- #1

StephenPrivitera

- 363

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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

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.