I am studying the alternative definition of the derivative, and I feel confused about it.
First, I know that the definition of the derivative is:
$$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\tag{1}$$
I also know that an alternative formula for the derivative is:
$$f'(x) = \lim\limits_{z \to x} \frac{f(z)-f(x)}{z-x}\tag{2}$$
However, I don't understand why the third formula below cannot be another alternative formula:
$$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}\tag{3}$$
I am aware of the fact that, if (3) exists, it is also the same derivative.
But I don't understand how (2) and (3) are different from each other. Please help me figure out this problem.
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abcd2022
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Where have you seen the 3rd formula? I have never seen this before. – Clyde Kertzer Mar 29 '22 at 21:50
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6@ClydeKertzer (and the OP) It's known as the symmetric derivative. – Theo Bendit Mar 29 '22 at 21:52
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Formula (1) does not depend on the sign of $h$. We ca write $f'(x)=\lim_{h\to 0} \frac{f(x-h)-f(x)}{-h}$. Average with (1) gives (3).
herb steinberg
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So, yes, if one wants to quibble, existence of the "symmetric derivative" is implied by, but does not imply, existence of the derivative. Not a big point, but a potential "gotcha" in some technical situations. :) – paul garrett Mar 29 '22 at 22:51
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It is possible the symmetric expression would not imply the existence of the derivative,. Example $f(x)=|x|$ at $x=0$. You need existence and then show equality. – herb steinberg Mar 30 '22 at 00:37