# A general e-d proof

1. Oct 12, 2003

### StephenPrivitera

What I tried to do with this proof is show that if a function is differentiable at a point, then the limit of the function at this point is the value of the function at that point. This seems like a pretty obvious point (since differentiability depends on continuity and continuity depends on the existence of the limit), so I tried to go a step further and identify a specific delta for each epsilon that would guarantee the existence of this limit. I was hopefully successful.
I spent about two hours working out the details (don't worry - it's not too long) and when I finally finished I was too exhausted to see if my proof stood up to a bunch of specific tests. Anyway, I was hoping someone more experience in math could check my proof. If it's wrong, show me where.
The purpose of this proof is to make finding deltas easy when I know that a function is differentiable. This is acceptable because finding delta is not a necessary part of a limit proof. If my first guess for delta work for all epsilons, then I have completed the proof.

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2. Oct 13, 2003

### HallsofIvy

Staff Emeritus
Have you had the "mean value theorem" in calculus yet?

That looks to me like what you are trying to do.

The mean value theorem says that if f is continuous on some interval [a,b]and differentiable on (a,b) then there exist some point c in (a,b) such that (f(b)- f(a))/(b-a)= f'(c). From that,
f(b)- f(a)= f'(c)(b-a). In particular, if f' is continuous on (a,b) (which will always be true if f is also differentiable at a and b), then |f'(c)| has a maximum value on [a,b], say M, and we can say
|f(b)- f(a)|<= M|b-a|. Finally, given epsilon> 0, we can take
delta= epsilon/M.

3. Oct 13, 2003

### StephenPrivitera

Yes, I have studied the mean value theorem before (although not in the form you presented), but I was not thinking of this proof in that context. Does the mean value theorem allow you to find the value of M?

4. Oct 19, 2003

### HallsofIvy

Staff Emeritus
NO, there is no simple formula for M but it doesn't matter- it's a constant and so as x-> a M|x-a| goes to 0.