题目

设 $f(x)$ 连续，且 $f(0) \ne 0$，则 $\lim\limits_{x\to 0} [1 + \displaystyle\int_0^x(x - t)f(t) dt]^{\dfrac{1}{x\int_0^xf(x - t)dt}}$

解答

幂指函数先取指对数，然后单独处理指数部分

变上限积分函数求极限，考虑求导去积分符号

而被积函数中如果含有积分上限变量，优先考虑分离

$\displaystyle\int_0^x(x - t)f(t) dt = x\displaystyle\int_0^xf(t) dt - \displaystyle\int_0^xtf(t) dt$

$\displaystyle x\int_0^xf(x - t)dt \xlongequal{\text{令}x-t=u}\displaystyle x\int_0^x f(u)du$

$$
\begin{aligned}
&
\lim\limits_{x\to 0} \dfrac{x\displaystyle\int_0^xf(t) dt - \displaystyle\int_0^xtf(t) dt}{x\displaystyle\int_0^xf(u)du}
\\
\xlongequal{L’} &
\lim_{x\to 0} \dfrac{\displaystyle\int_0^xf(t) dt + xf(x) - xf(x)}
{\displaystyle\int_0^xf(u)du + xf(x)}
\\
= &
\lim_{x\to 0} \dfrac{\displaystyle\int_0^xf(t) dt}
{\displaystyle\int_0^xf(t)dt + xf(x)}
\\
= &
\lim_{x\to 0} \dfrac{xf(\xi)}
{xf(\xi) + xf(x)}
\\
= &
\lim_{x\to 0} \dfrac{f(\xi)}
{f(\xi) + f(x)}
\\
= &
\lim_{x\to 0} \dfrac{f(0)}
{f(0) + f(0)}
\\
=& \dfrac{1}{2}
\end{aligned}
$$

倒数第四步用了 积分中值定理，然后又由于 $f(x)$ 连续，且 $f(0) \ne 0$ 故直接带入

故原极限 = $e^{\frac{1}{2}}$