We know that if $f$ is differentiable at $x=a$, then
$$f'(a)=\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$
Recently I knew that if $f$ is twice differentiable at $x=a$, then
$$f''(a)=\lim_{k \rightarrow 0} \bigg( \lim_{h \rightarrow 0} \frac{f(a+h+k)-f(a+h)-f(a+k)+f(a)}{hk} \bigg)$$
This can be re-expressed (with the indices of the limits being $h_1$ and $h_2$:
$$f''(a)=\lim_{h_2 \rightarrow 0} \bigg( \lim_{h_1 \rightarrow 0} \frac{f(a+h_1+h_2)-f(a+h_1)-f(a+h_2)+f(a)}{h_1h_2} \bigg)$$
I do not know how can we derive the above formula of $f''$, and how to derive higher order derivatives of a function which is $n$-times differentiable at $x=a$:
$$f^{(n)}(a)= \lim_{h_n \rightarrow 0} \big( \lim_{h_{n-1} \rightarrow 0} \big(\lim_{h_{n-2} \rightarrow 0} \big(\dots \big( \lim_{h_1 \rightarrow 0} \frac{\text{some expression involving }f \text{ only, with no derivatives}}{h_1h_2h_3\dots h_n} \big) \dots \big) \big) \big)$$
I saw this, the answer is right but is not in the required form.
Your help would be appreciated. THANKS!
