两种方法计算：$\lim_{x \rightarrow \infty}$ $($ $\sin \frac{2}{x}$ $+$ $\cos \frac{1}{x}$ $)^{x}$

一、题目

$$
\lim_{x \rightarrow \infty} \big( \sin \frac{2}{x} + \cos \frac{1}{x} \big)^{x} = ?
$$

难度评级：

解 题 思 路 简 图

 graph TB
	A(求极限) --> B(幂指函数)
	A --> C(1 的无穷次幂)
	A --> D(等价无穷小)
	B --> E(规划解题方法)
	C --> E
	D --> E
	E --> F(1 的无穷次幂等于 e)
	E --> G(e 抬起)

二、解析

方法一：$1^{\infty}$ $=$ $e$

$$
\lim_{x \rightarrow \infty} \big( \sin \frac{2}{x} + \cos \frac{1}{x} \big)^{x} =
$$

$$
\lim_{x \rightarrow \infty} \big[ (\sin \frac{2}{x} + \cos \frac{1}{x} – 1) + 1 \big]^{x} =
$$

Next

$$
\lim_{x \rightarrow \infty} \big[ (\sin \frac{2}{x} + \cos \frac{1}{x} – 1) + 1 \big]^{\frac{1}{\sin \frac{2}{x} + \cos \frac{1}{x} – 1} \cdot x \cdot \frac{\sin \frac{2}{x} + \cos \frac{1}{x} – 1}{1}} =
$$

$$
\lim_{x \rightarrow \infty} e^{x \cdot \frac{\sin \frac{2}{x} + \cos \frac{1}{x} – 1}{1}} =
$$

Next

$$
e^{\lim_{x \rightarrow \infty} x \cdot (\sin \frac{2}{x} + \cos \frac{1}{x} – 1)} =
$$

$$
e^{\lim_{x \rightarrow \infty} x \cdot [2 \sin \frac{1}{x} \cos \frac{1}{x} + (\cos \frac{1}{x} – 1)]} =
$$

$$
e^{\lim_{x \rightarrow \infty} x \cdot [\frac{2}{x} – \frac{1}{2} (\frac{1}{x})^{2}]} =
$$

$$
e^{\lim_{x \rightarrow \infty} (\frac{2x}{x} – \frac{x}{2x^{2}})} =
$$

$$
e^{\lim_{x \rightarrow \infty} (2 – \frac{1}{2x})} = e^{2 – 0} = e^{2}.
$$

方法二：$e$ 抬起

$$
\lim_{x \rightarrow \infty} \big( \sin \frac{2}{x} + \cos \frac{1}{x} \big)^{x} =
$$

Next

$$
\lim_{x \rightarrow \infty} e^{x \ln ( \sin \frac{2}{x} + \cos \frac{1}{x} )} =
$$

$$
\lim_{x \rightarrow \infty} e^{x \ln ( 2 \sin \frac{1}{x} \cos \frac{1}{x} + \cos \frac{1}{x} )} =
$$

Next

$$
\lim_{x \rightarrow \infty} e^{x \ln ( \frac{2}{x} + 1 )} =
$$

$$
\lim_{x \rightarrow \infty} e^{x \cdot \frac{2}{x}} = e^{2}.
$$

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