用乘除法连接的等价无穷小可以采取“逐个击破”的方式计算

题目 02

难度评级：

解析 02

*

$$e^{x^2}–1\sim \left(x^2\right)$$

**

$$\begin{array}{r}\sqrt{1+x}-\sqrt{1-x}\\ \\ & =(1+x{)}^{\frac{1}{2}}-(1-x{)}^{\frac{1}{2}}\\ \\ & =e^{\frac{1}{2}\ln (1+x)}-e^{\frac{1}{2}\ln (1-x)}\\ \\ & =e^{\frac{1}{2}x}-e^{\frac{1}{2}(-x)}\\ \\ & =(e^{\frac{1}{2}x}-1)-(e^{\frac{1}{2}(-x)}-1)\\ \\ & =\frac{1}{2}x+\frac{1}{2}x\\ \\ & =x\end{array}$$

***

$$\begin{array}{r}\ln (1-x)+\ln (1+x)\\ \\ & =\ln (1-x)(1+x)\\ \\ & =\ln (1-x^2)\\ \\ & =-x^2\end{array}$$

或者：

$$\begin{array}{r}\ln (1-x)+\ln (1+x)\\ \\ & =-[(-x-\ln (1-x))+(x-\ln (1+x)]\\ \\ & =-[\frac{1}{2}(-x{)}^2+\frac{1}{2}{x}^2]\\ \\ & =-x^2\end{array}$$

****

$$\sin \frac{x}{1+x}=\left(\frac{x}{1+x}\right)$$

合并

$$\begin{array}{r}I\\ \\ & =\frac{x^2\cdot x}{-x^2\cdot \frac{x}{1+x}}\\ \\ & =\frac{x^3}{\frac{-x^3}{1+x}}\\ \\ & =x^3\cdot \frac{1+x}{-x^3}\\ \\ & =-(1+0)\\ \\ & =-1\end{array}$$

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