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Homework Help: L'Hospital's Rule

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data

    lim -> 0 (sin(x)-x) / x^3

    2. Relevant equations

    L'Hospital's Rule

    3. The attempt at a solution

    I could just use L'Hospital's rule since it's 0/0. However, the answer is wrong when I do it that way. What am I missing? It states on the question that it must be rewritten before the ruel can be applied and that it has to be applied more than once.

    Please help.
  2. jcsd
  3. Nov 12, 2009 #2


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    will have a closer look, but one way to gain better understading about the problem would be to expand the sin in a taylor series about 0....
  4. Nov 12, 2009 #3


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    show your L'Hop working, i think it works fine, but has to be applied twice
  5. Nov 12, 2009 #4
    I don't see why you'd need to rewrite anything. You just have to argue that

    lim x->0 f/g = lim x->0 f'/g' = lim x->0 f''/g'' = ... is true for the nth derivate as long as the limit of n-1th derivate is either zero or infinite for both f and g.
  6. Nov 12, 2009 #5
    The first time applied gives me: (cosx - 1) / 3x^2

    The second: (-sinx) / 6x

    I definitely know this is wrong...
  7. Nov 12, 2009 #6
    ... and what would the third give you?
  8. Nov 12, 2009 #7
    Third would be -cosx/6
  9. Nov 12, 2009 #8

    Gib Z

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    Pretty much what applying L'Hopital's rule does really.
  10. Nov 12, 2009 #9


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    And the limit of that as x goes to 0 is ?
  11. Nov 12, 2009 #10
    Try expanding sin(x) using the Maclaurin series.
    You will get the answer directly.

    Attached Files:

  12. Nov 12, 2009 #11
    That's not the answer according to calculators.
  13. Nov 12, 2009 #12
    According to Mathematica it is.
  14. Nov 12, 2009 #13


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    Homework Helper

    statements like that, with very little information, are hard to help with...

    show your method & results and explain why you think there is a disconnect, and we can comment/help out...
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