# 这道题目看似很简单，但全身都是“坑”

## 一、题目

$$\lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + x f(x)} {x^{2}} = 3$$

$$\lim_{ x \rightarrow 0 } \frac{ 2+f(x) }{x} = ?$$

## 二、解析

### 错误的解法

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + x f(x)} {x^{2}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{\textcolor{orange}{ \ln (1 + 2x) }}{x ^{2}} + \lim_{ x \rightarrow 0 } \textcolor{magenta}{ \frac{f(x)}{x} } \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{\textcolor{orange}{2x}}{x ^{2}} + \lim_{ x \rightarrow 0 } \textcolor{magenta}{\frac{f(x)}{x}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{2}{x} + \lim_{ x \rightarrow 0 } \textcolor{magenta}{\frac{f(x)}{x}} \\ \\ \Rightarrow & \textcolor{springgreen}{\boldsymbol{ \lim_{ x \rightarrow 0 } \frac{2 + f(x)}{x} = 3 }} \end{aligned}

### 正确的解法 01：泰勒公式

$$\ln (1 + x) = x – \frac{1}{2} x ^{2} + o(x ^{2})$$

$$\textcolor{orange}{ \ln(1 + 2x) = 2x – 2x ^{2} + o(x ^{2}) }$$

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + x f(x)} {x^{2}} = 3 \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{2x – 2 x ^{2} + o(x ^{2}) + x f(x)}{x ^{2}} = 3 \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{2x + x f(x) – 2 x ^{2}}{x ^{2}} = 3 \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{2 + f(x)}{x} – 2 = 3 \\ \\ \Rightarrow & \textcolor{springgreen}{ \boldsymbol{ \lim_{ x \rightarrow 0 } \frac{2 + f(x)}{x} = 5 }} \end{aligned}

### 正确的解法 02：“凑”

$$\lim_{ x \rightarrow 0 } \frac{ 2+f(x) }{x} = A$$

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + x f(x)} {x^{2}} = 3 \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) \textcolor{orangered}{-2x + 2x} + x f(x)} {x^{2}} = 3 \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{\ln (1+2x) \textcolor{orangered}{-2x}}{x ^{2}} + \lim_{ x \rightarrow 0 } \frac{\textcolor{orangered}{2x} + x f(x)}{x ^{2}} = 3 \\ \\ \Rightarrow & \textcolor{pink}{ \lim_{ x \rightarrow 0 } \frac{- \frac{1}{2} (2x)^{2}}{x ^{2}} } + \textcolor{yellow}{ \lim_{ x \rightarrow 0 } \frac{2 + f(x)}{x} } = 3 \\ \\ \Rightarrow & \textcolor{pink}{-2} + \textcolor{yellow}{ A} = 3 \\ \\ \Rightarrow & A = 5 \\ \\ \Rightarrow & \textcolor{springgreen}{\boldsymbol{ \lim_{ x \rightarrow 0 } \frac{2 + f(x)}{x} = 5 }} \end{aligned}

### 正确的解法 03：“猜”

\begin{aligned} & \lim_{x \rightarrow 0} \left[ 2 + f(x) \right] \rightarrow 0 \\ \\ \Rightarrow & \lim_{x \rightarrow 0} \left[ 2 + f(x) \right] = \lim_{x \rightarrow 0} kx \\ \\ \Rightarrow & \textcolor{orange}{ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} kx – 2 } \end{aligned}

\begin{aligned} & \lim_{x \rightarrow 0} \left[ 2 + f(x) \right] = 0 \\ \\ \Rightarrow & \textcolor{orange}{ \lim_{x \rightarrow 0} f(x) = -2 } \end{aligned}

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

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + x f(x)} {x^{2}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) – 2x} {x^{2}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ \frac{2}{1+2x} – 2 }{2x} \\ \\ \Rightarrow & \lim_{x \rightarrow 0} \frac{-4x}{2x (1+2x)} \\ \\ \Rightarrow & \lim_{x \rightarrow 0} \frac{-2}{1+2x} \\ \\ \Rightarrow & -2 \textcolor{orangered}{ \neq 3 } \end{aligned}

$$\textcolor{springgreen}{ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} kx – 2 }$$

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+2x ) + \textcolor{magenta}{ x f(x) }} {x^{2}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ \ln( 1+ \textcolor{orange}{2x} ) + \textcolor{magenta}{k x ^{2} – 2x}} {x^{2}} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{\frac{2}{1+2x} + 2kx – 2}{2x} \\ \\ \Rightarrow & \lim_{x \rightarrow 0} \frac{\frac{-4}{(1 + 2x) ^{2}} + 2k}{2} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ -4 + 2k } {2} = 3 \\ \\ \Rightarrow & \textcolor{springgreen}{ k = 5 } \end{aligned}

\begin{aligned} & \lim_{ x \rightarrow 0 } \frac{ 2+f(x) }{x} \\ \\ \Rightarrow & \lim_{ x \rightarrow 0 } \frac{ 2 + \textcolor{springgreen}{5} x-2 }{x} = \textcolor{springgreen}{\boldsymbol{5}} \end{aligned}