TITLE: Using the same limit for a second derivative
QUESTION [5 upvotes]: I've been trying to answer the same question answered here: Second derivative "formula derivation"
And I'm stuck in a step that is not addressed both in the answer and in the comments of the question over there. In the original question he uses the fact that

$$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$
  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h}  }{h}$$

Which I basically see as taking the derivatives with the same limit 3 times. 
Shouldn't it be as follow?
$$f''(x) = \lim_{h\to0} \frac{  \lim_{h_1\to0}\frac{ f(x+h+h_1) - f(x+h)}{h_1} - \lim_{h_2\to0}\frac{ f(x+h_2) - f(x)}{h_2}  }{h}$$
How do you justify moving to the equation given in the original answer?

REPLY [4 votes]: The equation
$$ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \tag{1}$$
is clearly correct: that's indeed the very definition of the derivative (hence, the derivative exists iff the limit exists).
Now consider the equation
$$ f''(x) = \lim_{h \to 0} \dfrac{f(x+2h)-2f(x+h) + f(x)}{h^2} \tag{2a}$$
or the "centered" version:
$$ f''(x) = \lim_{h \to 0} \dfrac{f(x+h)-2f(x) + f(x-h)}{h^2} \tag{2b}$$
These are not correct in the same sense that $(1)$ is. They are true (as other answers have shown) only if the second derivative exists. Hence they are not  valid definitions of the second derivative. The correct definition corresponds to the multiple limit you wrote in the question body, and we cannot get from that to $(2a)$ or $(2b)$. In concrete: it can happen that the limits on the RHS of  $(2a)$ or $(2b)$ exist, but $f''(x)$ does not.
The problem is easy to spot in the case of eq. $(2b)$. If it were true, then for any odd function we'd have $f''(0)=0$ - which of course is not true (take for example $f(x)=\sqrt[3]{x}$ ; or any odd function discontinuous at $x=0$).
An example for eq $(2a)$: take $f(x)=x$ for $x$ irrational, $0$ otherwise. Again, the limit in $(2a)$ gives $0$, but $f''(0)$ does not exist.