Is the symmetric definition of the derivative equivalent? Is the symmetric definition of the derivative (below) equivalent to the usual one?
\begin{equation}
\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}
\end{equation}
I've seen it used before in my computational physics class. I assumed it was equivalent but it seems like it wouldn't matter if there were a hole at $x=h$ in the symmetric derivative, whereas with the usual one it wouldn't be defined. Which is kinda interesting...
If they're not equivalent - is there a good reason as to why we should use the common one? Or is the symmetric one actually more useful in some sense because it "doesn't care" about holes?

$$\lim_h \frac{f(x-h)-f(x+h)}{2h}= \frac12 \lim_h \frac{f(x-h)-f(x)+f(x)-f(x+h)}{h}$$
$$=\frac12(\lim_h \frac{f(x-h)-f(x)}h+\lim_h \frac{f(x)-f(x+h)}h)=\frac{1}{2}(-f'(x)-f'(x))=-f'(x)$$
However this definition can be "strange" sometimes (in the sense it does not catch the idea of "derivative as slope"): let $f(x):=1/x^2$, and
$$\lim_h \frac{f(-h)-f(h)}{2h}=0.$$
Even without asymptotes, you can have something like $g(x):=|x|$, and
$$\lim_h \frac{g(-h)-g(h)}{2h}=0.$$
