2022考研数二第17题解析：一点处导数的定义

一、题目

已知函数 $f\left(x\right)$ 在 $x = 1$ 处可导，且 $\lim\limits_{x \to 0} \frac{f\left(\mathrm{e}^{x^{2}}\right) – 3 f\left(1 + \sin^{2} x\right)}{x^{2}} = 2$，求 $f^{\prime}\left(1\right)$.

二、解析

由于，当 $x \to 0$ 的时候，$x^{2} \to 2$, 且极限 $\lim\limits_{x \to 0} \frac{f\left(\mathrm{e}^{x^{2}}\right) – 3 f\left(1 + \sin^{2} x\right)}{x^{2}}$ 存在，所以：

$$
\lim_{x \to 0} \left[ f \left( \mathrm{e}^{x^{2}} \right) – 3 f \left( 1 + \sin^{2} x \right) \right] = 0
$$

于是：

$$
\begin{aligned}
& \ \lim_{x \to 0} \left[ f \left( \mathrm{e}^{x^{2}} \right) – 3 f \left( 1 + \sin^{2} x \right) \right] = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f \left( 1 \right) – 3 f \left( 1 \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ f \left( 1 \right) = 0 }
\end{aligned}
$$

接着，有：

$$
\begin{aligned}
& \ \lim_{x \to 0} \frac{f\left(\mathrm{e}^{x^{2}}\right)-3f\left(1+\sin^{2}x\right)} {x^{2}} = 2 \\ \\
\textcolor{lightgreen}{\leadsto} & \ \lim_{x \to 0} \left[ \frac{f\left(\mathrm{e}^{x^{2}}\right)-f\left( 1 \right)}{x^{2}} -3\frac{f\left(1+\sin^{2} x\right)-f\left( 1 \right)}{x^{2}} \right] = 2 \\ \\
\textcolor{lightgreen}{\leadsto} & \ \lim_{x \to 0} \frac{f\left(\mathrm{e}^{x^{2}}\right)-f\left( 1 \right)} {\mathrm{e}^{x^{2}}-1} \cdot \frac{\mathrm{e}^{x^{2}}-1}{x^{2}} -3\lim_{x \to 0} \frac{f\left(1+\sin^{2} x\right)-f\left( 1 \right)} {\sin^{2} x} \cdot \frac{\sin^{2} x}{x^{2}} = 2 \\ \\
\textcolor{lightgreen}{\leadsto} & \ \textcolor{gray}{ \mathrm{e}^{x^{2}}-1\sim x^{2}, \ \sin^{2} x \sim x^{2} } \\ \\
\textcolor{lightgreen}{\leadsto} & \ f ^{\prime} \left( 1 \right)-3f ^{\prime} \left( 1 \right)=2 \\ \\ \textcolor{lightgreen}{\leadsto} & \ -2f ^{\prime} \left( 1 \right)=2 \\ \\
\textcolor{lightgreen}{\leadsto} & \ \textcolor{lightgreen}{ f ^{\prime} \left( 1 \right) = -1 }
\end{aligned}
$$

一、题目

二、解析

三、相关习题