Hi all, need some help please!
The question is as follows:
Suppose f(x) is differentiable at x = a, show that
lim as h->0 of (f(a+h)-f(a-h)) / 2h =f'(a)
I tried working from the difference quotient,
i.e: lim as h->0 of f(a+h) - f(a) / h, and i took this +h as the limit from the right side, and f(a-h) - f(a) / -h as limit from left side...is that along the right lines?
basically, you take the negative from that h and multiply the top part of the expression, and get -f(a-h) + f(a) / h for the left side, and since we know both are equal o f'(a), we can add them and say they, when added, are equal to 2f'(a), and the f(a) and -(fa) simply cancel out and you get:
2f'(a) = f(a+h) - f(a-h) / h
Then take the 2 across to get:
f'(a) = f(a+h) - f(a-h) / 2h
Does that make sense? Would i get any marks i wonder, is the logic 'followable'?...Please let me know! I'm sure there must be other ways too...