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Passing the limit through the derivative of a differentiable function

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose f is differentiable on an open interval I and let x* [tex]\in[/tex] I. Show that there exists a sequence {x_n}[tex]\subset[/tex] I such that lim[n->inf]{x_n}=x* and lim[n->inf]{f'(x_n)}=f'(x*).

    2. Relevant equations

    We know that a function g is continuous iff for any sequence {x_n} with lim[n->inf]{x_n}=x*, lim[n->inf]{g(x_n)}=g(x*).

    3. The attempt at a solution

    I think I need to show that since f is differentiable on I, then its derivative is continuous on I, and since its derivative is continuous on I, then there exists a sequence {x_n} with lim[n->inf]{x_n}=x* for which lim[n->inf]{f'(x_n)}=f'(x*).

    But I am not sure how to show this, or even if its right.
  2. jcsd
  3. Jul 20, 2010 #2
    No, f differentiable on I does not imply its derivative is continuous. The canonical counterexample is f(x) = (x^2)*sin(1/x) if x =/= 0 and f(x) = 0 if x = 0.

    In the conclusion, you have a double limit. There is a certain procedure to apply to these limits that would make the problem very simple. The continuity of f would be essential in this case.

    But I think it's probably more illuminating to work from first principles. Write out the definition of the derivative of f at x* (epsilon-delta definition of limit), pick a specific sequence converging to x*, and make the appropriate modifications to your definition to show the conclusion.
  4. Jul 23, 2010 #3
    I did what snipez90 said, but I got stuck. I know that it is necessary to apply the Mean Value Theorem, but I don't really understand how. Maybe you can figure out how the MVT can be applied.
  5. Jul 23, 2010 #4
    Take your [tex] x \in I [/tex]. You know that there is a sequence of positive numbers, [tex] h_n [/tex] such that [tex] \lim_{n \rightarrow \infty} h_n = 0 [/tex]. Thus [tex] \lim_{n \rightarrow \infty} \frac{f(x + h_n) - f(x)}{h_n} = f'(x) [/tex].

    Now look at the difference quotient [f(x + h_n) - f(x)] / h_n . By the MVT [f(x + h_n) - f(x)] / h_n = f'(c_n) for x < c_n < x + h_n

    As n goes to infinity that difference quotient gets closer and closer to the derivative of f at x, right? So what can you say about the sequence f'(c_n)?
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