题目

当 $x\to0^+$ 时,求出下列无穷小量的最高阶:

$$
\begin{matrix}
\displaystyle\int_0^x(e^{t^2} - 1)dt & &\displaystyle\int_0^x\ln(1+\sqrt{t^3})dt\\
\displaystyle\int_0^{\sin x}\sin t^3dt & &\displaystyle\int_0^{1-\cos x}\sqrt{\sin^3t}dt\
\end{matrix}
$$

解答

当 $x\to0^+$ 时:

$$
\displaystyle\int_0^x(e^{t^2} - 1)dt \sim
\displaystyle\int_0^xt^2dt = \dfrac{1}{3}x^3
$$

$$
\displaystyle\int_0^x\ln(1+\sqrt{t^3})dt \sim
\displaystyle\int_0^x t^{\frac{3}{2}}dt =
\dfrac{2}{5} x^{\frac{5}{2}}
$$

$$
\displaystyle\int_0^{\sin x}\sin t^3dt \sim
\displaystyle\int_0^{x}t^3dt = \dfrac{1}{4} x^4
$$

$$
\displaystyle\int_0^{1-\cos x}\sqrt{\sin^3t}dt \sim
\displaystyle\int_0^{\frac{1}{2}x^2}x^{\frac{3}{2}}dt =
\dfrac{\sqrt{2}}{20} x^5
$$

故最高阶的为 $\displaystyle\int_0^{1-\cos x}\sqrt{\sin^3t}dt$