荒原之梦·考研数学：2025 考研每日一题（002）

题目

$$I=\underset{x\to 1}{\lim }\frac{(1-x)(1-\sqrt{x})\cdots (1-\sqrt[n]{x})}{(1-x{)}^n}=?$$

解析

I = lim ⁡ x → 1 ( 1 − x ) ( 1 − x ) ⋯ ( 1 − x n ) ( 1 − x ) n = lim ⁡ ( x − 1 ) → 0 ( 1 − x ) [ 1 − 1 + ( x − 1 ) ] ⋯ ( 1 − 1 + ( x − 1 ) n ) ( 1 − x ) n = lim ⁡ ( x − 1 ) → 0 ( 1 − x ) [ − 1 2 ( x − 1 ) ] ⋯ [ − 1 n ( x − 1 ) ] ( 1 − x ) n = lim ⁡ ( 1 − x ) → 0 ( 1 − x ) 1 2 ( 1 − x ) ⋯ 1 n ( 1 − x ) ( 1 − x ) n = lim ⁡ ( 1 − x ) → 0 ( 1 − x ) n ⋅ 1 1 ⋅ 1 2 ⋯ 1 n ( 1 − x ) n = 1 n !

$$\begin{aligned} I = & \lim_{x \rightarrow 1} \frac{(1-x) (1-\sqrt{x}) \cdots (1- \sqrt[n]{x})}{ (1-x)^{n} } \\ = & \lim_{(x-1) \rightarrow 0} \frac{(1-x) [1-\sqrt{1 + (x-1)}] \cdots (1- \sqrt[n]{1 + (x-1)})}{ (1-x)^{n} } \\ = & \lim_{(x-1) \rightarrow 0} \frac{(1-x) [-\frac{1}{2} (x-1)] \cdots [-\frac{1}{n} (x-1)]}{(1-x)^{n}} \\ = & \lim_{(1-x) \rightarrow 0} \frac{(1-x) \frac{1}{2} (1-x) \cdots \frac{1}{n} (1-x)}{(1-x)^{n}} \\ = & \lim_{(1-x) \rightarrow 0} \frac{(1-x)^{n} \cdot \frac{1}{1} \cdot \frac{1}{2} \cdots \frac{1}{n}}{(1-x)^{n}} = \frac{1}{n!} \end{aligned}$$ I=====​x→1lim​(1−x)n(1−x)(1−x ​)⋯(1−nx ​)​(x−1)→0lim​(1−x)n(1−x)[1−1+(x−1) ​]⋯(1−n1+(x−1) ​)​(x−1)→0lim​(1−x)n(1−x)[−21​(x−1)]⋯[−n1​(x−1)]​(1−x)→0lim​(1−x)n(1−x)21​(1−x)⋯n1​(1−x)​(1−x)→0lim​(1−x)n(1−x)n⋅11​⋅21​⋯n1​​=n!1​​

详细解析：当分子中包含无穷多个因式的时候，该怎么计算极限？ - 荒原之梦

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