# 2016年考研数二第15题解析：无穷小、$e$ 抬起、两个重要无穷小

## 题目

$$\lim_{x \rightarrow 0} (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}}.$$

## 解析

### 方法一

$$\lim_{x \rightarrow 0} (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}} \Rightarrow$$

$$\lim_{x \rightarrow 0} [1 + (\cos 2x + 2x \sin x – 1)]^{\frac{1}{x^{4}}} \Rightarrow$$

$$\lim_{x \rightarrow 0} [1 + (\cos 2x + 2x \sin x – 1)]^{\frac{1}{\cos 2x + 2x \sin x – 1} \cdot \frac{\cos 2x + 2x \sin x – 1}{1} \cdot \frac{1}{x^{4}}} \Rightarrow$$

$$\lim_{x \rightarrow 0} e^{\frac{\cos 2x + 2x \sin x – 1}{x^{4}}} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{\cos 2x + 2x \sin x – 1}{x^{4}} }.$$

$$\cos 2x = 1 – \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} \Rightarrow$$

$$\cos 2x = 1 – 2x^{2} + \frac{2}{3}x^{4} + o(x^{4}).$$

$$x \sin x = x(x – \frac{1}{6}x^{3}) \Rightarrow$$

$$x \sin x = x^{2} – \frac{1}{6} x^{4} + o(x^{4}).$$

$$\lim_{x \rightarrow 0} \frac{\cos 2x + 2x \sin x – 1}{x^{4}} =$$

$$\lim_{x \rightarrow 0} \frac{1 – 2x^{2} + \frac{2}{3}x^{4} + 2x^{2} – \frac{1}{3} x^{4} – 1}{x^{4}} =$$

$$\lim_{x \rightarrow 0} \frac{\frac{1}{3} x^{4}}{x^{4}} = \frac{1}{3}.$$

$$\lim_{x \rightarrow 0} (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}} =$$

$$\lim_{x \rightarrow 0} e^{\frac{\cos 2x + 2x \sin x – 1}{x^{4}}} = e^{\frac{1}{3}}.$$

### 方法二

$$\lim_{x \rightarrow 0} (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}} \Rightarrow$$

$$e^{\ln\lim_{x \rightarrow 0} (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}}} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \ln (\cos 2x + 2x \sin x)^{\frac{1}{x^{4}}}} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{1}{x^{4}} \ln (\cos 2x + 2x \sin x)} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{\ln (\cos 2x + 2x \sin x)}{x^{4}}} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{\ln (1 + \cos 2x + 2x \sin x – 1)}{x^{4}}} \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{\cos 2x + 2x \sin x – 1}{x^{4}}}.$$

### 方法三

$$\lim_{x \rightarrow 0} \frac{\cos 2x + 2x \sin x – 1}{x^{4}} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{-2\sin 2x + 2 \sin x + 2 x \cos x}{4x^{3}} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{-4\cos 2x + 2 \cos x + 2 \cos x – 2x \sin x}{12x^{2}} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{8\sin 2x – 2 \sin x – 2 \sin x – 2 \sin x – 2x \cos x}{24x} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{8\sin 2x – 6 \sin x – 2x \cos x}{24x} \Rightarrow 洛必达 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{16\cos 2x – 6 \cos x – 2 \cos x + 2 x \sin x}{24} \Rightarrow 洛必达 \Rightarrow$$

$$\frac{16-6-2}{24} = \frac{1}{3}.$$