Derivative Proof: Prove f'(x)=lim x→0 (f(x+h)-f(x-h))/2h

[v0id19]

Homework Statement

if f' is continuous, show that:
\mathop{\lim}\limits_{x \to 0}(\frac{f(x+h)-f(x-h)}{2h})=f'(x)
be sure to explain why f' must be continuous

Homework Equations

not really any equations, this is for AP Calc BC and we've just done L'Hospital's theorem and the derivatives/integrals of logs and inverse trig functions.

The Attempt at a Solution

I know that as x-->0 it becomes \frac{f(h)-f(-h)}{2h} . I thought about proving that the difference quotient can be manipulated into the above formula, but haven't had any success.

Any pointers?

 

[jgens]

Are you certain that it's supposed to be as x -> 0 and not h? Aside from that, my only suggestion is to add and subtract f(x)/2h and apply some limit lemmas.

 

[v0id19]

yeah it says x-->0
what is a 'limit lemma'?

 

[snipez90]

I think jgens is right, it's likely h -> 0.

Hmmm, I'm surprised this is Calc BC. I think I've seen this problem in Spivak before.

The trick is to consider the difference quotient and replace h with -h, and then take the limit as h -> 0. Then you should be able to work with the expression given.

EDIT: Ok, this is a slightly different problem. Hmmm, see if you can make use of the definition of continuity, and the equivalence of the statements lim x -> a f(x) and lim h -> 0 f(a+h).

 

[jgens]

I think this approach may work:

Derivative of a function: y = f(x) than y - dy = f(x - h), hence dy/dx = lim h -> 0 (f(x) - f(x - h))/h

Given (f(x + h) - f(x - h))/2h = (f(x +h) - f(x) + f(x) - f(x - h))/2h. Take the limit as h -> 0 and I think that should yield a solution.

A limit lemma would be lim x-> c kf(x) = klim x -> c f(x) where k is a constant.

 

[v0id19]

thanks i'll give that a try--my only question is how do i get it from x-->0 to h-->0?

 

[v0id19]

So it turns out my math teacher actually did write the question wrong... it really is h-->0. it's solved with a simple l'hospital. thanks guys