# Math Help - limit question using power series

1. ## limit question using power series

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

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

2. Originally Posted by fastcarslaugh
$\lim_{x \to 0} \frac{e^{-3x^3} - 1 + 3x^3 - \frac{9}{2}x^6}{2 x^9}$
$
$

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

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

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