# 披着数列极限外衣的函数无穷小问题：但是不能直接用等价无穷小公式哦

## 二、解析

### 解法一：转为函数极限后用洛必达求解

$$I = \lim \limits_{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{n}-e}{\frac{1}{n}} \Rightarrow x=\frac{1}{n} \Rightarrow$$

$$\lim \limits_{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=\lim \limits_{x \rightarrow 0} \frac{e^{\frac{1}{x} \ln (1+x)}-e}{x} \Rightarrow \frac{0}{0} \Rightarrow$$

$$\lim \limits_{x \rightarrow 0} \frac{\left[\frac{\ln (1+x)}{x}\right]^{\prime} e^{\frac{1}{x} \ln (1+x)}}{1}$$

$$\lim \limits_{x \rightarrow 0} e^{\frac{1}{x} \ln (1+x)}=e$$

$$\lim \limits_{x \rightarrow 0}\left[\frac{\ln (1+x)}{x}\right]^{\prime}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}}=\frac{x-(1+x) \ln (1+x)}{(1+x) x^{2}} \Rightarrow$$

$$\lim \limits_{x \rightarrow 0} \frac{[x-\ln (1+x)]-x \ln (1+x)}{x^{2}+x^{3}}=\frac{\frac{1}{2} x^{2}-x^{2}}{x^{2}}=\frac{-1}{2}$$

$$I=\frac{-e}{2}.$$

### 方法二：先用等价无穷小的变体，再用洛必达

$$\lim \limits_{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{n}-e}{\frac{1}{n}} \Rightarrow e \cdot {\lim \limits_{n \rightarrow \infty}} \frac{\frac{\left(1+\frac{1}{n}\right)^{n}}{e}-1}{\frac{1}{n}}.$$

$$\lim \limits_{x \rightarrow 1}(x-1) \sim \lim \limits_{x \rightarrow 1} \ln [(x-1)+1] \sim \lim \limits_{x \rightarrow 1} \ln x.$$

Tips

$$\lim \limits_{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{n}}{e}=\frac{e}{e} \Rightarrow 1 \Rightarrow$$

$$e \lim \limits_{n \rightarrow \infty} \frac{\frac{\left(1+\frac{1}{n}\right)^{n}}{e}-1}{\frac{1}{n}}=e \lim \limits_{n \rightarrow \infty} \frac{\ln \left[\frac{\left(1+\frac{1}{n}\right)^{n}}{e}\right]}{\frac{1}{n}} \Rightarrow$$

$$e \lim \limits_{n \rightarrow \infty} \frac{n \ln \left(1+\frac{1}{n}\right)-1}{\frac{1}{n}} \Rightarrow$$

$$e \lim \limits_{n \rightarrow \infty} \frac{\frac{n \ln \left(1+\frac{1}{n}\right)}{n}-\frac{1}{n}}{\frac{1}{n^{2}}} = e \lim \limits_{n \rightarrow \infty} \frac{\ln \left(1+\frac{1}{n}\right)-\frac{1}{n}}{\frac{1}{n^{2}}}.$$

$$e \lim \limits_{x \rightarrow 0} \frac{\ln (1+x)-x}{x^{2}} \Rightarrow$$

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

$$e \lim \limits_{x \rightarrow 0} \frac{-x}{2 x}=\frac{-e}{2}.$$

### 方法三：先用等价无穷小的变体，再用泰勒公式

$$\ln (1+x)=x-\frac{1}{2} x^{2}+o\left(x^{2}\right).$$

$$\lim \limits_{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{n}-e}{\frac{1}{n}} \Rightarrow x=\frac{1}{n} \Rightarrow \lim \limits_{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=$$

$$e \cdot \lim \limits_{x \rightarrow 0} \frac{\frac{(1+x)^{\frac{1}{x}}}{e}-1}{x}=e \cdot \lim \limits_{x \rightarrow 0} \frac{\ln \left[\frac{(1+x)^{\frac{1}{x}}}{e}\right]}{x}=$$

$$e \lim \limits_{x \rightarrow 0} \frac{\frac{1}{x} \ln (1+x)-1}{x}=$$

$$e^{\lim \limits_{x \rightarrow 0} \frac{\frac{1}{x}\left(x-\frac{1}{2} x^{2}\right)-1}{x}}=$$

$$e \lim \limits_{x \rightarrow 0} \frac{1-\frac{1}{2} x-1}{x}=\frac{-e}{2}.$$

### 特别专题

Copyright © 2017-2024 ZhaoKaifeng.com 版权所有 All Rights Reserved.

Copyright © 2024   zhaokaifeng.com   All Rights Reserved.