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Math Help - solving limit using taylor series..

  1. #1
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    solving limit using taylor series..

    <br />
\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\<br />
    <br />
e^x=1+x+O(x^2)\\<br />
    <br />
e^{-x}=1-x+O(x^2)\\<br />
    <br />
xe^x=x+O(x^2)\\<br />
    <br />
cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\<br />
    <br />
\lim_{x->0} \frac{1-\frac{1}{2!}(x+O(x^2))^2+O(x^3)-1+\frac{1}{2!}(x+O(x^2))^2+O(x^3)}{x^3}<br />

    but i dont know how to deal with the remainders
    there squaring of them etc..

    ??
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  2. #2
    Member Mentia's Avatar
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    Apply L'Hopital a few times.

    You will get limit x-> 0

    (1/6)*((-3*E^x*(-2 + x)*(-1 + x)*Cos[x/E^x] -
    3*E^(5*x)*(1 + x)*(2 + x)*Cos[E^x*x] -
    E^(2*x)*(-3 + x)*Sin[x/E^x] + (-1 + x)^3*Sin[x/E^x] +
    E^(6*x)*(1 + x)^3*Sin[E^x*x] - E^(4*x)*(3 + x)*Sin[E^x*x])/E^(3*x))

    = -2
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  3. #3
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    i know that i can solve it with lhopital..
    but i want to solve it by the way i showed..
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  4. #4
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    The nice thing about taking a limit at zero in this case is you don't need to worry about the remainders because as you approach zero the series is exact to any order you want.

    So, take 3rd order:

     \lim _{x->0} cos(xe^x) = \lim _{x->0} 1-\frac{x^2}{2}-x^3

    No remainder because this is exact at limit x-> 0

    Similarly:

     \lim _{x->0} cos(xe^{-x}) = \lim _{x->0} 1-\frac{x^2}{2}+x^3

    Then, \lim _{x->0} cos(xe^x) - cos(xe^{-x}) = \lim _{x->0} -2x^{3}

    Then,

    <br />
\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3} = \lim _{x->0} \frac {-2x^{3}} {x^{3}} = -2<br />
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  5. #5
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    i dont want to use the x->0 in the middle of process

    i want to turn the last expression that i got
    <br />
\lim_{x->0} \frac{1-\frac{1}{2!}(x+O(x^2))^2+O(x^3)-1+\frac{1}{2!}(x+O(x^2))^2+O(x^3)}{x^3}<br />
    into 1 O() expression..
    Last edited by transgalactic; March 9th 2009 at 10:18 PM.
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  6. #6
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    i tried to solve it again
    <br />
\lim _{x->0} \frac{cos(xe^x)-cos(xe^{-x})}{x^3}\\<br />
    <br />
e^x=1+x+O(x^2)\\<br />
    <br />
e^{-x}=1-x+O(x^2)\\<br />
    <br />
xe^x=x+x^2+O(x^2)<br />
    <br />
xe^{-x}=x-x^2+O(x^2)<br />
    <br />
cos(x)=1-\frac{1}{2!}x^2+O(x^3)\\<br />
    <br />
\lim_{x->0} \frac{1-\frac{1}{2!}(x+x^2+O(x^2))^2+O(x^3)-1+\frac{1}{2!}(x-x^2+O(x^2))^2+O(x^3)}{x^3}=\\<br />
    <br />
=\lim_{x->0} \frac{1-\frac{1}{2!}(x^2+O(x^2))+O(x^3)-1+\frac{1}{2!}(x^2+O(x^2))+O(x^3)}{x^3}=0\\<br />

    the answer is 1/2
    why i got 0??
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  7. #7
    Member Mentia's Avatar
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    I'm sorry to disagree, but the answer is -2.

    Notice I didn't actually evaluate the limit until the very end. The remainders are unnecessary since we are taking a limit at zero.
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  8. #8
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    thanks i solved it
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