等价无穷小的本质：$x = 0$ 处斜率相等

一、前言

等价无穷小公式是考研数学中一个非常常用的工具。

但是，这些等价无穷小公式都是怎么来的呢？

如果说 $\lim_{x \rightarrow 0} \frac{\alpha(x)}{\beta(x)} = 1$ 就意味着 $\lim_{x \rightarrow 0} \alpha(x)$ 和 $\lim_{x \rightarrow 0} \beta(x)$ 是等价无穷小，但是，为什么式子 $\frac{\lim_{x \rightarrow 0} \alpha(x)}{\lim_{x \rightarrow 0} \beta(x)}$ 就等于 $1$ 呢？

在本文中，「荒原之梦考研数学」将借助“一点处的斜率”这一概念，为同学讲清楚等价无穷小公式的来龙去脉。当然，同学们也可以借助本文中使用的方法，来推导和记忆等价无穷小公式。

二、正文

我们知道，当 $x \rightarrow 0$ 时，下面的式子一定成立：

$$
\lim_{x \rightarrow 0} \frac{\mathit{\textcolor{black}{\colorbox{orange}{x}}}}{\mathit{\textcolor{white}{\colorbox{green}{x}}}} = 1
$$

由于当 $x \rightarrow 0$ 时，变量的斜率对应了变量趋于 $0$ 的“速度”——也就是无穷小的阶。同时，我们知道：

$$
\left( \mathit{\textcolor{black}{\colorbox{orange}{x}}} \right)^{\prime} = 1 = \left( \mathit{\textcolor{white}{\colorbox{green}{x}}} \right) ^{\prime}
$$

由于 $\mathit{\textcolor{black}{\colorbox{orange}{x}}}^{2}$ 和 $\mathit{\textcolor{white}{\colorbox{green}{x}}}^{2}$ 经过两次求导之后，得到的斜率都是 $2$:

$$
\left( \mathit{\textcolor{black}{\colorbox{orange}{x}}}^{2} \right)^{\prime \prime} = 2 = \left( \mathit{\textcolor{white}{\colorbox{green}{x}}}^{2} \right) ^{\prime \prime}
$$

所以，$\mathit{\textcolor{black}{\colorbox{orange}{x}}}^{2}$ 和 $\mathit{\textcolor{white}{\colorbox{green}{x}}}^{2}$ 也是等价无穷小：

$$
\lim_{x \rightarrow 0} \frac{\mathit{\textcolor{black}{\colorbox{orange}{x}}}^{2}}{\mathit{\textcolor{white}{\colorbox{green}{x}}}^{2}} = 1
$$

推广可知，一般情况下，只要 $x \rightarrow 0$ 时下式成立，那么，$\alpha(x)$ 与 $\beta(x)$ 就是等价无穷小，其中 $(n)$ 表示求导直到为常数的次数，$k$ 为常数：

$$
\textcolor{lightgreen}{
\lim_{x \rightarrow 0} \left[ \alpha(x) \right]^{(n)} = \lim_{x \rightarrow 0} \left[ \beta(x) \right] ^{(n)} = k
}
$$

因此，我们接下来就可以利用求导和判断斜率的方式，对常用的等价无穷小公式做验证：

1. $\textcolor{gray}{ \boxed{ \textcolor{white}{ \tan x \textcolor{orange}{\sim} x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \tan x \right)^{\prime} = \frac{1}{\cos^{2} x} = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right)^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\tan x \sim x
}
$$

2. $\textcolor{gray}{ \boxed{ \textcolor{white}{ \sin x \textcolor{orange}{\sim} x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \sin x \right) ^{\prime} = \cos x = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\sin x \sim x
}
$$

3. $\textcolor{gray}{ \boxed{ \textcolor{white}{\arcsin x \textcolor{orange}{\sim} x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \arcsin x \right) ^{\prime} = \left( \frac{1}{\sqrt{1-x^{2}}} \right) = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\arcsin x \sim x
}
$$

4. $\textcolor{gray}{ \boxed{ \textcolor{white}{\arctan x \textcolor{orange}{\sim} x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \arctan x \right) ^{\prime} = \frac{1}{1+x^{2}} = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\arctan x \sim x
}
$$

5. $\textcolor{gray}{ \boxed{ \textcolor{white}{\ln(1 + x) \textcolor{orange}{\sim} x}}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left[ \ln (1 + x) \right] ^{\prime} = \frac{1}{1 + x} = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\ln (1 + x) \sim x
}
$$

6. $\textcolor{gray}{ \boxed{ \textcolor{white}{ \mathrm{e}^{x} – 1 \textcolor{orange}{\sim} x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \mathrm{e}^{x} – 1 \right) ^{\prime} = \mathrm{e}^{x} = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\mathrm{e}^{x} – 1 \sim x
}
$$

