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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".

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    I saw basically the same question already asked today, but can't find it any more. Regardless, there's numerous instances where this was already asked. An example: https://math.stackexchange.com/a/210273/399334 (By the way, yes it is correct) – John Doe Apr 24 '18 at 19:44
  • A limit with respect to $\Delta x$ of limits with respect to $\Delta x$ (line (2)) is ... problematic. – Blue Apr 25 '18 at 00:20
  • @Blue I edited. Please see again. –  Apr 25 '18 at 05:00
  • Nested limits with the same variable aren't sensical. In line (2), assuming that the individual limits exist, the numerator is their difference, a numerical value. No "$\Delta x$"s remains up there, so the "outer" limit is on "$\text{number}/\Delta x$"; it's either $0$ (if "number" is $0$) or else it's undefined (otherwise). What you "want" is more like $$\lim_{\Delta x\to 0}\frac{\lim_{\Delta u\to 0} \dfrac{f(x+\Delta x+\Delta u)-f(x+\Delta x)}{\Delta u} - \cdots}{\Delta x}$$ coupled with an argument about why you can take $\Delta u = \Delta x$ (which, in general, you can't). – Blue Apr 25 '18 at 05:20
  • @Blue Yes you are right, but there is no contradiction here: If $f(x)$ is differentiable this is true $\lim_{x\to 0} \frac {\lim_{x \to 0} f(x+\Delta x)-\lim_{x\to 0}f(x)}{\Delta x} =\lim_{x \to 0} \frac {f(x+\Delta x)-f(x)}{\Delta x}$ $$\\$$ which that $\lim_{x \to 0} f(x+\Delta x)$ and $\lim_{x \to 0} f(x)$ are finite. –  Apr 25 '18 at 05:43
  • @Blue I mean $\Delta x \to 0$ not $x \to 0$. –  Apr 25 '18 at 05:59
  • @Blue I forgot to thank you, for editing the question. :) If the solution is still problematic, I would like to learn the right solution. Thank you. –  Apr 25 '18 at 06:19
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    My previous comment stands: Nested limits with the same variable aren't sensical. The "inner" limit changes the variable into a number, so that it's no longer a variable for the "outer" limit to handle. That's about all I have to say here. You're welcome for the edit. :) – Blue Apr 25 '18 at 06:58
  • @Blue but if the second derivative true,my manipulation is correct. :) –  Apr 25 '18 at 07:10
  • @Beginner: The fact that $16/64 = 1/4$ doesn't make "just cancel the $6$s" a correct manipulation. ;) – Blue Apr 25 '18 at 07:13
  • @Blue I never really saw such coincidence :). By the way Please show that, $$\lim_{x\to 0}(\lim_{x\to 0} f(x)-\lim_{x \to 0} g(x))=\lim_{x\to 0}(f(x)-g(x))$$ is wrong. –  Apr 25 '18 at 07:41
  • @Beginner: It's not wrong (if the lims on $f$ and $g$ are individually defined), but it's not helpful. After all, if the lims are defined, then already $\lim f - \lim g = \lim(f-g)$; that's a rule. An extra $\lim$ on the left-hand side does nothing (no $x$s remain), so doing nothing to the right-hand side is equivalent, and your statement is vacuously true. But this has nothing to do with your argument above. In going from (2) to (3), you aren't merely absorbing a do-nothing limit; there's a denominator that still has a varying quantity, so the outer limit does something. – Blue Apr 25 '18 at 09:29
  • @Blue Yes, now I understand you. You are right. Well, What can we do? –  Apr 25 '18 at 09:57
  • @Blue You are right again...I looked/found at wikipedia. (Second derivative) And really, This "solution" is not rigorous. And there is a counterexample, which that the formula is "wrong". –  Apr 25 '18 at 10:20

2 Answers2

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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.

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

szw1710
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