I've just got this general rule for the $n^{th}$ order derivative of a function to exist:
A function $f(x)$ is differentiable $n$ times only if: $$\lim_{h\rightarrow 0^+}\frac{\sum_{r=0}^{n}(-1)^r\cdot \binom{n}{r}\cdot f(x-(n-r)h)}{(-h)^n}=\lim_{h\rightarrow 0^+}\frac{\sum_{r=0}^n(-1)^r\cdot \binom{n}{r}\cdot f(x+(n-r)h)}{h^n}$$ Please not that $h$ itself is assumed to be positive in both the limits. That's why I've written that $h\rightarrow 0^+$ in both. I call the LHS the Left-hand $n^{th}$ derivative and the RHS, the right hand $n^{th}$ order derivative.
Have I got this result correct or Can counter-examples be given against it (examples in which the $n^{th}$ order derivative of a function does not exist but still these two limits are equal or examples in which the $n^{th}$ derivative of a function exists but these two limits are unequal)?
EDIT: Here, $x$ can be replaced by $c$ to check the differentiability at a particular point $x=c$.
UPDATE: After the answer by mjqxxxx, I guess that the requirement is that the first $n-1$ derivatives should exist, only then the formula can be used.
For $n=2$, your requirement is that $$ \lim_{h\rightarrow 0^+}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}=\lim_{h\rightarrow 0^+}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2} $$ (where, presumably, you require both limits to exist). This is necessary but not sufficient for the second derivative of $f$ to exist. For instance, consider $f(x)=|x|$ and apply your definition at $x=0$. Then $|-2h|-2|-h|+|0|=0$ and $|2h|-2|h|+|0|=0$, so both limits exist and are equal to zero; but $f(x)$ is not, in fact, twice-differentiable at $x=0$, because it's not even once-differentiable.