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If f is twice differentiable, prove that $$ \lim_{h \to 0} \frac{f(c+h)- 2 f(c) + f(c-h)}{h^2}=f''(c)$$ for $f:(a,b) \to \Bbb R$ , $c \in (a,b)$.

I am trying to start from the RHS and get to the LHS , I was able to prove LHS=RHS using L hospital. This was my progress so far: $$\lim_{a\to 0}{\frac{f'(c+a)-f'(c)}{a}}=\lim_{a\to 0}{\frac{\lim_{h \to 0}\left[\frac{f(c+a+h)-f(c+a)}{h}-\frac{f(c+h)-f(c)}{h}\right]}{a}}$$ But I don't really know if you could form them into a single limit , or anything else. And what to do after.

PS: I'm in highschool, so I won't be able to understand solutions involving undergraduate math, like real analysis, etc.

Edit 1: replaced h with $h^2$.

Edit 2: was able to start from lhs to go towards rhs but not the other way around.

user170231
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q123LsaB
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    Doesn’t this look like $$\lim_{h\to 0}\dfrac{[f(c+h)-f(c)]+[f(c-h)-f(c)]}{h}=2f’(c)?$$ – insipidintegrator Jul 27 '22 at 16:47
  • $$\lim_{h \to 0} \frac{ f{c+h)- f(c-h)}{h} = 2 f'(c)$$ if i recall correctly. – q123LsaB Jul 27 '22 at 16:49
  • Do you know L'Hospital's Rule? If so, apply it to to the limit expression that appears in the title (you need $h^2$ in the denominator, not $h$). What do you find? Now apply once more. – Mark Viola Jul 27 '22 at 16:50
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    The statement is incorrect. The limit is $2f'(c)$, as pointed out in the first comment. – uniquesolution Jul 27 '22 at 16:51
  • See: "Taylor polynomial" in your calculus textbook. – GEdgar Jul 27 '22 at 16:51
  • @insipidintegrator It can and does happen twice since the correct denominator is $h^2$, not $h$. – Mark Viola Jul 27 '22 at 16:52
  • $$\lim_{h \to 0} \frac{f(c-h)- f(c)}{-h} = f'(c)$$, @insipidintegrator so the formula you used was for:$$\lim_{h \to 0} \frac{f(c+h)- f(c-h)}{h} = 2 \cdot f'(c)$$ – q123LsaB Jul 27 '22 at 16:53
  • replaced, h with $h^2$. sorry for the inconvinience. – q123LsaB Jul 27 '22 at 16:56
  • Does the alleged duplicate really answer this question? – GEdgar Jul 27 '22 at 16:57
  • @GEdgar How does the new "alleged duplicate" sound? – Mark Viola Jul 27 '22 at 17:03
  • Okay, i could use the Left hand side expression, ie : $$\lim_{h \to 0} \frac{f(c+h)- 2 f(c) + f(c-h)}{h^2}$$ to equate to f''(x) by Lhospital. But if i as to derive if from first principle, how do i get f''(x) in the form of that, ie start from right hand side, and equate it towards the left. – q123LsaB Jul 27 '22 at 17:10
  • @MarkViola ... The new one is better. But not perfect. The "asymptotic expansion" used by user21820 is a good way to do this. [But the proof of it given there requires $f''(x)$ exists for $x$ near $c$. Instead, you could cite Taylor's theorem, requiring only that $f''(c)$ exist at that one point.] – GEdgar Jul 27 '22 at 17:12
  • @GEdgar Hi Edgar. I believe that L'Hospital's Rule is the better way to go, especially given that this is a high school student. – Mark Viola Jul 27 '22 at 17:52
  • @q123LsaB Applying LHR twice, we find that $$\begin{align}\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}&=\lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}\\&=\lim_{h\to 0}\frac{f''(x+h)+f''(x-h)}{2}\\&=f''(x)\end{align}$$. – Mark Viola Jul 27 '22 at 17:56
  • @MarkViola mhm, i was able to use L hospital to from the expression, after the comments suggested so. I wanted to know if you could start from f"(x) and end up to the lhs expression – q123LsaB Jul 27 '22 at 17:58
  • I think one can do$$\begin{align}f''(c)&=\lim_{h\to0}\frac{f'(c+h)-f'(c)}h\[1ex]&=\lim_{h\to0}\frac1h\left(\lim_{h\to0}\frac{f(c+2h)-f(c+h)}h-\lim_{h\to0}\frac{f(c+h)-f(c)}h\right)\[1ex]&=\lim_{h\to0}\frac{f(c+2h)-2f(c+h)+f(c)}{h^2}\[1ex]&=\lim_{h\to0}\frac{f(c^+h)-2f(c^)+f(c^-h)}{h^2}\end{align}$$where $c^=c+h$ in the last line. I'd like to hear any reasons why this doesn't work if one cannot. – user170231 Jul 27 '22 at 19:15
  • @user170231 your second equation is wrong. The term in brackets does not depend on $h$ anymore, so you have something like $\lim_{h\to 0}\frac{1}{h}\cdot c$ – Gono Jul 27 '22 at 19:51
  • @user170231 - Your big mistake is $$\lim_{h\to0}\frac{f'(c+h)-f'(c)}h=\lim_{h\to0}\frac1h\left(\lim_{h_1\to0}\frac{f(c+h+h_1)-f(c+h)}{h_1}-\lim_{h_2\to0}\frac{f(c+h_2)-f(c)}{h_2}\right)$$ The three separate limits each have their own variable. You have to justify combining the separate limits into a single limit. Just using the same letter for the dummy variables is not enough. – Paul Sinclair Jul 28 '22 at 17:05
  • Right, thank you both for pointing that out. I suppose we can invoke the property $D(f+g)=Df+Dg$ to write$$f'(c+h)-f'(c)=(f(c+h)-f(c))'=\lim_{h_1\to0}\frac{(f(c+h+h_1)-f(c+h_1))-(f(c+h)-f(c))}{h_1}$$to eliminate one of the limit variables but that doesn't seem too helpful. – user170231 Jul 28 '22 at 17:26

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