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This exercise is from my textbook:

Let $I\subseteq \mathbb{R}$ be an open interval, let $f : I\rightarrow \mathbb{R}$ be differentiable on $I$, and suppose $f''(a)$ exists at $a\in I$. Show that:

$$f''(a) = \lim_{h\to 0} \frac{f(a+h) - 2f(a) + f(a-h)}{h^2}$$

which I solved using L'Hospital's rule, but I wasn't satisfied with that so I started to looking for an alternative proof and I found the following which uses the definition of derivative:

\begin{equation*} \begin{split} f''(a) &= \lim_{h\to 0} \frac{f'(a) - f'(a-h)}{h}\\ \\ &= \lim_{h\to 0} \frac{\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} - \lim_{h\to 0}\frac{f(a)-f(a-h)}{h}}{h}\\ \\ &= \lim_{h\to 0}\frac{\frac{f(a+h)-f(a)}{h} - \frac{f(a)-f(a-h)}{h}}{h}\\ \\ &= \lim_{h\to 0} \frac{f(a+h) - 2f(a) + f(a-h)}{h^2} \end{split} \end{equation*}

Is this valid? It looks more like an abuse of notation.

Jorge S.
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    It's not exactly rigorous. You should use different variables for each of your limits so that you avoid conflating them as you're doing here. – jgon May 18 '22 at 03:56
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    I agree with "abuse of notation." The standard method is to apply the Cauchy Mean Value theorem on the interval $[0,h]$ to obtain $\tau \in (0,h)$ with $$ \frac{f(x+h) + f(x-h) -2 f(x)}{h^2} = \frac{f'(x+\tau) - f'(x-\tau)}{2\tau} \to f''(x)$$ as $h\to 0$. Similar in spirit to your idea. – B. S. Thomson May 18 '22 at 04:03
  • There are a number of other methods given before on this site. Note, in particular, that this expression is called the second symmetric derivative and plays an important role in many studies, convexity certainly: https://math.stackexchange.com/questions/1809060/proof-of-the-second-symmetric-derivative – B. S. Thomson May 18 '22 at 04:12
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    Let me try something similar in a different circumstance. Assume $f$ is continuous at $x$. Then:\begin{align}f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x + h) - \lim_{h \to 0}f(x + h)}{h} \&= \lim_{h \to 0} \frac{f(x + h) - f(x + h)}{h} = \lim_{h \to 0} 0 = 0.\end{align}So, every continuous function is differentiable, with derivative $0$. This conclusion seems highly suspicious... – Theo Bendit May 18 '22 at 04:13
  • @TheoBendit Maybe I am missing something but I think the result you got should be an indeterminate form $0/0$ to be precise..instead of $0$. –  May 18 '22 at 04:37
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    @RamanujanXXV The numerator of $\frac{f(x + h) - f(x + h)}{h}$ is does not just approach $0$ as $h \to 0$, it is zero, constantly. Yes, it is an indeterminate form, but at the same time (and without contradiction), it evaluates to $0$, because it's the limit of a function that is constantly $0$. Indeterminate forms can still be $0$ too! That's not where the problem lies. The problem lies in the steps where the inner limit is introduced, with the same dummy variable as the outer limit, and where that inner limit just disappears. The rest is sound. – Theo Bendit May 18 '22 at 05:09

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No, you haven't justified any of the steps. In your chain of four equalities, I'd say the second is an abuse of notation (absolutely terrible abuse of notation; never overuse dummy variables in several spots, That's one of the easiest ways to get confused and give incorrect answers/justifications) while the third is completely unjustified.

If all you do is apply the definitions (and some minor theorems), then this is what you should end up with: \begin{align} f''(a)&=\lim_{h\to 0}\frac{f'(a)-f'(a-h)}{h}\\ &=\lim_{h\to 0}\frac{\left(\lim\limits_{k\to 0}\frac{f(a+k)-f(a)}{k}\right) - \left(\lim\limits_{z\to 0}\frac{f(a-h+z)-f(a-h)}{z}\right)}{h}. \end{align} Each occurrence of a derivative has a limit of difference quotients. Now, you have to justify why we are allowed to use one and the same variable $h$ in all three cases and why we can dispense with the inner two limits. This is not a trivial matter at all. You have to use some further techniques, all involving some form of the mean-value theorem (which tells us how the derivative of a function is related to differences in the original function) in the guise of L'Hopital's rule or Taylor's theorem.

peek-a-boo
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