From 73c8fff5757ecf712a9435fbd44f3624e7bca467 Mon Sep 17 00:00:00 2001
From: unknown <2974730459@qq.com>
Date: Fri, 26 Dec 2025 00:37:55 +0800
Subject: [PATCH] vault backup: 2025-12-26 00:37:55

---
 编写小组/课前测/课前测4.md | 24 ++++++++++++++++++++++--
 1 file changed, 22 insertions(+), 2 deletions(-)

diff --git a/编写小组/课前测/课前测4.md b/编写小组/课前测/课前测4.md
index 5955394..d9fd563 100644
--- a/编写小组/课前测/课前测4.md
+++ b/编写小组/课前测/课前测4.md
@@ -1,14 +1,34 @@
-1. 设 $x_n = \frac{\cos\left(\frac{2n\pi}{3}\right)}{n} + 1$，证明 $x_n \to 1$。
+---
+tags:
+  - 编写小组
+---
 
+1.设 $x_n = \frac{\cos\left(\frac{2n\pi}{3}\right)}{n} + 1$，证明 $x_n \to 1$。
 
+```
 
 
 
-2. 函数 $f(x) = \begin{cases} x^2, & \text{若 } x \text{ 为有理数} \\ -x^2, & \text{若 } x \text{ 为无理数} \end{cases}$
+
+
+
+```
+
+
+2.函数 $f(x) = \begin{cases} x^2, & \text{若 } x \text{ 为有理数} \\ -x^2, & \text{若 } x \text{ 为无理数} \end{cases}$
 求
 $$
 \lim_{x \to 0} f(x) 
 $$
+```
+
+
+
+
+
+
+
+```
 
 
 3.设$f(x)=\begin{cases} \frac{1}{|x|^{\alpha}}sin\frac{1}{x} \ \ ,x\neq0 \\ 0,\ \ \ x=0\end{cases}$在$x=0$处可导，则$\alpha$的取值范围是[   ].