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L'Hopital's Rule: Advanced Analysis

  1. Oct 26, 2008 #1
    Suppose f is defined in a neighborhood of x, and suppose f '' (x) exists. Show that:

    lim [f(x+h)+f(x-h)-2f(x)] / h^2 = f''(x).
    h->0

    Show by an example that that the limit may exist even if f '' (x) may not. (hint: use lHopital's Theorem).

    Proof:

    f '' (x) exists implies f ' (x) exists and is differentiable which implies f(x) exists and is differentiable.

    Thus,

    f '' (x) =

    lim [f ' (x) - f ' (t)] / (x-t) (def of derivative)
    x->t

    Doing a substitution h = x - t

    lim [f ' (x+h) - f ' (x)] / h (def of derivative)
    h->0

    = lim [f(x+h) - f(x) - (f(x) - f(x-h)] / h / h
    h->0

    = lim [f(x+h) + f(x-h) - 2f(x)] / h^2 (*)
    h->0

    QED

    What I'm having trouble with is the example of where the limit exists but f '' (x) does not. Either using the original limit/derivative definition or the end of the proof (*) I get 0/0 or another indeterminate form.
     
  2. jcsd
  3. Oct 26, 2008 #2

    HallsofIvy

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    The very simplest function I know that does not have a derivative at one point is |x|, which is not differentiable at x=0. The anti-derivative of that, i.e. f(x) such that f'(x)= |x|, does not have a second derivative at x= 0. Will that work?
     
  4. Oct 26, 2008 #3
    I tried both absolute value of x and square root of x and still get 0/0 or if I apply l'Hopital's Rule I get infinity. Which also means the limit does not exist.
     
  5. Oct 26, 2008 #4

    HallsofIvy

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    So you have not tried what I suggested? Why not?
     
  6. Oct 26, 2008 #5
    I did.

    The problem is if f ' (x) = |x| then f '' (0) does not exist, but neither does:

    lim [f (0 + h) + f (0 - h) - 2 f (0)] / h^2, unless I am missing something here???
    h->0
     
  7. Oct 26, 2008 #6

    HallsofIvy

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    That limit does indeed exist. What is f(x) in this case?
     
  8. Oct 26, 2008 #7
    I see my mistake.

    Thanks!
     
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