极限运算中的高阶无穷小 $o(x^{n})$ 该怎么参与运算？

一、题目

二、解析

首先，对于高阶无穷小（$x \rightarrow 0$），我们有以下等式：

$$
\begin{aligned}
o(x^{n+k}) + o(x^{n}) = o(x^{n}) & \textcolor{blue}{ \leadsto } o(x^{n+1}) + o(x^{n}) = o^{n} \\ \\
x^{k} o(x^{n}) = o(x^{n+k}) & \textcolor{blue}{ \leadsto } x o(x^{n}) = x^{n+1} \\ \\
\begin{rcases}
– mx^{k} o(x^{n}) = o(x^{n+k}) \\ \\
– mx^{k} o(x^{n}) = o(x^{n+k})
\end{rcases} & \textcolor{blue}{ \leadsto } \begin{cases}
o(x^{n}) + o(x^{n}) = o(x^{n}) \\ \\
– o(x^{n}) = o(x^{n}) \\ \\
– \frac{1}{16} o(x^{n}) = o(x^{n})
\end{cases}
\end{aligned}
$$

其中，$m$ 和 $k$ 为常数。

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式子 $I$ 的分子和分母大概是和 $x^{3}$ 同阶的无穷小量，所以，我们首先根据泰勒公式，将式子 $I$ 的分子和分母展开到 $x^{3}$:

$$
\begin{aligned}
\textcolor{orange}{ \mathrm{e}^{x} } – 1 – x – \textcolor{yellow}{ \frac{x}{2} \sin x } & = \textcolor{orange}{ 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + o(x^{3}) } – 1 – x \\
& \textcolor{yellow}{ – \frac{x}{2} \left[x – \frac{1}{6}x^{3} + o(x^{3})\right] } \\ \\
& = \frac{1}{6}x^{3} + \frac{1}{12}x^{4} + o(x^{3}) \textcolor{magenta}{ – \frac{1}{2} x o(x^{3}) } \\ \\
& = \frac{1}{6}x^{3} + \frac{1}{12}x^{4} + o(x^{3}) \textcolor{magenta}{ + \frac{1}{2} x o(x^{3}) } \\ \\
& = \frac{1}{6}x^{3} + \frac{1}{12}x^{4} + o(x^{3}) \textcolor{magenta}{ + x o(x^{3}) } \\ \\
& = \frac{1}{6}x^{3} + \frac{1}{12}x^{4} + o(x^{3}) \textcolor{magenta}{ + o(x^{4}) } \\ \\
& = \frac{1}{6}x^{3} + \frac{1}{12}x^{4} + + o(x^{3}) \textcolor{magenta}{\cancel{ \textcolor{white}{ + o(x^{4}) }}} \\ \\
& = \textcolor{lightgreen}{ \frac{1}{6}x^{3} + o(x^{3}) } \\ \\ \\
\textcolor{orange}{ \sin x } – \textcolor{yellow}{ x \cos x } & = \textcolor{orange}{ x – \frac{1}{6}x^{3} + o(x^{3}) } – \textcolor{yellow}{ x \left( 1 – \frac{1}{2}x^{2} + o(x^{3}) \right) } \\ \\
& = \frac{1}{3} x^{3} + o(x^{3}) \textcolor{magenta}{+ x o(x^{3})} \\ \\
& = \frac{1}{3} x^{3} + o(x^{3}) \textcolor{magenta}{+ o(x^{4})} \\ \\
& = \frac{1}{3} x^{3} + o(x^{3}) \textcolor{magenta}{\cancel{ \textcolor{white}{ + o(x^{4}) }}} \\ \\
& = \textcolor{lightgreen}{ \frac{1}{3} x^{3} + o(x^{3}) }
\end{aligned}
$$

综上可得：

$$
\begin{aligned}
I & = \lim_{x \rightarrow 0} \frac{\mathrm{e}^{x}-1 – x-\frac{x}{2} \sin x}{\sin x – x \cos x} \\ \\
& = \frac{ \textcolor{lightgreen}{ \frac{1}{6}x^{3} + o(x^{3}) } }{ \textcolor{lightgreen}{ \frac{1}{3} x^{3} + o(x^{3}) } } \\ \\
& = \frac{1}{6} \cdot \frac{3}{1} \\ \\
& = \textcolor{springgreen}{\frac{1}{2}}
\end{aligned}
$$

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