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Derivatives and Continuity

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    If the continuous function f(x) has a derivative f'(x) at each point x in the neighborhood of [tex]x=\xi[/tex], and if [tex] f'(x)[/tex] approaches a limit L as [tex] x \rightarrow \xi[/tex], then show [tex]f'(\xi)[/tex] exists and is equal to L.


    2. Relevant equations



    3. The attempt at a solution

    Since the derivative exists at each point x in the neighborhood of [tex]x = \xi[/tex] and f'(x) tends to a limit L as [tex]x \rightarrow \xi[/tex], we have [tex] |f'(x) - L| < \epsilon [/tex] whenever [tex]|x-\xi| < \delta[/tex]. Since we can take x arbitrarily close to [tex]\xi[/tex], we can write [tex] x = \xi + h[/tex] for any real h. In this case, we will take h to be positive. We then have [tex] |f'(\xi + h) - L| < \epsilon[/tex] whenever [tex] |h| < \delta [/tex].

    Now, choose a positive quantity h* such that [tex] 0 < h < h* < \delta [/tex]. The following inequality is then true: [tex] \xi < \xi + h < \xi + h* [/tex]. We have formed an open neighborhood about the point [tex] x = \xi + h [/tex]. Then by the Mean Value Theorem, we have [tex]f'(\xi + h) = \frac{f(\xi + h) - f(\xi)}{h*}[/tex]. So we then have [tex] |\frac{f(\xi + h) - f(\xi)}{h*} - L| < \epsilon[/tex] whenever [tex] |h*| < \delta[/tex]. This inequality says that the limit of the quotient [tex] \frac{f(\xi + h) - f(\xi)}{h*}[/tex] as h* approaches 0 through positive values exists, and is equal to L. Using a similar argument it is easy to see that for h* tending to 0 through negative values, the limit is again L. Thus the derivative at [tex]f'(\xi)[/tex] exists and is equal to L.


    Is this proof correct?
     
    Last edited: Jan 14, 2009
  2. jcsd
  3. Jan 15, 2009 #2

    Dick

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    There's definitely some problems though I think you've got the right idea. The MVT doesn't say
    [tex]
    f'(\xi + h) = \frac{f(\xi + h) - f(\xi)}{h*}
    [/tex]
    It says
    [tex]
    f'(\xi + h) = \frac{f(\xi + h*) - f(\xi)}{h*}
    [/tex]
    And I think you should be able to write this much more simply, e.g. the limit h->0 of (f(xi+h)-f(xi))/h is f'(xi). The MVT says there is a point y_h in [xi,xi+h] such that f'(y_h) is the same as the difference quotient. As h->0, y_h->xi so f'(y_h)->L. And there's really no need to split it into sides, you don't have any one sided limits or derivatives.
     
  4. Jan 15, 2009 #3
    Oops, I mean't to write [tex] f(\xi + h) = \frac{f(\xi+h*) - f(\xi)}{h*}[/tex] for h* > h. I chose h* > h because if I chose h* < h, then I have xi < xi + h* < xi + h, and I cannot write out the derivative of xi + h as the difference quotient using xi and xi + h* (I believe the MVT says that xi + h must be contained in the open interval (xi, xi+h*).) But choosing h* so that h < h* < delta, h* goes to 0 and the quotient [f(xi + h) - f(xi)]/[h*] will go to L since the quotient - L is less than epsilon.


    I see what you're saying though. I have made the problem more complicated than I had to. But with my correction above, is my proof correct?
     
    Last edited: Jan 15, 2009
  5. Jan 15, 2009 #4

    Dick

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    I believe (f(x+h)-f(x))/h approaches f'(x). And I believe there is an 0<h*<h such that f'(x+h*)=(f(x+h)-f(x))/h. But I don't think (f(x+h)-f(x))/h* has to approach anything.
     
  6. Jan 15, 2009 #5
    But if you have 0 < h < h* then you have [tex]f'(\xi + h) = \frac{f(\xi + h*) - f(\xi)}{h*} [/tex] which approaches L as h* approaches 0. Since L is a definite value, then the limit exists. Isn't this the definition of a derivative or am I missing something here...
     
    Last edited: Jan 15, 2009
  7. Jan 15, 2009 #6

    Dick

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    That's fine. As h*->0 the left side approaches L because the limit of the derivative exists and since h*>h>0 that means h->0 which means the right side approaches f'(xi) since f is differentiable at xi. That's the proof all right. I was just worried because you were writing expressions like (f(xi+h)-f(xi))/h*, mixing h and h* in the difference quotient.
     
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