2

(1)

等价于

\begin{aligned} lim_{x \to 1} \frac{x^{n} - 1}{x^{m} - 1} \\ = & lim_{t \to 0} \frac{(t + 1)^{n} - 1}{(t + 1)^{m} - 1} \\ = & lim_{t \to 0} \frac{t^{n} + n t + o (t)}{t^{m} + m t + o (t)} \\ = & lim_{t \to 0} \frac{t^{n - 1} + n + o (1)}{t^{m - 1} + m + o (1)} \\ = & \frac{n}{m} \end{aligned}

(2)

令 $1 + α x = t^{m}$ 可得

lim_{t \to 1} \frac{α (t - 1)}{t^{m} - 1} = \frac{α}{m}

(3)

lim_{x \to 0} \frac{\sqrt[m]{1 + α x} - \sqrt[n]{1 + β x}}{x} = lim_{x \to 0} \frac{\sqrt[m]{1 + α x} - 1}{x} - lim_{x \to 0} \frac{\sqrt[n]{1 + β x} - 1}{x} = \frac{α}{m} - \frac{β}{n}

(4)

原式等价于

\prod_{i = 1}^{n} lim_{x \to 1} \frac{1 - \sqrt[i]{x}}{1 - x} = \prod_{i = 1}^{n} \frac{1}{i} = \frac{1}{n!}

3

(7)

在我手上这个版本中是 $x \to + \infty$ , 实际上应该是 $n \to + \infty$ .

lim_{n \to \infty} n (\cos (\frac{x}{\sqrt{n}}) - 1)

等价于

lim_{t \to 0} x^{2} \frac{\cos (t) - 1}{t^{2}}

由于 $\cos (t) - 1 \sim - \frac{1}{2} t^{2}$ , 结果为 $\frac{- x^{2}}{2}$ .

4

lim_{n \to \infty} n \frac{\sum_{i = 1}^{n} x^{i}}{n} = lim_{n \to \infty} \sum_{i = 1}^{n} x^{i} = lim_{n \to \infty} \frac{x - x^{n + 1}}{1 - x} = \frac{x}{1 - x}

2

(1)