怎么表示切线在 X 轴上的截距？

一、题目

已知 $f(x)$ 有二阶连续导数, 且 $f(0)=f^{\prime }(0)=0$, $f^{\prime \prime }(x)>0$, 又设 $u=u(x)$ 是曲线 $y=f(x)$ 在点 $(x,f(x))$ 处的切线在 $x$ 轴上的截距, 则 $\underset{x\to 0}{lim}\frac{x}{u(x)}=?$

难度评级：

二、解析

根据题意，写出截距的表达式：

$$x_0\ne 0\Rightarrow (x_0,f\left(x_0\right))\Rightarrow$$

$$y-f\left(x_0\right)=f^{\prime }\left(x_0\right)(x-x_0)\Rightarrow$$

$$y=0\Rightarrow$$

$$-f\left(x_0\right)=x{f}^{\prime }\left(x_0\right)-x_0{f}^{\prime }\left(x_0\right)\Rightarrow$$

$$x_0=u\left(x_0\right)=\frac{x_0{f}^{\prime }\left(x_0\right)-f\left(x_0\right)}{f^{\prime }\left(x_0\right)}\Rightarrow$$

于是：

$$\begin{array}{}\text{(1)}& \underset{x_0\to 0}{lim}\frac{x}{u(x)}=\frac{x{f}^{\prime }\left(x_0\right)}{x_0{f}^{\prime }\left(x_0\right)-f\left(x_0\right)}\end{array}$$

对于 $(1)$ 式，有如下两种解法：

方法 1：洛必达

$$\underset{x\to 0}{lim}\frac{x}{u(x)}=\frac{x{f}^{\prime }(x)}{x{f}^{\prime }(x)-f(x)}\Rightarrow$$

$$\frac{0}{0}\Rightarrow \text{ 洛必达运算 }\Rightarrow$$

$$\frac{f^{\prime }\left(x_0\right)+x{f}^{\prime \prime }(x)}{f^{\prime }(x)+x{f}^{\prime \prime }(x)-f^{\prime }(x)}=\frac{x{f}^{\prime \prime }(x)+f^{\prime }(x)}{x{f}^{\prime \prime }(x)}=$$

$$1+\frac{f^{\prime }(x)}{x{f}^{\prime \prime }(x)}=1+\frac{f^{\prime }(x)-f^{\prime }(0)}{x-0}\cdot \frac{1}{f^{\prime \prime }(x)}=$$

$$1+\frac{f^{\prime \prime }(x)}{f^{\prime \prime }(x)}=1+1=2$$

方法 2：泰勒公式

$$\underset{x\to 0}{lim}\frac{x}{u(x)}=\frac{x{f}^{\prime }(x)}{x{f}^{\prime }(x)-f(x)}\Rightarrow$$

又：

$$x{f}^{\prime }(x)=x[f^{\prime }(0)+f^{\prime \prime }(0)\cdot x+o(x)]=$$

$$x^2{f}^{\prime \prime }(0)+o\left(x^2\right)$$

又：

$$f(x)=f(0)+f^{\prime }(0)\cdot x+\frac{1}{2}{f}^{\prime \prime }(0)\cdot x^2+0\left(x^2\right)=$$

$$\frac{1}{2}{x}^2{f}^{\prime \prime }(0)+0\left(x^2\right)$$

于是：

$$\underset{x\to 0}{lim}\frac{x}{u(x)}=\frac{x^2{f}^{\prime \prime }(0)+o\left(x^2\right)}{x^2{f}^{\prime \prime }(0)-\frac{1}{2}{x}^2{f}^{\prime \prime }(0)+o\left(x^2\right)}=2$$

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