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 $\lim \limits_{x \rightarrow 0} \frac{x}{u(x)}=?$

难度评级：

二、解析

根据题意，写出截距的表达式：

$$
x_{0} \neq 0 \Rightarrow\left(x_{0}, f\left(x_{0}\right)\right) \Rightarrow
$$

$$
y-f\left(x_{0}\right)=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) \Rightarrow
$$

$$
y=0 \Rightarrow
$$

$$
-f\left(x_{0}\right)=x f^{\prime}\left(x_{0}\right)-x_{0} f^{\prime}\left(x_{0}\right) \Rightarrow
$$

$$
x_{0}=u\left(x_{0}\right)=\frac{x_{0} f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \Rightarrow
$$

于是：

$$
\textcolor{springgreen}{
\lim \limits_{x_{0} \rightarrow 0} \frac{x}{u(x)}=\frac{x f^{\prime}\left(x_{0}\right)}{x_{0} f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right)} } \tag{1}
$$

对于 $(1)$ 式，有如下两种解法：

方法 1：洛必达

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

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

$$
\frac{f^{\prime}\left(x_{0}\right)+x f^{\prime \prime}(x)}{f^{\prime}(x)+x f^{\prime \prime}(x)-f^{\prime}(x)}=\frac{x f^{\prime \prime}(x)+f^{\prime}(x)}{x f^{\prime \prime}(x)} =
$$

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

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

方法 2：泰勒公式

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

又：

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

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

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：

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

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