TITLE: prove 3rd derivative of a function in a point
QUESTION [7 upvotes]: Let $f$ be twice differentiable and $f'''$ exists in one point $x\in D$. I want to show for this one point $$f'''(x)=\lim_{h\rightarrow0}\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$$
I did the following: I know $\lim_{h_1\rightarrow0}\frac{f(x+h_1)-f(x)}{h_1}=f'(x)$ and so follows by pluging in $$f''(x)=\lim_{h_2\rightarrow0}\frac{f'(x+h_2)-f'(x)}{h_2}=...=\lim_{h_1\rightarrow0}\frac{f(x+h_1)+f(x-h_1)-2f(x)}{h_1^2}$$
Doing this one more time I'll get ($f'''$ exists in $x$)
$$f'''(x)=\lim_{h_3\rightarrow0}\frac{f''(x+h_3)-f''(x)}{h_3}=...=\lim_{h_1\rightarrow0}\frac{f(x+3h_1)-3f(x+2h_1)+3f(x+h_1)-f(x)}{h_1^3}$$
First question: Can you do it like that?
And my second question: Why can we always choose 'the same $h$' for the limits? Formally for e.g. $f''$ it's
$$\lim_{h_2\rightarrow0}\frac{\lim_{h_1\rightarrow0}\frac{f(x+h_1+h_2)-f(x+h_2)}{h_1}-\lim_{h_1\rightarrow0}\frac{f(x+h_1)-f(x)}{h_1}}{h_2}$$ and I have $h_1$ and $h_2$ in it.
Edit
Using L'Hospital twice on the right side I get
$$\dots=\lim_{h\rightarrow0}\frac{3f''(x+3h)-4f''(x+2h)+f''(x+h)}{2h}$$
And so we get $\begin{align}&\dots\\&=\lim_{h\rightarrow0}\left\{\frac16\frac{3f''(x+h)-3f''(x)}{h}-\frac13\frac{12f''(x+2h)-12f''(x)}{2h}+\frac12\frac{9f''(x+3h)-9f''(x)}{3h}\right\}\\&=\frac36\lim_{h\rightarrow0}\frac{f''(x+h)-f''(x)}{h}-\frac{12}3\lim_{h\rightarrow0}\frac{f''(x+2h)-f''(x)}{2h}+\frac92\lim_{h\rightarrow0}\frac{f''(x+3h)-f''(x)}{3h}\\&=f'''(x)\end{align}$

REPLY [4 votes]: This is not a direct answer to your questions but a simple answer to the problem
By Taylor's theorem  we have
$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{6}f'''(x)+o(h^3)$$
$$f(x+2h)=f(x)+2hf'(x)+{2h^2}f''(x)+\frac{4h^3}{3}f'''(x)+o(h^3)$$
$$f(x+3h)=f(x)+3hf'(x)+\frac{9h^2}{2}f''(x)+\frac{9h^3}{2}f'''(x)++o(h^3)$$
so we can see easily (it's a simple calculation) that 
$$f'''(x)=\lim_{h\rightarrow0}\frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}$$