keeps taking appropriate linear combinations of a symmetric difference for some $h$ and the symmetric difference at half $h$ to zero out successive powers of $h^2$; this is the famous Richardson extrapolation.
A: Lemma: Let $f$ be a convex function on an open interval $I$. For all $x \in I$,
$$
g(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}
$$
exists and $f(y) \geq f(x) + g(x) (y-x)$ for all $y \in I$.
In particular, $g$ is a subderivative of $f$. 
A: $$
\begin{eqnarray*}
\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} &=& 
\frac12 \lim_{h\to 0}\left(\frac{f(x+h)-f(x)}h+\frac{f(x)-f(x-h)}h\right)
\\ &=& \frac12 (f'(x)+f'(x)) = f'(x)
\end{eqnarray*}
$$
Assuming, of course that $f$ is differentiable at $x$.
A: This is a wrong way of thinking, the comment by Jesse Madnick explains why.
This is symmetric differentiation and is useful when both left and right hand limits exists, as oppose to usual derivative definition when only it is sufficient that right hand limit should exist.
The only place that I have seen that to be of use to make things nicer was in Fractional Calculus, providing some nice formulas to generalise differentiation.
To see why, try using the above limit definition of differentiation to get 2nd , or 3rd order definitions of a derivative, you should be able to recognise the resulting pascal/binomial like looking formulas, from there one might just try to generalise that to get a derivative definition that would say something about $\tfrac{1}{2}^{\text {th}}$ derivative.
In short : there is nothing more to this definition than the other usual one, except that this one shows an extra symbol (2), why use one more extra symbol when there is no need? maybe just to reap the benefits of it being symmetric.
Edit : Is somebody going to point the mistake? What am I missing here?
