# 一个看似不可能的等价无穷小代换的应用

## 一、题目

$$\alpha = \lim_{x \rightarrow 0} \Big[ x^{2} – \ln^{2}(1+x) \Big]$$

$$\beta = \lim_{x \rightarrow 0} \frac{e^{\sqrt{1+x^{3}}} – e^{\sqrt{1-x^{3}}}}{e}$$

## 二、解析

Next

$$\alpha = \lim_{x \rightarrow 0} \Big[ x^{2} – \ln^{2}(1+x) \Big] =$$

$$\alpha = \lim_{x \rightarrow 0} \Big[ x + \ln(1+x) \Big] \cdot \Big[ x – \ln(1+x) \Big] =$$

$$\alpha = \lim_{x \rightarrow 0} (x + x) \cdot \frac{1}{2}x^{2} =$$

$$\alpha = \lim_{x \rightarrow 0} 2x \cdot \frac{1}{2}x^{2} = x^{3}.$$

Next

$$\beta = \lim_{x \rightarrow 0} \frac{e^{\sqrt{1+x^{3}}} – e^{\sqrt{1-x^{3}}}}{e} =$$

Next

$$\beta = \lim_{x \rightarrow 0} \frac{e^{\sqrt{1+x^{3}}}(1 – e^{\sqrt{1-x^{3}}- \sqrt{1+x^{3}}})}{e} =$$

$$\beta = \lim_{x \rightarrow 0} \frac{e^{\sqrt{1+x^{3}}}}{e} \cdot (1 – e^{\sqrt{1-x^{3}}- \sqrt{1+x^{3}}}) =$$

$$\beta = \lim_{x \rightarrow 0} \frac{e^{1}}{e} \cdot (\sqrt{1+x^{3}} – \sqrt{1-x^{3}}) =$$

$$\beta = \lim_{x \rightarrow 0} \frac{\sqrt{1+x^{3}} – \sqrt{1-x^{3}}}{1} =$$

Next

$$\beta = \lim_{x \rightarrow 0} \frac{(\sqrt{1+x^{3}} – \sqrt{1-x^{3}})(\sqrt{1+x^{3}} + \sqrt{1-x^{3}})}{\sqrt{1+x^{3}} + \sqrt{1-x^{3}}} =$$

$$\beta = \lim_{x \rightarrow 0} \frac{(1 + x^{3} – 1 + x^{3})}{\sqrt{1+x^{3}} + \sqrt{1-x^{3}}} =$$

$$\beta = \lim_{x \rightarrow 0} \frac{2 x^{3}}{2} = x^{3}.$$

Next

$$\frac{\alpha}{\beta} = 1.$$