calculus - General rule for the $n^{th}$ order derivative of a function to exist

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.