第224题 | 知识点：强化极限训练（十四）

题目

设函数 $f(x)$ 一阶连续可导,且 $f(0)=0$, $f’(0)\ne0$ ,求 $\lim\limits_{x\to0}\dfrac{\displaystyle\int_0^{x^2}f(t)dt}{x^2\displaystyle\int_0^xf(t)dt}$

解答

连续可导：函数可导，且导函数连续

$f(x)$ 一阶连续可导 $\quad\Rightarrow\quad$ $\lim\limits_{x \to 0} f’(x_0 + x) = f’(x_0)$

$$
\begin{aligned}
\lim\limits_{x\to0}\dfrac{\displaystyle\int_0^{x^2}f(t)dt}{x^2\displaystyle\int_0^xf(t)dt}
&\xlongequal{L’}
\lim\limits_{x\to0}\dfrac{2x f(x^2)}{2x\displaystyle\int_0^xf(t)dt + x^2f(x)}
\\
&=
\lim\limits_{x\to0}\dfrac{2f(x^2)}{2\displaystyle\int_0^xf(t)dt + xf(x)}
\\
&\xlongequal{L’}
\lim\limits_{x\to0}\dfrac{4xf’(x^2)}{3f(x) + xf’(x)}
\\
&=
\lim\limits_{x\to0}\dfrac{4f’(x^2)}{3 \cdot \dfrac{f(x)}{x} + f’(x)}
\\
&=
\frac{4f’(0)}{3 \cdot \lim\limits_{x\to0} \dfrac{f(x)}{x} + f’(0)}
\\
&=
\frac{4f’(0)}{3f’(0) + f’(0)}
\\
&=
1
\\
\end{aligned}
$$