一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
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 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
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:
$$
\begin{aligned}
\lim_{x \rightarrow 0} \frac{f(x) – x}{x^{3}} & = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – \sin x + \sin x – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot e^{x^{2}} – \sin x}{x^{3}} + \lim_{x \rightarrow 0} \frac{\sin x – x}{x^{3}} \\ \\
& = \lim_{x \rightarrow 0} \frac{\sin x \cdot (e^{x^{2}} – 1)}{x^{3}} + \lim_{x \rightarrow 0} \frac{ – \frac{1}{6} x^{3} }{x^{3}} \\ \\
& = 1 – \frac{1}{6} = \frac{5}{6}
\end{aligned}
$$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
线性代数![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
以独特的视角解析线性代数,让繁复的知识变得直观明了。
特别专题![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。