容易想的思路不容易做，容易做的思路不容易想

一、题目

二、解析

未定式分析

由于：

$$
\begin{aligned}
& \lim_{x \rightarrow 0} \frac{\arctan x}{x} = 1 \\ \\
& \lim_{x \rightarrow 0} \frac{1}{x^{2}} = \infty
\end{aligned}
$$

所以，式子 $I$ 是一个 $1^{\infty}$ 型的未定式.

思路容易想的解法：$\mathrm{e}$ 抬起

$$
\begin{aligned}
I & = \lim_{x \rightarrow 0} \left( \frac{\arctan x}{x} \right)^{\frac{1}{x^{2}}} \\ \\
& = \lim_{x \rightarrow 0} \mathrm{e}^{\ln \left( \frac{\arctan x}{x} \right)^{\frac{1}{x^{2}}}} \\ \\
& = \lim_{x \rightarrow 0} \mathrm{e}^{\textcolor{lightgreen}{ \frac{1}{x^{2}} \ln \left( \frac{\arctan x}{x} \right) }}
\end{aligned}
$$

其中，对于 $\textcolor{lightgreen}{ \frac{1}{x^{2}} \ln \left( \frac{\arctan x}{x} \right) }$, 有：

$$
\begin{aligned}
& \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{1}{x^{2}} \ln \left( \frac{\arctan x}{x} \right) } \\ \\
= \ & \lim_{x \rightarrow 0} \frac{1}{x^{2}} \left( \ln \arctan x – \ln x \right) \\ \\
= \ & \lim_{x \rightarrow 0} \frac{\ln \arctan x – \ln x}{x^{2}} \\ \\
\leadsto \ & \textcolor{gray}{洛必达运算} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{1}{2x} \left[ \frac{1}{(1 + x^{2}) \arctan x} – \frac{1}{x} \right] \\ \\
= \ & \lim_{x \rightarrow 0} \frac{1}{2x} \left[ \frac{x-(1 + x^{2}) \arctan x}{x(1 + x^{2}) \arctan x} \right] \\ \\
= \ & \lim_{x \rightarrow 0} \frac{x – (1 + x^{2}) \arctan x}{2x^{2} (1 + x^{2}) \arctan x} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{\textcolor{orangered}{x} \times \textcolor{orange}{ \left[ x – (1 + x^{2}) \arctan x \right] } }{\textcolor{orange}{ x^{3} } \times \textcolor{orangered}{ 2 (1 + x^{2}) \arctan x} } \\ \\
= \ & \lim_{x \rightarrow 0} \textcolor{orangered}{ \frac{x}{2(1 + x^{2}) \arctan x} } \times \lim_{x \rightarrow 0} \textcolor{orange}{ \frac{x – (1 + x^{2}) \arctan x}{x^{3}} } \\ \\
= \ & \lim_{x \rightarrow 0} \textcolor{orangered}{ \frac{x}{2(1 + x^{2}) x} } \times \lim_{x \rightarrow 0} \textcolor{orange}{ \frac{x – (1 + x^{2}) \arctan x}{x^{3}} } \\ \\
= \ & \lim_{x \rightarrow 0} \textcolor{orangered}{ \frac{1}{2(1 + x^{2}) } } \times \lim_{x \rightarrow 0} \textcolor{orange}{ \frac{x – (1 + x^{2}) \arctan x}{x^{3}} } \\ \\
= \ & \textcolor{orangered}{ \frac{1}{2} } \lim_{x \rightarrow 0} \textcolor{orange}{ \frac{x – (1 + x^{2}) \arctan x}{x^{3}} } \\ \\
\leadsto \ & \textcolor{gray}{洛必达运算} \\ \\
= \ & \frac{1}{2} \lim_{x \rightarrow 0} \frac{1 – 1 – 2x \arctan x}{3x^{2}} \\ \\
= \ & \frac{1}{2} \lim_{x \rightarrow 0} \frac{-2 \arctan x}{3x} \\ \\
= \ & \frac{1}{2} \cdot \frac{-2}{3} \\ \\
= \ & \textcolor{lightgreen}{ -\frac{1}{3} }
\end{aligned}
$$

综上可知：

$$
I = \mathrm{e}^{\textcolor{lightgreen}{\frac{-1}{3}}}
$$

思路不容易想的解法：凑等价无穷小

根据等价无穷小公式，当 $k \rightarrow 0$ 时，有 $\ln (1 + k) \sim k$.

又因为，当 $x \rightarrow 0$, 有：

$$
\left( \frac{\arctan x}{x} – 1 \right) \rightarrow 0
$$

于是：

$$
\begin{aligned}
& \ln \left( \frac{\arctan x}{x} \right) \\ \\
= \ & \ln \left[ 1 + \left( \frac{\arctan x}{x} – 1 \right) \right] \sim \left( \frac{\arctan x}{x} – 1 \right)
\end{aligned}
$$

又因为：

$$
\begin{aligned}
I & = \lim_{x \rightarrow 0} \left( \frac{\arctan x}{x} \right)^{\frac{1}{x^{2}}} \\ \\
& = \lim_{x \rightarrow 0} \mathrm{e}^{\ln \left( \frac{\arctan x}{x} \right)^{\frac{1}{x^{2}}}} \\ \\
& = \lim_{x \rightarrow 0} \mathrm{e}^{\textcolor{lightgreen}{ \frac{1}{x^{2}} \ln \left( \frac{\arctan x}{x} \right) }}
\end{aligned}
$$

所以：

$$
\begin{aligned}
& \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{1}{x^{2}} \ln \left( \frac{\arctan x}{x} \right) } \\ \\
= \ & \lim_{x \rightarrow 0} \frac{1}{x^{2}} \left( \frac{\arctan x}{x} – 1 \right) \\ \\
= \ & \lim_{x \rightarrow 0} \frac{1}{x^{2}} \left( \frac{\arctan x – x}{x} \right) \\ \\
= \ & \lim_{x \rightarrow 0} \frac{\arctan x – x}{x^{3}} \\ \\
\leadsto \ & \textcolor{gray}{洛必达运算} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{\frac{1}{1 + x^{2}} – 1}{3x^{2}} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{\frac{-x^{2}}{1 + x^{2}}}{3x^{2}} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{-1}{3(1 + x^{2})} \\ \\
= \ & \textcolor{lightgreen}{ -\frac{1}{3} }
\end{aligned}
$$

综上可知：

$$
I = \mathrm{e}^{\textcolor{lightgreen}{\frac{-1}{3}}}
$$

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