# 2011年考研数二第09题解析

## 题目

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

## 解析

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

$$x \rightarrow 0 时, \ln(1+x) \sim x.$$

### 方法一

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

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

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

$$\lim_{x \rightarrow 0} (1+\frac{2^{x}-1}{2})^{\frac{2}{2^{x}-1} \cdot \frac{2^{x}-1}{2} \cdot \frac{1}{x}} =$$

$$e^{\lim_{x \rightarrow 0} \frac{2^{x}-1}{2x}} \Rightarrow 洛必达 \Rightarrow$$

$$e^{\lim_{x \rightarrow 0} \frac{2^{x} \ln 2}{2}} =$$

$$e^{\frac{1}{2} \lim_{x \rightarrow 0} 2^{x} \ln 2} =$$

$$e^{\frac{1}{2} \ln 2} =$$

$$e^{\ln \sqrt{2}} = \sqrt{2}.$$

### 方法二

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

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

$$\lim_{x \rightarrow 0} \frac{1}{x} \ln (1+ \frac{1+2^{x}}{2} – 1) =$$

$$\lim_{x \rightarrow 0} \frac{1}{x} \ln (1+\frac{2^{x}-1}{2}) \Rightarrow 等价无穷小 \Rightarrow$$

$$\lim_{x \rightarrow 0} \frac{1}{x} \cdot \frac{2^{x}-1}{2} =$$

$$\frac{1}{2} \lim_{x \rightarrow 0} \frac{2^{x}-1}{x} \Rightarrow 洛必达 \Rightarrow$$

$$\frac{1}{2} \lim_{x \rightarrow 0} \frac{2^{x} \ln 2}{1} = \frac{1}{2} \ln2 = \ln \sqrt{2}.$$

$$\lim_{x \rightarrow 0} \ln (\frac{1+2^{x}}{2})^{\frac{1}{x}} = \ln \sqrt{2}.$$

$$\lim_{x \rightarrow 0} (\frac{1+2^{x}}{2})^{\frac{1}{x}} = \sqrt{2}.$$