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Is function of convergent sequence rate of change equal to derivative?

  1. Sep 16, 2010 #1
    Given a convergent sequence [tex]x_n \rightarrow x[/tex] and a function f, is
    [tex] lim_{n \rightarrow \infty} \frac{f(x_n) - f(x_{n-1})}{x_n - x_{n-1}} = f'(x) [/tex] ?

    I believe it it is, but I haven't been able to figure out how to prove it. Does anyone know of a proof or counter-example?

    And probably should add [tex] x_n \not = x_{n-1} \forall n [/tex]
     
    Last edited: Sep 16, 2010
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  3. Sep 16, 2010 #2

    HallsofIvy

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    Saying that
    [tex]\lim_{x\to a} f(x)= F[/tex]
    is exactly the same as saying
    [tex]\lim_{n\to\infty} f(x_n)= F[/tex]
    where
    [tex]\lim_{n\to\infty} x_n= a[/tex]

    so, yes, your equation is correct.
     
  4. Sep 16, 2010 #3

    Office_Shredder

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    Certainly if the derivative is not continuous this isn't going to work. For the function f(x)=x2sin(1/x) when x is not 0, f(0)=0, this is differentiable with a derivative of 0 at 0. But its derivative is 2xsin(1/x)-cos(1/x) is not continuous (see wolfram alpha for a graph) and you can find a sequence of points converging to zero such that the difference quotient evaluates to whatever you want it to be really
     
  5. Sep 16, 2010 #4
    We may assume f and f' are continuous, since that is not the main question I am concerned with.


    It is not quite the same thing, since that is not how the derivative is defined - all definitions I've seen have one value fixed. I.e. just plugging in [tex] x_n [/tex] to [tex] f'(x) [/tex] does not give the same function as I described, so sequential characterization doesn't work directly:

    [tex] f'(x_n) = \lim_{x \rightarrow x_n} \frac{f(x) - f(x_n)}{x-x_n} [/tex]

    or we could define it as

    [tex] f'(x) = \lim_{n \rightarrow \infty} \frac{f(x) - f(x_n)}{x-x_n} [/tex]

    However neither of these are the same as the limit I gave, in which both [tex] x_n [/tex] and [tex] x_{n-1} [/tex] are changing sequences, and aren't fixed. Even if accepted as another definition, then can you offer a proof the definitions are equivalent?


    The problem I was stuck with when I tried to use something like
    [tex] f'(x_n) = \lim_{x \rightarrow x_n} \frac{f(x) - f(x_n)}{x-x_n} [/tex]
    to get a bound, is that then any delta requirement depends on the specific value of n, but I may need to choose n large enough so that [tex] x_n [/tex] and [tex] x_{n-1} [/tex] meet the delta requirement, but that ends up potentially changing [tex] x_n [/tex] and the delta needed again since it depends on the specific x_n, so it's sort of like a race condition.
     
    Last edited: Sep 16, 2010
  6. Sep 16, 2010 #5
    Well if f' is continuous at x and f' is defined in a neighborhood of x, you should be able to apply the mean value theorem to the interval consisting of the endpoints x_(n-1) and x_n (taking n to be large enough to begin with). Then letting n approach infinity and using the continuity of f' at x seems to establish the desired equality.

    Note that I did in fact consider Office_Shredder's function (since it is more or less the canonical example of a function whose derivative exists everywhere but does not have a continuous derivative), and for obvious sequences tending to x = 0 I think the equality still holds.
     
  7. Sep 16, 2010 #6

    Ah, thanks! I didn't even think of using the mean-value theorem...
     
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