# 极限乘法运算中，极限非零的因子的极限可以直接代入

## 一、题目

$$I = \lim \limits_{x \rightarrow 0}\left(\frac{1+\sin x \cos \alpha x}{1+\sin x \cos \beta x}\right)^{\cos t^{3} x} = \ ?$$

## 二、解析

\begin{aligned} I = & \lim \limits_{x \rightarrow 0}\left(\frac{1+\sin x \cos \alpha x}{1+\sin x \cos \beta x}\right)^{\cos t^{3} x} \\ = & \lim \limits_{x \rightarrow 0}\left(1+\frac{1+\sin x \cos 2 x}{1+\sin x \cos \beta x}-1\right)^{\cos t^{3} x} \\ = & \lim \limits_{x \rightarrow 0}\left[1+\frac{\sin x \cos 2 x-\sin x \cos \beta x}{1+\sin x \cos \beta x}\right]^{\operatorname{cost}^{3} x} \end{aligned}

$$K = \frac{\sin x \cos 2 x-\sin x \cos \beta x}{1+\sin x \cos \beta x}$$

\begin{aligned} I = & \lim \limits_{x \rightarrow 0}[1 + K]^{\cos t^{3} x} \\ = & \lim \limits_{x \rightarrow 0}[1 + K]^{\frac{1}{K}} \cdot K \cos ^{3} x \\ = & e^{\lim \limits_{x \rightarrow 0} K \cos t^{3} x} \end{aligned}

\begin{aligned} \lim \limits_{x \rightarrow 0} K \cos t^{3} x = & \lim \limits_{x \rightarrow 0} \frac{\sin x(\cos \alpha x-\cos \beta x)}{1+\sin x \cos \beta x} \cdot \frac{\cos ^{3} x}{\sin ^{3} x} \\ = & \lim \limits_{x \rightarrow 0} \frac{\textcolor{springgreen}{\cos ^{3} x} (\cos \alpha x-\cos \beta x) }{\sin ^{2} x(1+\sin x \cos \beta x)} \\ = & \lim \limits_{x \rightarrow 0} \frac{\textcolor{springgreen}{1} \cdot (\cos 2 x-\cos \beta x)}{\sin ^{2} x(\textcolor{orangered}{1+\sin x \cos \beta x})} \\ = & \lim \limits_{x \rightarrow 0} \frac{(1-\cos \beta x)-(1-\cos \alpha x) }{x^{2}(\textcolor{orangered}{1+0})} \\ = & \lim \limits_{x \rightarrow 0} \frac{\frac{1}{2}\left[(\beta x)^{2}-(\alpha x)^{2}\right]}{x^{2}}=\frac{\beta^{2}-\alpha^{2}}{2} \end{aligned}

$$I=e^{\frac{\beta^{2}-2^{2}}{2}}$$