Introductory Real Analysis Problem?

Question

Let f: (a,b) --> R (the set of real numbers) be a differentiable function at x belongs to (a,b). Prove that the limit as h goes to zero of [f(x+h)-f(x-h)]/h exists and is equal to f ' (x). Is the converse statement true?

Answer 1

Actually, it's not equal to f'(x) unless f'(x) = 0. It is equal to 2f'(x)

[h→0]lim (f(x+h) - f(x-h))/h
[h→0]lim (f(x+h) - f(x) + f(x) - f(x-h))/h
[h→0]lim (f(x+h) - f(x))/h + [h→0]lim (f(x) - f(x-h))/h
f'(x) + [h→0]lim (f(x) - f(x-h))/h
f'(x) + [k→0]lim (f(x) - f(x+k))/(-k) (letting k=-h)
f'(x) + [k→0]lim (f(x+k) - f(x))/k
f'(x) + f'(x)
2f'(x)

The converse is not true -- [h→0]lim (|0+h| - |0-h|)/h exists, but |x| is not differentiable at zero.

Incidentally Gabe, stop deleting your questions after people answer them. Getting rid of the evidence of people's hard work in answering them is extremely rude.

Answer 2

i could say the two are advantageous. They pass hand in hand. Algebra is a device very like a hammer is to a wood worker. analysis is it is techniques. with out one or the different, no longer something would be built.

Answer 3

it depends how you define differentiability on first place
you can take what you wrote as definition already