A general formula for the second degree derivative?

Question

Is the method I wrote for the second degree derivative formula is correct? Can you verify?

$$\begin{align} f''(x) &=\lim_{\Delta x \to 0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x} \tag{1}\\[4pt] &= \lim_{\Delta x \to 0} \left[\frac{\lim_{\Delta x\to 0}\frac{f(x+2 \Delta x)-f(x+\Delta x)}{\Delta x}-\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x} \right] \tag{2}\\[4pt] &= \lim_{\Delta x \to 0} \left[\frac{\frac{f(x+2 \Delta x)-f(x+\Delta x)}{\Delta x}-\frac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x} \right] \tag{3}\\[4pt] &=\lim_{\Delta x \to 0}\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{{\Delta x}^2} \tag{4}\\[4pt] \Longrightarrow \quad f''(x) &=\lim_{\Delta x \to 0}\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{{\Delta x}^2} \tag{5}\\[4pt] \Longrightarrow \quad f''(x-\Delta x) &=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-2f(x)+f(x-\Delta x)}{{\Delta x}^2} \tag{6}\\[4pt] \Longrightarrow \quad f''(x) &=\lim_{\Delta x \to 0}\frac{f(x-\Delta x)+f(x+\Delta x)-2f(x)}{{\Delta x}^2} \tag{7} \end{align}$$

REMARK Q:

My solution is different (especially the last steps) than the one shown here. Therefore, it can not be evaluated under the category "Duplicate".

Answer 1

You are assuming that $\lim_{x\to 0}(f(x)-g(x)) =\lim_{x\to 0}f(x)-\lim_{x\to 0}g(x) $.

This is not necessarily true.

Answer 2

By such definition any additive function is twice differentable with the 2nd derivative equal to zero. But there are discontinuous additive functions.