同阶无穷小：次幂相等，系数可以不相等

一、题目

当 $x \rightarrow 0$ 时，下列无穷小与 $x^{3}$ 为同阶无穷小的是哪一个？

(A) $x^{3}+x^{2}$.

(B) $\frac{1-\cos x}{x}$.

(C) $\int_{0}^{\ln (1+x)}\left(\mathrm{e}^{t^{2}}-1\right) \mathrm{d} t$.

(D) $(1+\sin x)^{\ln (1+x)}-1$.

难度评级：

二、解析

A

$$
\lim \limits_{x \rightarrow 0}\left(x^{3}+x^{2}\right)=\lim \limits_{x \rightarrow 0}\left(0+x^{2}\right)=\lim \limits_{x \rightarrow 0} x^{2}
$$

B

$$
\lim \limits_{x \rightarrow 0} \frac{1-\cos x}{x} \sim \lim \limits_{x \rightarrow 0} \frac{\frac{1}{2} x^{2}}{x}=\lim \limits_{x \rightarrow 0} \frac{1}{2} x
$$

C

$$
\lim \limits_{x \rightarrow 0} \int_{0}^{\ln (1+x)}\left(e^{t^{2}}-1\right) d t \Rightarrow 求导 \Rightarrow
$$

$$
\lim \limits_{x \rightarrow 0} \frac{1}{1+x}\left[e^{[\ln (1+x)]^{2}}-1\right]=
$$

$$
\lim \limits_{x \rightarrow 0} \frac{1}{1+0}\left[e^{[\ln (1+x)]^{2}}-1\right]=\lim \limits_{x \rightarrow 0}[\ln (1+x)]^{2}=\lim \limits_{x \rightarrow 0} x^{2} \Rightarrow
$$

由于求一次导会降一阶，因此：

$$
\lim \limits_{x \rightarrow 0} \int_{0}^{\ln (1+x)}\left(e^{t^{2}}-1\right) d t \sim \lim \limits_{x \rightarrow 0} x^{3}
$$

D

$$
\lim \limits_{x \rightarrow 0}(1+\sin x)^{\ln (1+x)}-1=\lim \limits_{x \rightarrow 0} \sin x \ln (1+x) \sim \lim \limits_{x \rightarrow 0} x^{2}
$$

或者：

$$
\lim \limits_{x \rightarrow 0}(1+\sin x)^{\ln (1+x)}-1=\lim \limits_{x \rightarrow 0} e^{\ln (1+x) \ln (1+\sin x)}-1=
$$

$$
\lim \limits_{x \rightarrow 0} \ln (1+x) \ln (1+\sin x) \sim \lim \limits_{x \rightarrow 0} x^{2}.
$$

---