式子复杂不要怕，先分析其“型”，再确定求解之“法”

一、题目

已知：

$$
f(x) = \lim_{t \rightarrow x} \sin x \cdot \left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}}
$$

则：

$$
\lim_{x \rightarrow 0} \frac{f(x) – x}{x^{3}} = ?
$$

难度评级：

二、解析

首先：

$$
\begin{aligned}
f(x) & = \lim_{t \rightarrow x} \sin x \cdot \left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}} \\ \\
& = \sin x \cdot \lim_{t \rightarrow x} \left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}}
\end{aligned}
$$

观察可知，当 $t \rightarrow x$ 的时候，$\left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}}$ 是一个 $1^{\infty}$ 型极限，因此：

$$
\begin{aligned}
f(x) & = \lim_{t \rightarrow x} \sin x \cdot \left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}} \\ \\
& = \sin x \cdot \lim_{t \rightarrow x} \left( \frac{t}{x} \right)^{\frac{t^{3}}{t – x}} \\ \\
& = \sin x \cdot \lim_{t \rightarrow x} \left( 1 + \frac{t}{x} – 1 \right)^{\frac{t^{3}}{t – x}} \\ \\
& = \sin x \cdot \lim_{t \rightarrow x} \left( 1 + \frac{t – x}{x} \right)^{\frac{t^{3}}{t – x}} \\ \\
& = \sin x \cdot \lim_{t \rightarrow x} \left( 1 + \frac{t – x}{x} \right)^{\frac{x}{t – x} \cdot \frac{t^{3}}{t – x} \cdot \frac{t – x}{x}} \\ \\
& = \sin x \cdot \lim_{x \rightarrow x} e^{\frac{t^{3}}{x}} \\ \\
& = \sin x \cdot e^{x^{2}}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\lim_{x \rightarrow 0} \frac{f(x) – x}{x^{3}} & = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – \sin x + \sin x – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – \sin x}{x^{3}} + \lim_{x \rightarrow 0} \frac{\sin x – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot (e^{x^{2}} – 1)}{x^{3}} + \lim_{x \rightarrow 0} \frac{ – \frac{1}{6} x^{3} }{x^{3}} \\ \\
& = 1 – \frac{1}{6} = \frac{5}{6}
\end{aligned}
$$

---