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Math Help - Find limit using taylor series

  1. #1
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    Find limit using taylor series

    I am supposed to find the limit as x -> 0 of (e^(x^2) - 1 - x^2) / (x sinx - x^2).

    I do not know if my process or answer is at all correct, but here is what I have done:

    I found the taylor series for e^(x^2) = 1 + x^2 + x^4/2! + x^6/3! ...

    then I added in the rest of the numerator (-1 - x^2) to get:
    x^4/2

    for the denominator, I found the taylor series for sinx and then added and multiplied in the rest of the denominator to get:
    -x^4/6 + x^6/5! -x^8/7! ....

    taking the ratio of the x^4 terms give me 1/2 * -6/1 = -3

    is this correct?

    if so, is it just normal practice to ignore the rest of the terms in the series and just focus on the first one? Thanks!!
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    1,458
    Hi

    Your work is OK

    You have found an equivalent to \frac{e^{x^2} - 1 - x^2}{x \sin x - x^2} in the neighborhood of 0 at the first order which is the same as the limit of the function at 0

    If you want to find an equivalent at the second order

    \frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}+\frac{x^6}{120}} = \frac{\frac{x^4}{2}+\frac{x^6}{6}}{-\frac{x^4}{6}\left(1-\frac{x^2}{20}\right)} = \frac{-3-x^2}{1-\frac{x^2}{20}}

    which gives (-3-x^2)\left(1+\frac{x^2}{20}\right) = -3-\frac{23}{20}\:x^2
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