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(a)Find the first three non-zero terms of the Maclaurin series for $f(x)=e^{-x^{2}} \sin x$.

(b) Hence or otherwise show that $\lim _{x \rightarrow 0} \frac{f(x)-x}{x^{3}}=-\frac{7}{6}$.
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We use

$$\begin{array}{l} e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \quad \Rightarrow e^{-x^{2}}=1-x^{2}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\cdots \\ \sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots \end{array}$$
so that
\begin{aligned} e^{-x^{2}} \sin x &=\left(1-x^{2}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\cdots\right)\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right) \\ &=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots-x^{2}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right)+\frac{x^{4}}{2 !}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right) \\ &=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots-x^{3}+\frac{x^{5}}{3 !}+\frac{x^{5}}{2 !}+\cdots \\ &=x-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots \end{aligned}
Hence
$$\lim _{x \rightarrow 0} \frac{f(x)-x}{x^{3}}=\lim _{x \rightarrow 0} \frac{x-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots-x}{x^{3}}$$
$$\begin{array}{l} =\lim _{x \rightarrow 0} \frac{-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots}{x^{3}} \\ =\lim _{x \rightarrow 0}\left(-\frac{7}{6}+\frac{27 x^{2}}{40}+\cdots\right) \\ =-\frac{7}{6} \end{array}$$
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