Proof of Differentiability at a Point Implies Limit Exists

[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.

 

[HallsofIvy]

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.

 

[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?

 

[HallsofIvy]

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.