要计算 $\infty-\infty$ 型未定式，一般要先转为 $\infty / \infty$ 或 $0/0$ 型未定式

一、题目

$$
\begin{aligned}
I_{1} & = \lim_{x \rightarrow 0} \left( \frac{1}{\ln (x + \sqrt{1+x^{2}})} – \frac{1}{\ln (1+x)} \right) \\ \\
I_{2} & = \lim_{x \rightarrow 0} \left( \frac{1}{x^{2}} – \frac{1}{\sin^{2} x} \right)
\end{aligned}
$$

二、解析

$I_{1}$

由于 $x \rightarrow 0$ 的时候，式子 $I_{1}$ 是一个 $\infty – \infty$ 型的未定式，因此，需要先通分，将原式转换成 $\frac{\infty}{\infty}$ 型的未定式：

$$
\begin{aligned}
\textcolor{lightgreen}{ I_{1} } & = \lim_{x \rightarrow 0} \left( \frac{1}{\ln \left(x + \sqrt{1 + x^{2}} \right)} – \frac{1}{\ln (1 + x)} \right) \\ \\
& = \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{\ln (1 + x) – \ln \left(x + \sqrt{1 + x^{2}} \right) }{ \ln \left(x + \sqrt{1 + x^{2}} \right) \cdot \ln (1 + x)} } \\ \\
\end{aligned}
$$

于是，通过洛必达运算，可得：

$$
\begin{aligned}
\textcolor{lightgreen}{ I_{1} } & = \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{\ln (1 + x) – \ln \left(x + \sqrt{1 + x^{2}} \right) }{ \ln \left(x + \sqrt{1 + x^{2}} \right) \cdot \ln (1 + x)} } \\ \\
& = \lim_{x \rightarrow 0} \frac{\frac{1}{1 + x} – \frac{1}{\sqrt{1 + x^{2}}}}{\frac{1}{\sqrt{1 + x^{2}}} \ln(1 + x) + \frac{1}{1 + x} \ln \left(x + \sqrt{1 + x^{2}} \right) } \\ \\
& = \lim_{x \rightarrow 0} \frac{\sqrt{1+x^{2}} – (1+x)}{(1 + x)(\sqrt{1+x^{2}})} \cdot \frac{1}{\frac{1}{\sqrt{1 + x^{2}}} \ln(1 + x) + \frac{1}{1 + x} \ln \left(x + \sqrt{1 + x^{2}} \right) } \\ \\
& = \lim_{x \rightarrow 0}\frac{\sqrt{1 + x^{2}}−\left(1 + x\right)}{\left(1 + x\right)\ln \left(1 + x\right) + \sqrt{1 + x^{2}} \cdot \ln \left(x + \sqrt{1 + x^{2}}\right)} \\ \\
& \leadsto \textcolor{gray}{洛必达运算} \\ \\
& = \lim_{x \rightarrow 0} \frac{\frac{x}{\sqrt{1 + x^{2}}} – 1}{\ln(1 + x) + 1 + \frac{x}{\sqrt{ 1 + x^{2}}} \ln \left(x + \sqrt{ 1 + x^{2}} \right) + 1 } \\ \\
& = \textcolor{lightgreen}{ – \frac{1}{2} }
\end{aligned}
$$

$I_{2}$

由于 $x \rightarrow 0$ 的时候，式子 $I_{2}$ 是一个 $\infty – \infty$ 型的未定式，因此，需要先通分，将原式转换成 $\frac{0}{0}$ 型的未定式：

$$
\begin{aligned}
\textcolor{lightgreen}{ I_{2} } = \lim_{x \rightarrow 0} \left(\frac{1}{x^{2}} – \frac{1}{\sin^{2} x} \right) = \ & \lim_{x \rightarrow 0} \frac{\sin^{2} x – x^{2}}{ x^{2} \sin^{2} x } \\ \\
= \ & \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{ \sin^{2} x – x^{2}}{ x^{4} } }
\end{aligned}
$$

于是，通过洛必达运算，可得：

$$
\begin{aligned}
\textcolor{lightgreen}{ I_{2} } = \ & \lim_{x \rightarrow 0} \textcolor{lightgreen}{ \frac{ \sin^{2} x – x^{2}}{ x^{4} } } \\ \\
= \ & \lim_{ x \rightarrow 0 } \frac{ \sin 2x – 2x } { 4 x^{3}} \\ \\
= \ & \lim_{x \rightarrow 0} \frac{2 \cos 2 x – 2}{1 2 x^{2}} \\ \\
= \ & \frac{1}{6} \lim_{x \rightarrow 0} \frac{-2 \sin 2x }{ 2x } \\ \\
= \ & \textcolor{lightgreen}{ – \frac{1}{3} }
\end{aligned}

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