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Math Help - limit question using power series

  1. #1
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    limit question using power series

    \lim_{x \to 0} \frac{e^{-3x^3} - 1 + 3x^3 - \frac{9}{2}x^6}{2 x^9}<br />


    It's telling me to use power series to find the limit, and I am lost.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by fastcarslaugh View Post
    \lim_{x \to 0} \frac{e^{-3x^3} - 1 + 3x^3 - \frac{9}{2}x^6}{2 x^9}
    <br />



    It's telling me to use power series to find the limit, and I am lost.

    \begin{aligned}e^{-3x^3}&=\sum_{n=0}^{\infty}\frac{(-3x^3)^n}{n!}\\<br />
&=\sum_{n=0}^{\infty}\frac{(-3)^nx^{3n}}{n!}\\<br />
&=1-3x^3+\frac{9x^6}{2}-\frac{9x^9}{2}+\cdots<br />
\end{aligned}

    \therefore~e^{-3x^3}-1+3x^3-\frac{9x^6}{2}=\frac{9x^9}{2}+\cdots

    \implies\lim_{x\to{0}}\frac{e^{-3x^3}-1+3x^3-\frac{9x^6}{2}}{2x^9}=\frac{-9}{4}
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