2022考研数二第11题解析：极限的计算、三角函数、常用的等价无穷小

一、题目

二、解析

$$
\begin{aligned}
& \ \lim\limits_{x \to 0} \left(\frac{1 + \mathrm{e}^{x}}{2}\right)^{\cot x} \\ \\
= & \ \lim\limits_{x \to 0} \left(\frac{1 + \mathrm{e}^{x}}{2}\right)^{\frac{\cos x}{\sin x}} \\ \\
= & \ \lim\limits_{x \to 0} \mathrm{e}^{\frac{\cos x}{\sin x} \cdot \ln \left(\frac{1 + \mathrm{e}^{x}}{2}\right)} \\ \\
= & \ \lim\limits_{x \to 0} \mathrm{e}^{\frac{\cos x}{\sin x} \cdot \ln \left(\frac{1 + \mathrm{e}^{x}}{2} – 1 + 1\right)} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{ \lim\limits_{x \to 0} \ln \left( x+1 \right) \sim \lim\limits_{x \to 0} x } \\ \\
= & \ \mathrm{e}^{\lim\limits_{x \to 0} \frac{\cos x}{\sin x} \left(\frac{1 + \mathrm{e}^{x}}{2} – 1\right)} \\ \\
= & \ \mathrm{e}^{\lim\limits_{x \to 0} \frac{\cos x \left(\mathrm{e}^{x} – 1\right)}{2 \sin x}} \\ \\
= & \ \mathrm{e}^{\lim\limits_{x \to 0} \frac{\mathrm{e}^{x} – 1}{2 x}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{ \lim\limits_{x \to 0} \mathrm{e}^{x} – 1 \sim \lim\limits_{x \to 0} x } \\ \\
= & \ \mathrm{e}^{\lim\limits_{x \to 0} \frac{x}{2 x}} \\ \\
= & \ \mathrm{e}^{\frac{1}{2}}
\end{aligned}
$$

三、补充

[1]. 常用的等价无穷小公式汇总

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