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I have the limit $\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h}$. Using the mean value theorem, we have $f''(c)$ for some $c\in[a-h,a+h]$. However, we can also use L'Hospital's rule, which reduces the limit to $\lim_{h\rightarrow0}\frac{f''(a+h)+f''(a-h)}{2}$, or $f''(a)$. I'm not sure why there is a discrepancy, and I'm almost completely certain I didn't make any careless mistakes. We can assume that the first and second derivatives are continuous.

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    $\begingroup$ For $h$ tends to zero, then $c$ tends to $a$ $\endgroup$ Jun 28, 2019 at 17:54

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As $h\rightarrow0 , c\rightarrow a $ [By squeeze Theorem or Sandwich Theorem ] So there is no discrepancy and they both point to the same thing ie

$$\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} = f''(a) = f''(c)$$

Also there is a method 3 which can be employed ie $$\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} =$$

$$\lim_{h\rightarrow0}\frac{(f'(a+h)-f'(a))-(f'(a-h)-f'(a))}{2h} =$$

$$\frac{\lim_{h\rightarrow0}\frac{(f'(a+h)-f'(a))}{h} +\lim_{h\rightarrow0}\frac{(f'(a-h)-f'(a))}{-h}}{2}= $$

$$\frac{f''(a)+ f''(a)}{2}=f''(a)$$

As $\lim_{h\rightarrow0}\frac{(f(a+h)-f(a))}{h} = f'(a)$

$\lim_{h\rightarrow0}\frac{(f(a-h)-f(a))}{-h} = f'(a)$

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The mean value theorem does not say that the limit is $f''(c)$, only that there exists $c\in[a-h,a+h]$ with the expression equivalent to $f''(c)$. The resolution of the problem comes from noticing that $c$ and $f''(c)$ depend on $h$, and then $\lim_{h\to0}c=a$ by the squeeze theorem.

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