同济高数 第1章 第1-6-1题

[AI解答]

[AI解答]

**（1）** $\displaystyle{\lim} _{x \rightarrow 0} \frac{\sin \omega x}{x}$ 利用重要极限 $\displaystyle{\lim}_{t \to 0} \frac{\sin t}{t} = 1$，令 $t = \omega x$，则 $$ \lim_{x \to 0} \frac{\sin \omega x}{x} = \lim_{x \to 0} \frac{\sin(\omega x)}{\omega x} \cdot \omega = 1 \cdot \omega = \omega. $$

**（2）** $\displaystyle{\lim} _{x \rightarrow 0} \frac{\tan 3 x}{x}$ 因为 $\displaystyle \tan 3x = \frac{\sin 3x}{\cos 3x}$，所以 $$ \lim_{x \to 0} \frac{\tan 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{x} \cdot \frac{1}{\cos 3x}. $$ 而 $\displaystyle{\lim}_{x \to 0} \frac{\sin 3x}{x} = 3$，$\displaystyle{\lim}_{x \to 0} \frac{1}{\cos 3x} = 1$，故极限为 $3$。

**（3）** $\displaystyle{\lim} _{x \rightarrow 0} \frac{\sin 2 x}{\sin 5 x}$ 利用 $\displaystyle{\lim}_{t \to 0} \frac{\sin t}{t} = 1$，有 $$ \lim_{x \to 0} \frac{\sin 2x}{\sin 5x} = \lim_{x \to 0} \frac{\frac{\sin 2x}{2x} \cdot 2x}{\frac{\sin 5x}{5x} \cdot 5x} = \frac{1 \cdot 2}{1 \cdot 5} = \frac{2}{5}. $$

**（4）** $\displaystyle{\lim} _{x \rightarrow 0} x \cot x$ 因为 $\displaystyle \cot x = \frac{\cos x}{\sin x}$，所以 $$ \lim_{x \to 0} x \cot x = \lim_{x \to 0} \frac{x \cos x}{\sin x} = \lim_{x \to 0} \frac{x}{\sin x} \cdot \cos x = 1 \cdot 1 = 1. $$

**（5）** $\displaystyle{\lim} _{x \rightarrow 0} \frac{1-\cos 2 x}{x \sin x}$ 利用 $1 - \cos 2x = 2\sin^2 x$，得 $$ \lim_{x \to 0} \frac{2\sin^2 x}{x \sin x} = \lim_{x \to 0} \frac{2\sin x}{x} = 2 \cdot 1 = 2. $$

**（6）** $\displaystyle{\lim} _{n \rightarrow \infty} 2^{n} \sin \frac{x}{2^{n}}$（$x \neq 0$） 令 $\displaystyle t = \frac{x}{2^n}$，则当 $n \to \infty$ 时 $t \to 0$，且 $\displaystyle 2^n = \frac{x}{t}$，于是 $$ \lim_{n \to \infty} 2^n \sin \frac{x}{2^n} = \lim_{t \to 0} \frac{x}{t} \sin t = x \cdot \lim_{t \to 0} \frac{\sin t}{t} = x \cdot 1 = x. $$

难度：★☆☆☆☆