Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  2. jcsd
  3. Sep 16, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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]
    [tex]\lim_{n\to\infty} x_n= a[/tex]

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Is function of convergent sequence rate of change equal to derivative?