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L'Hospital's Rule

  1. Jul 26, 2004 #1
    Having a very hard time solving this problem:

    Use L'Hospital's rule to show that

    [tex] \lim_{h\rightarrow0}\frac{f(x+h)-f(x-h)}{2h}=f^{'}(x)[/tex]

    Thanks.
     
    Last edited: Jul 26, 2004
  2. jcsd
  3. Jul 26, 2004 #2
    I can prove it without using l'hopital rule

    we know that f'(x) =lim as h->0 (f(x+h) - f(x) / h) by definition
    and also f'(x) = lim as h-> 0 (f(x) - f(x-h) /h )

    if we add both of them we get 2(f'(x)) = (f(x+h) - f(x) + f(x) - f(x-h) ) / h

    2(f'(x)) = (f(x+h) - f(x-h) ) / h
    divide by 2 on both sides
    f'(x) = (f(x+h) - f(x-h) ) / 2h
     
  4. Jul 26, 2004 #3
    Thanks for the solution. In fact, I did the same thing, except in the other way. But the signs were not working for some dumb reason, dumb stuff kept cancelling out.
     
  5. Jul 27, 2004 #4

    Galileo

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    Just use the chain rule:
    [tex]\frac{d}{dh}f(x\pm h)=\pm f'(x \pm h)[/tex]
    so you get
    [tex]\frac{f'(x+h)+f'(x-h)}{2}[/tex]
    then take the limit as h goes to zero.
     
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