掌握这个拆分法，再复杂的极限题也能各个击破！

题目 1

$$
I = \lim_{x \to 0} \frac{\cos(\tan x) – 1 – \ln(\cos x)}{x(x – \arctan x)}
$$

难度评级：

解析 1

需要用到的公式（$x \to 0$）：

$$
\begin{aligned}
& x – \arctan x \sim \frac{1}{3}x^{3} \\ \\
& f(b) – f(a) = (b – a) \cdot f^{\prime}(\xi) \\ \\
& \ln(1+x) = x – \frac{1}{2}x^{2} + \frac{1}{3}x^{3} + \cdots \\ \\
& u – \ln(1+u) \sim \frac{1}{2} u^{2}, \ u \to 0 \\ \\
& \cos x – 1 \sim -\frac{1}{2}x^{2}
\end{aligned}
$$

先处理分母部分，根据麦克劳林公式：

$$
\begin{aligned}
& \ \arctan x = x – \frac{1}{3}x^{3} + o(x^{3}) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ x – \arctan x = \frac{1}{3}x^{3} + o(x^{3}) \sim \frac{1}{3}x^{3} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ x(x – \arctan x) \sim x \cdot \frac{1}{3}x^{3} \sim \frac{1}{3}x^{4}
\end{aligned}
$$

接着，为了在原式中使用拉格朗日中值定理，我们在分子中加减一项 $\cos x$, 即：

$$
\begin{aligned}
I & = \lim_{x \to 0} \frac{\cos(\tan x) – 1 – \ln(\cos x)}{\frac{1}{3}x^{4}} \\ \\
& = 3 \lim_{x \to 0} \frac{\cos(\tan x) – \cos x + \cos x – 1 – \ln(\cos x)}{x^{4}} \\ \\
& = 3 \lim_{x \to 0} \frac{\cos(\tan x) – \cos x}{x^{4}} + 3 \lim_{x \to 0} \frac{\cos x – 1 – \ln(\cos x)}{x^{4}} \\ \\
\end{aligned}
$$

于是，由拉格朗日中值定理，可知：

$$
\begin{aligned}
I_{1} & = \lim_{x \to 0} \frac{\cos(\tan x) – \cos x}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{(\tan x – x) \cdot (-\sin \xi)}{x^{4}}
\end{aligned}
$$

其中，$\xi$ 介于 $x$ 和 $\tan x$ 之间，当 $x \to 0$ 时，$\tan x \sim x$, 因此，$\xi \sim x$, 于是：

$$
I_{1} = \lim_{x \to 0} \frac{(\frac{1}{3}x^{3}) \cdot (-x)}{x^{4}} = \frac{-1}{3}
$$

接着，由等价无穷小替换与泰勒展开可知：

$$
\begin{aligned}
I_{2} & = \lim_{x \to 0} \frac{\cos x – 1 – \ln(\cos x)}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{(\cos x – 1) – \ln(1 + (\cos x – 1))}{x^{4}}
\end{aligned}
$$

于是，令 $u = \cos x – 1$, 则由前面的补充知识点可知，$u – \ln(1+u) \sim \frac{1}{2}u^{2}$, 因此：

$$
\begin{aligned}
I_{2} & = \lim_{x \to 0} \frac{\frac{1}{2}(\cos x – 1)^{2}}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{\frac{1}{2} \left( \frac{-1}{2}x^{2} \right)^{2}}{x^{4}} \\ \\
& = \frac{1}{8}
\end{aligned}
$$

综上可知：

$$
\begin{aligned}
I & = 3(I_{1} + I_{2}) \\ \\
& = 3 \left(\frac{-1}{3} + \frac{1}{8}\right) \\ \\
& = 3 \left(\frac{-5}{24}\right) \\ \\
& = \frac{-5}{8}
\end{aligned}
$$

题目 2

$$
I = \lim_{x \to 0} \frac{\mathrm{e}^{\sin^{2} x} – \sqrt{1 + 2x^{2}}}{\ln(1 + x^{2}) – \sin^{2} x}
$$

