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Evaluating Limit using Maclaurin series

  1. Aug 3, 2010 #1
    gif.latex?\lim_{x&space;\to&space;0}\frac{\sin&space;x-x}{2e^x-2-2x-x^2}.gif

    This is an example given in my textbook. The final answer is 1/6. I know that sinx and 2e^x have to be replaced with their corresponding Maclaurin series. However, I'm having trouble understanding the steps they took to get the limit in a form in which it could be evaluated by substituting x=0.

    If anyone could help, I'd appreciate it!
     
  2. jcsd
  3. Aug 3, 2010 #2

    Dick

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    Substitute the series. What's the lowest power of x that doesn't cancel in numerator and denominator? Cancel it as a common factor.
     
  4. Aug 3, 2010 #3

    vela

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    The limit is actually equal to -1/2, not 1/6, so if you think you don't understand it because you're getting the "wrong" answer, that might be why.
     
  5. Aug 3, 2010 #4

    Mark44

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    I concur with vela on the value of the limit.

    If you don't actually have to use Maclaurin series to evaluate this limit, it can be found by applying L'Hopital's Rule a few times.
     
  6. Aug 4, 2010 #5
    Yes, the answer is -1/2!

    However, I'm still having some problems.

    2k+1&space;\right&space;)!}}{\frac{2x^3}{6}+x^4\sum_{k=4}^{\infty}\frac{2x^{k-4}}{k!}}.gif

    I'm not understanding how they went from the limit on the left to the one on the right.
     
  7. Aug 4, 2010 #6

    Mark44

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    Don't write sin(x) and 2ex as summations. Instead, write sin(x) as x - x3/3! + x5/5! -+ ...
    and write 2ex as 2(1 + x + x2/2! + x3/3! + ...)
     
  8. Aug 4, 2010 #7

    vela

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    They just wrote the first term of the series explicitly and then factored out the common power of x from the remaining terms. It's easier to see what's going on in this type of problems if you just write out the series like Mark44 suggested.
     
  9. Aug 5, 2010 #8
    Thanks guys!
     
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