I've just got this general rule for the nth 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.
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