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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.

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    What do you mean by $\cdot^n$? – 2017-02-28
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    @MichaelBurr It's the number of combinations of $n$ things taken $r$ at a time. Is there some way to make it more clear? – 2017-02-28
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    Just use \binom{n}{r} . I'm more used to seeing $C_r^n$ instead of $n$ to the left of $C$. As written (due to spacing), it appears that the $n$ applies to the $\cdot$ and not the $C$. – 2017-02-28
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    Typically we see the notation $\binom {n}{r}$ here. – 2017-02-28
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    It appears that you don't understand what others have mentioned for your previous questions. There is no way to define higher order derivatives in terms of original function. You are just trying to use some limit which equals $f^{(n)} (x) $ if the latter exists and then creating two versions with $h$ and $-h$ with $h\to 0$. Note that $\lim_{h\to 0}f(h)=\lim_{h\to 0}f(-h)$ for any function $f$ just via definition of limit and hence I see your exercise of using equality of such limits of $f(h) $ and $f(-h) $ a bit pointless. What do you really want to achieve here? – 2017-02-28
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    Your question will make some sense if you replace $h\to 0$ with $h\to 0^{+}$. Is this what you intend? – 2017-02-28
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    @ParamanandSingh Just take any function whose $n^{th}$ derivative doesn't exist, and check by my rule, the limits will be unequal. And, take another function whose $n^{th}$ derivative exists, then these limits will be equal. It appears that there is a way to define higher order derivatives in terms of the original function. I derived it so it must be true. If you get a counter example, write it as an answer. I'll accept your answer. – 2017-02-28
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    If you really think they must be true "because you derived them", then it's fine. I don't need to state anything further. – 2017-02-28
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    @ParamanandSingh By 'I derived it' I didn't mean I'm an expert and I can't make mistakes, so it must be true. That wasn't what I implied. I meant that something which is proved must be true. And, I've checked that it works. Please give a counter example if you don't think it's true. – 2017-02-28
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    Ok sorry for misunderstanding. Right now I can't think of a counter-example (because even I am not that an expert) but I am certain that people more knowledgeable will find a counter-example and also mention it here. Meanwhile please change $h\to 0$ to $h\to 0^{+}$. – 2017-02-28
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    @ParamanandSingh's point is that you could, equivalently, say that $f$ is $n$ times differentiable at $x$ if and only if the (two-sided) limit $\lim_{h\rightarrow 0}\left(\sum_{r=0}^{n}(-1)^r{{n}\choose{r}}f(x+(n-r)h)\right) / h^n$ exists; and $f^{(n)}(x)$ is equal to the value of that limit. – 2017-02-28
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    And that statement would also be incorrect, because you need to require the lower derivatives to exist as well. – 2017-02-28

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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.

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    I never bothered to check the simple example $|x|$. +1 – 2017-03-01
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    I guess the requirement is that the first $n-1$ derivatives should exist. I think only then the formula is true. – 2017-03-01