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

本题至少有两种解法,一种是通过凑如下重要无穷小的方式解:

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

另一种是凑如下等价无穷小的方式解:

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

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

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

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

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

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

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

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

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

由上知:

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

于是:

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

综上可知,本题的答案为 $\sqrt{2}$.