Proof of Differentiability at a Point Implies Limit Exists

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Discussion Overview

The discussion revolves around the proof that differentiability at a point implies the existence of a limit at that point. Participants explore the relationship between differentiability, continuity, and the conditions under which limits can be established, with a focus on finding specific delta values for given epsilon values in limit proofs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to prove that if a function is differentiable at a point, then the limit of the function at that point equals the function's value at that point, and seeks feedback on their proof.
  • Another participant references the mean value theorem, suggesting it relates to the proof being discussed and outlines how it can be used to establish a relationship between the function's values and its derivative.
  • A participant acknowledges familiarity with the mean value theorem but questions whether it provides a method to determine the maximum value M used in the limit argument.
  • Another participant clarifies that while there is no simple formula for M, it is a constant that approaches zero as x approaches a.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the mean value theorem to the proof. There is no consensus on the specific application of the theorem or the determination of the constant M.

Contextual Notes

The discussion includes assumptions about the continuity and differentiability of functions, and the implications of these properties on limit existence. The relationship between delta and epsilon is also a point of exploration, with no definitive resolution on the method for finding delta.

Who May Find This Useful

Readers interested in mathematical proofs, particularly in calculus and analysis, may find this discussion relevant, especially those exploring the concepts of differentiability and limits.

StephenPrivitera
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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 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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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.
 
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?
 
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.
 

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