7. $\textcolor{gray}{ \boxed{ \textcolor{white}{1 – \cos x \textcolor{orange}{\sim} \frac{1}{2} x^{2} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( 1 – \cos x \right)^{\prime \prime} = \cos x = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( \frac{1}{2}x^{2} \right) ^{\prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
1 – \cos x \sim \frac{1}{2} x^{2}
}
$$

8. $\textcolor{gray}{ \boxed{ \textcolor{white}{x – \ln(1 + x) \textcolor{orange}{\sim} \frac{1}{2} x^{2} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left[ x – \ln(1 + x) \right] ^{\prime \prime} = \frac{1}{(1+x)^{2}} = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( \frac{1}{2}x^{2} \right) ^{\prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
x – \ln (1+x) \sim \frac{1}{2} x^{2}
}
$$

9. $\textcolor{gray}{ \boxed{ \textcolor{white}{\tan x – \sin x \textcolor{orange}{\sim} \frac{1}{2} x^{3} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \tan x – \sin x \right) ^{\prime \prime \prime} = 4\sec^2 x \tan^2 x + 2\sec^4 x + \cos x = \ & \textcolor{black}{\colorbox{yellow}{3}} \\ \\
\left( \frac{1}{2} x^{3} \right) ^{\prime \prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{3}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\tan x – \sin x \sim \frac{1}{2} x^{3}
}
$$

10. $\textcolor{gray}{ \boxed{ \textcolor{white}{\arcsin x – \arctan x \textcolor{orange}{\sim} \frac{1}{2} x^{3} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \arcsin x – \arctan x \right) ^{\prime \prime \prime} = \frac{1 + 2x^2}{(1 – x^2)^{5/2}} + \frac{2(1 – 3x^2)}{(1 + x^2)^3} = \ & \textcolor{black}{\colorbox{yellow}{3}} \\ \\
\left( \frac{1}{2} x^{3} \right) ^{\prime \prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{3}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\arcsin x – \arctan x \sim \frac{1}{2} x^{3}
}
$$

11. $\textcolor{gray}{ \boxed{ \textcolor{white}{\tan x – x \textcolor{orange}{\sim} \frac{1}{3} x^{3} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( \tan x – x \right) ^{\prime \prime \prime} = 4\sec^2 x \tan^2 x + 2\sec^4 x = \ & \textcolor{black}{\colorbox{yellow}{2}} \\ \\
\left( \frac{1}{3}x^{3} \right) ^{\prime \prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{2}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
\tan x – x \sim \frac{1}{3} x^{3}
}
$$

12. $\textcolor{gray}{ \boxed{ \textcolor{white}{x – \arctan x \textcolor{orange}{\sim} \frac{1}{3} x^{3} }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( x – \arctan x \right) ^{\prime \prime \prime} = \frac{2(1-3x^2)}{(1+x^2)^3} = \ & \textcolor{black}{\colorbox{yellow}{2}} \\ \\
\left( \frac{1}{3} x^{3} \right) ^{\prime \prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{2}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
x – \arctan x \sim \frac{1}{3} x^{3}
}
$$

13. $\textcolor{gray}{ \boxed{ \textcolor{white}{x – \sin x \textcolor{orange}{\sim} \frac{1}{6} x^{3}}}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( x – \sin x \right)^{\prime \prime \prime} = \cos x = \ & \textcolor{black}{\colorbox{yellow}{1}} \\ \\
\left( \frac{1}{6} x^{3} \right) ^{\prime \prime \prime} = \ & \textcolor{black}{\colorbox{yellow}{1}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
x – \sin x \sim \frac{1}{6} x^{3}
}
$$

14. $\textcolor{gray}{ \boxed{ \textcolor{white}{(1 + x)^{a} – 1 \textcolor{orange}{\sim} ax }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left[ (1 + x)^{a} – 1 \right] ^{\prime} = a (1+x)^{a-1} = \ & \textcolor{black}{\colorbox{yellow}{a}} \\ \\
\left( ax \right)^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{a}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
(1+x)^{a} – 1 \sim ax
}
$$

15. $\textcolor{gray}{ \boxed{ \textcolor{white}{a^{x} – 1 \textcolor{orange}{\sim} \ln a \times x }}}$

因为，当 $x \rightarrow 0$ 时：

$$
\begin{aligned}
\left( a^{x} – 1 \right) ^{\prime} = a^{x} \ln a = \ & \textcolor{black}{\colorbox{yellow}{ln a}} \\ \\
\left( \ln a \times x \right) ^{\prime} = \ & \textcolor{black}{\colorbox{yellow}{ln a}}
\end{aligned}
$$

所以：

$$
\textcolor{lightgreen}{
a^{x} – 1 \sim \ln a \times x
}
$$

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