Definition of the nth derivative? [First post] If the definition of the derivative is 
$$
f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}
$$
Would it make sense that the nth derivative would be (I know that the 'n' in delta x to the nth power is useless) 
$$
f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n}
$$
I came to this conclusion using this method
$$
f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}
$$
(this is correct right?)
$$
f^{\prime\prime}(x) = \lim_{\Delta x \to 0} \dfrac{f^\prime(x+\Delta x) - f^\prime(x)}{\Delta x}=$$$$\lim_{\Delta x \to 0}\dfrac{\dfrac{f((x+\Delta x)+\Delta x)-f(x+\Delta x)}{\Delta x}-\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x}=$$$$\lim_{\Delta x \to 0}\dfrac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2}
$$
After following this method a couple of times(I think I used it to the 5th derivative) I
noticed the pattern of
$$(a-b)^n$$
And that is how i arrived at 
$$
f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n}
$$
Have I made a fatal error somewhere or does this definition actually follow through?
Thanks for your time I really appreciate it.
P.S. Any input on using tags will be appreciated. 
 A: This is probably not a good definition of the $n$th derivative.  To see this, consider the case  $n = 2$:
$$
f''(x) = \lim_{h \to 0}  \frac{f(x + 2h) - 2f(x + h) + f(x)}{h^2}
$$
Define $f: \mathbb{R} \to \mathbb{R}$ as follows.  First, define $f(0) = 0$.
Now define $f$ on the intervals $\left[-1, -\tfrac12\right)$ and $\left(\tfrac12, 1\right]$ to be your favorite unbounded function, for instance $\frac{1}{x^2 - 1/4}$ is a good choice.
Now, for any $x$, let $k$ be the unique integer such that $2^k x$ is contained in one of these intervals, and define $f(x) = 2^{-k} f(2^k x)$.
This construction satisfies $f(2h) = 2f(h)$ for all $h \in \mathbb{R}$, so the derivative formula above gives
$$
f''(0)
= \lim_{h \to 0} \frac{f(2h) - 2f(h) + f(0)}{h^2}
= \lim_{h \to 0} \frac{0}{h^2}
= 0
$$
However, $f$ is wildly discontinuous at $0$, and is in fact unbounded in any neighborhood containing $0$.
A: when you did for f''(x) you are taking both limits as a single limit for delta x. what I did is take $h_1$ for first limit and $h_2$ for second limit, so that nth derivative is limit $(h_1, h_2, h_3,.... h_n)$ --> $(0,0,0,....,0)$ and some function.
