# 给定一个无穷大量，怎么转为无穷小量？

## 二、解析

### 解法一：直接按照无穷大量计算

$$\lim \limits_{x \rightarrow \infty}\left(\sqrt[3]{1-x^{6}}-a x^{2}-b\right)=0 \Rightarrow$$

$$\sqrt[3]{1-x^{6}}-a x^{2}=b \Rightarrow \sqrt[3]{-x^{6}}-a x^{2}=b \Rightarrow$$

$$-x^{2}-a x^{2}=b \Rightarrow$$

$$\left\{\begin{array}{l} a=-1 \\ b=0 \end{array}\right.$$

### 解法二：转为无穷小量计算

$$\sqrt[3]{1-x^{6}}-a x^{2}=b \Rightarrow \sqrt[3]{x^{6}\left(\frac{1}{x^{6}}-1\right)}-a x^{2}=b \Rightarrow$$

$$\sqrt[3]{-x^{6}\left(1-\frac{1}{x^{6}}\right)}-a x^{2}=b \Rightarrow$$

$$-x^{2}\left[1-\frac{1}{x^{6}}\right]^{\frac{1}{3}}-a x^{2}=b \Rightarrow$$

$$-x^{2}\left[\left(1-\frac{1}{x^{6}}\right)^{\frac{1}{3}}+a\right]=b \Rightarrow$$

$$x \rightarrow 0 \Rightarrow (1+x)^{a}-1 \sim a x \Rightarrow$$

$$\textcolor{springgreen}{ x \rightarrow 0 \Rightarrow (1+x)^{a} \sim 1+a x } \Rightarrow$$

$$-x^{2}\left[\left(1-\frac{1}{x^{6}}\right)^{\frac{1}{3}}+a\right]=b \Rightarrow$$

$$-x^{2}\left[\frac{1}{3} \cdot \frac{-1}{x^{6}}+1+a\right]=b \Rightarrow$$

$$\left\{\begin{array} { l } { 1 + a = 0 } \\ { b = 0 } \end{array} \quad \Rightarrow \quad \left\{\begin{array}{l} a=-1 \\ b=0 \end{array}\right.\right.$$