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Math Help - limit problem 4_34

  1. #1
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    limit problem 4_34

    Help me to solve this problem

    Use Taylor's theorem to compute :

     lim_{x \to 0} \frac{sin(sinx)-x(1-x^2)^{\frac{1}{3}}}{x^3}
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  2. #2
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    \frac{\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{10}-\frac{8x^{7}}{315}+\frac{13x^{9}}{2520}-............\right)-\left(x-\frac{x^{3}}{3}-\frac{x^{5}}{9}-\frac{5x^{7}}{81}-\frac{10x^{9}}{243}-......\right)}{x^{3}}

    =\frac{19x^{2}}{90}+\frac{103x^{4}}{2835}+\frac{31  51x^{6}}{68040}+................

    \lim_{x\to 0}\left[\frac{19x^{2}}{90}+\frac{103x^{4}}{2835}+\frac{315  1x^{6}}{68040}+................\right]=0
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  3. #3
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    Thanks , but how you know how todo Taylor's theorem on this term :
    -x(1-x^2)^\frac{1}{3}
    ?
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  4. #4
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    Quote Originally Posted by gilyos View Post
    Thanks , but how you know how todo Taylor's theorem on this term :
    -x(1-x^2)^\frac{1}{3}
    ?
    You don't need to use Taylor's Theorem to do that

    you should expand it by binomial theroem

     (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 +  \frac{n(n-1)(n-2)(n-3)}{4!} x^4 +  \frac{n(n-1)(n-2)(n-3)(n-4)}{5!}x^5 + ...

    In this case  n= 1/3  , x <-> -x^2


     ( 1- x^2)^{1/3} = 1 - \frac{x^2}{3} + ....

    the next terms are left to you since I don't get a pencil and a piece of paper right now .

    You will see they are exactly what galactus has written for us
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