Real Analysis Question involving the definition of the derivative

philbein
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Homework Statement



Suppose f:(a,b) to R(Real numbers), is differentiable at any point x that is in the interval (a,b). Prove that lim(as h goes to 0) of (f(x+h)- f(x-h))/(2h) exists and equals the derivative of f(x) ( which is f '(x)). Give an example of a function where the limit exists, but the function is not differentiable.


Homework Equations



The main equation that we can use is
f '(x)= the limit as h goes to 0 of (f(x+h)-f(x))/h (definition of the derivative)
Probably also need the fact that x is in the domain, and that x+h is in the domain.

The Attempt at a Solution



I played around with the lim (h goes to 0) (f(x+h)-(fx-h))/2h, and got
(1/2)((f(x+h)-f(x))/h) + (1/2)((f(x)-f(x-h))/h) but I am stuck on where to go next. Any ideas would be great. Thanks
 
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Change h to -h in the second difference quotient. h can go to zero positively or negatively, can't it? Aren't they both (1/2)*f'(x)?
 
Look at
\frac{f(x+h)- f(x)+ f(x)- f(x-h)}{h}[/itex]<br /> How is that related to your problem and to the derivative?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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