难度评级：

解析 2

需要用到的公式（$x \to 0$）：

$$
\begin{aligned}
& \ln(1+x) = x – \frac{1}{2}x^{2} + \cdots \\ \\
& (1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^{2} + \cdots \\ \\
& \sqrt{1+x} = 1 + \frac{1}{2}x – \frac{1}{8}x^{2} + \cdots \\ \\
& \sin x = x – \frac{1}{6}x^{3} + \cdots \\ \\
& \mathrm{e}^{x} = 1 + x + \frac{1}{2}x^{2} + \cdots
\end{aligned}
$$

首先处理分母部分，通过麦克劳林公式展开到 $x^{4}$ 阶：

$$
\begin{aligned}
& \ \begin{cases} \ln(1 + x^{2}) = x^{2} – \frac{1}{2}x^{4} + o(x^{4}) \\ \\
\sin^{2} x = (x – \frac{1}{6}x^{3})^{2} = x^{2} – \frac{1}{3}x^{4} + o(x^{4}) \end{cases} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \ln(1 + x^{2}) – \sin^{2} x = \left(-\frac{1}{2} + \frac{1}{3}\right)x^{4} + o(x^{4}) \sim \frac{-1}{6}x^{4}
\end{aligned}
$$

接着，对原式进行拆分：

$$
\begin{aligned}
I & = \lim_{x \to 0} \frac{\mathrm{e}^{\sin^{2} x} – \sqrt{1 + 2x^{2}}}{-\frac{1}{6}x^{4}} \\ \\
& = -6 \lim_{x \to 0} \frac{\mathrm{e}^{\sin^{2} x} – \mathrm{e}^{x^{2}} + \mathrm{e}^{x^{2}} – \sqrt{1 + 2x^{2}}}{x^{4}} \\ \\
& = -6 \left[ \lim_{x \to 0} \frac{\mathrm{e}^{\sin^{2} x} – \mathrm{e}^{x^{2}}}{x^{4}} + \lim_{x \to 0} \frac{\mathrm{e}^{x^{2}} – \sqrt{1 + 2x^{2}}}{x^{4}} \right]
\end{aligned}
$$

于是，由拉格朗日中值定理，可知：

$$
\begin{aligned}
I_{1} & = \lim_{x \to 0} \frac{\mathrm{e}^{\sin^{2} x} – \mathrm{e}^{x^{2}}}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{(\sin^{2} x – x^{2}) \cdot \mathrm{e}^\xi}{x^{4}}
\end{aligned}
$$

由于 $\xi$ 介于 $\sin^{2} x$ 和 $x^{2}$ 之间，当 $x \to 0$ 时，$\xi \to 0$, 所以 $\mathrm{e}^\xi \to 1$, 于是：

$$
\begin{aligned}
\sin^{2} x – x^{2} & = (\sin x – x)(\sin x + x) \\ \\
& \sim \left(-\frac{1}{6}x^{3}\right) \cdot (2x) \\ \\
& \sim -\frac{1}{3}x^{4}
\end{aligned}
$$

所以：

$$
I_{1} = \lim_{x \to 0} \frac{-\frac{1}{3}x^{4} \cdot 1}{x^{4}} = -\frac{1}{3}
$$

接着，令：

$$
I_{2} = \lim_{x \to 0} \frac{\mathrm{e}^{x^{2}} – \sqrt{1 + 2x^{2}}}{x^{4}}
$$

又由泰勒展开，分别将两项展开至 $x^{4}$ 阶，可得：

$$
\begin{aligned}
\mathrm{e}^{x^{2}} & = 1 + x^{2} + \frac{1}{2}(x^{2})^{2} + o(x^{4}) \\ \\ & = 1 + x^{2} + \frac{1}{2}x^{4} + o(x^{4}) \\ \\
\sqrt{1 + 2x^{2}} & = 1 + \frac{1}{2}(2x^{2}) – \frac{1}{8}(2x^{2})^{2} + o(x^{4}) \\ \\ &= 1 + x^{2} – \frac{1}{2}x^{4} + o(x^{4})
\end{aligned}
$$

于是：

$$
\begin{aligned}
\mathrm{e}^{x^{2}} – \sqrt{1 + 2x^{2}} & = \left(1 + x^{2} + \frac{1}{2}x^{4}\right) – \left(1 + x^{2} – \frac{1}{2}x^{4}\right) + o(x^{4}) \\ \\ & \sim x^{4}
\end{aligned}
$$

