题目

设 $f(x)$ 连续, $\lim\limits_{x\to0}\dfrac{f(x)}{x}=1$ ,求极限 $\lim\limits_{x\to0}\Big[
1+\displaystyle\int_0^xtf(x^2-t^2)dt
\Big]^{\dfrac{1}{(\tan x - x)\ln(1+x)}}$

解答

$f(x)$ 连续 $\&$ $\lim\limits_{x\to0}\dfrac{f(x)}{x}=1 $ $\quad \Rightarrow \quad$ $f(0) = 0$ 且 $f’(0) = 1$

幂指函数，先取指对数，然后单独处理指数部分

$$
\begin{aligned}
&
\lim_{x\to0} \dfrac{\ln(1 + \displaystyle\int_0^x tf(x^2 - t^2)dt)}{(\tan x - x) \ln(1 + x)}
\\
=&
\lim_{x\to0} \dfrac{\displaystyle\int_0^x tf(x^2 - t^2)dt}{\dfrac{1}{3}x^4}
\\
& \text{令} x^2 - t^2 = u, \text{则} -2tdt = du
\\
=&
\dfrac{3}{2}
\lim_{x\to0} \dfrac{\displaystyle\int_0^{x^2} f(u)du}{x^4}
\\
\xlongequal{L’}&
\dfrac{3}{4}
\lim_{x\to0} \dfrac{f(x^2)}{x^2}
\\
=&
\dfrac{3}{4}
\lim_{x\to0} \dfrac{f(x^2) - f(0)}{x^2 - 0}
\\
=&
\dfrac{3}{4}
f_+’(0)
\\
=&
\dfrac{3}{4}
\\
\end{aligned}
$$

故 $\lim\limits_{x\to0}\Big[
1+\displaystyle\int_0^xtf(x^2-t^2)dt
\Big]^{\dfrac{1}{(\tan x - x)\ln(1+x)}} = e^{\frac{3}{4}}$