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Limit problem

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate limx->0 (e^x - 1- x - (x^2/2))/x^3

    3. The attempt at a solution

    I can't remember how to solve this limit. Do I need to evaluate each part seperately? I plugged in the 0 to find that the limit does exist. I just can't seem to figure out what to do next.
     
  2. jcsd
  3. Jan 5, 2012 #2
    Why not use l'Hôpital's rule?
     
  4. Jan 5, 2012 #3

    micromass

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    Or use the Taylor series of the exponential function.
     
  5. Jan 5, 2012 #4
    If I use l'Hopital's rule won't I end up with 3x^2 at the bottom?
     
  6. Jan 5, 2012 #5

    micromass

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    You will want to use l'Hopital multiple times.
     
  7. Jan 6, 2012 #6
    The limit of a sum is the sum of limits - Lim[a+b]=Lim[a]+Lim
    Then as everyone before has said, you're going to want to use l'hopitals rule for the last term
     
  8. Jan 7, 2012 #7
    I used L'Hopital's rule three times and ended up with limx->0(e^x)/5

    Is this correct? It still gives me 0 on top.
     
  9. Jan 7, 2012 #8
    Here is how I did it in case it might help

    lim->0 (e^x)' - 1' - x' - (x^2/2)' /x^3'

    = lim->0 (e^x' - 1' - x')/ 3x^2'

    = lim->0 (e^x' - 1') / 5x'

    = lim x->0 (e^x) /5
     
  10. Jan 7, 2012 #9
    Very close, but look at your second to thrid line in the denominator.

    What is: $$\frac{d}{dx}(3x^{2})$$

    After you fix that, evaluate it at 0.

    [tex]
    \lim_{x\to 0}~ e^{x}
    [/tex] should not be 0.
     
    Last edited: Jan 7, 2012
  11. Jan 7, 2012 #10
    Thanks!
    It should be 6x right?

    So is 1/6 the correct answer?
     
  12. Jan 7, 2012 #11

    SammyS

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    1/6 is the correct result.
     
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