因此：

$$
I_{2} = \lim_{x \to 0} \frac{x^{4}}{x^{4}} = 1
$$

综上可得：

$$
\begin{aligned}
I & = -6(I_{1} + I_{2}) = -6\left(-\frac{1}{3} + 1\right) \\ \\
& = -6 \cdot \frac{2}{3} \\ \\
& = -4
\end{aligned}
$$

题目 3

$$
I = \lim_{x \to 0} \frac{\cos(\sin x) – \mathrm{e}^{\cos x – 1}}{\tan^{2} x – \sin^{2} x}
$$

难度评级：

解析 3

需要用到的公式（$x \to 0$）：

$$
\begin{aligned}
& x + x^{3} \sim x \\ \\
& f(b) – f(a) = (b – a) \cdot f^{\prime}(\xi) \\ \\
& \mathrm{e}^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3!} + \cdots \\ \\
& \mathrm{e}^{x} – 1 \sim x \\ \\
& \mathrm{e}^{x} – 1 – x \sim \frac{x^{2}}{2} \\ \\
& \mathrm{e}^{\square} – 1 – \square \sim \frac{\square^{2}}{2}, \ \square \to 0
\end{aligned}
$$

首先对分母进行处理，得：

$\tan^{2} x – \sin^{2} x = (\tan x – \sin x)(\tan x + \sin x)$

接着， 通过泰勒公式（麦克劳林公式）展开，得：

$$
\begin{cases}
\tan x = x + \frac{1}{3}x^{3} + o(x^{3}) \\
\sin x = x – \frac{1}{6}x^{3} + o(x^{3})
\end{cases}
\Rightarrow
\begin{cases}
\tan x – \sin x = \frac{1}{2}x^{3} + o(x^{3}) \sim \frac{1}{2}x^{3} \\
\tan x + \sin x = 2x + o(x) \sim 2x
\end{cases}
$$

于是：

$$
\begin{aligned}
\tan^{2} x – \sin^{2} x & \sim \frac{1}{2}x^{3} \cdot 2x \\ \\
& = x^{4}
\end{aligned}
$$

于是，原式可化简为：

$$
\begin{aligned}
I & = \lim_{x \to 0} \frac{\cos(\sin x) – \mathrm{e}^{\cos x – 1}}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{\cos(\sin x) – \cos x + \cos x – \mathrm{e}^{\cos x – 1}}{x^{4}} \\ \\
& = \lim_{x \to 0} \frac{\cos(\sin x) – \cos x}{x^{4}} + \lim_{x \to 0} \frac{\cos x – \mathrm{e}^{\cos x – 1}}{x^{4}} \\ \\
\end{aligned}
$$

接着，由拉格朗日中值定理，可知：

$$
I_{1} = \lim_{x \to 0} \frac{\cos(\sin x) – \cos x}{x^{4}} = \lim_{x \to 0} \frac{(\sin x – x) \cdot (-\sin \xi)}{x^{4}}
$$

由于 $\xi$ 介于 $x$ 和 $\sin x$ 之间，当 $x \to 0$ 时，$\sin x \sim x$, 所以 $\xi \sim x$, 于是：

$$
I_{1} = \lim_{x \to 0} \frac{\left( \frac{-1}{6}x^{3} \right) \cdot (-x)}{x^{4}} = \frac{1}{6}
$$

又由泰勒公式展开，得：

$$
\begin{aligned}
I_{2} & = \lim_{x \to 0} \frac{\cos x – \mathrm{e}^{\cos x – 1}}{x^{4}} \\ \\
& = -\lim_{x \to 0} \frac{\mathrm{e}^{\cos x – 1} – \cos x}{x^{4}} \\ \\
& = -\lim_{x \to 0} \frac{\mathrm{e}^{\cos x – 1} – (\cos x – 1) – 1}{x^{4}} \\ \\
& = -\lim_{x \to 0} \frac{\frac{1}{2}(\cos x – 1)^{2}}{x^{4}} \\ \\
& = -\lim_{x \to 0} \frac{\frac{1}{2}(-\frac{1}{2}x^{2})^{2}}{x^{4}} = -\frac{1}{8}
\end{aligned}
$$

综上可得：

$$
\begin{aligned}
I & = I_{1} + I_{2} \\ \\
& = \frac{1}{6} – \frac{1}{8} \\ \\
& = \frac{4 – 3}{24} \\ \\
& = \frac{1}{24}
\end{aligned}
$$

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