From 57dab8600aae0f122911e28613e8fc660320086c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com>
Date: Sat, 27 Dec 2025 08:30:03 +0800
Subject: [PATCH] vault backup: 2025-12-27 08:30:03

---
 .../试卷/期中考前押题卷解析版.md  | 24 +++++++------------
 1 file changed, 9 insertions(+), 15 deletions(-)

diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md
index ea19c16..dc622a1 100644
--- a/编写小组/试卷/期中考前押题卷解析版.md
+++ b/编写小组/试卷/期中考前押题卷解析版.md
@@ -130,22 +130,16 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$
 
 6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。
 
-**解析：**  
-这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为：  
-$$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$
-利用等价无穷小和泰勒展开：
-
-- $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$；
-    
-- $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。  
-    因此：  
-    $$\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2).$$
-    取对数：  
-    $$\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1).$$
-    所以原极限为 $e^{7/2}$。
-    
+**解析：**  $$\begin{aligned}
+\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}
+&=\lim\limits_{x\to\infty}(1+\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)^{\frac{1}{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}\cdot x^2\cdot(\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)}
+\\&=e^{\lim\limits_{x\to\infty}\frac{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}{\frac{1}{x^2}}}
+\\&\overset{t=\frac{1}{x}}{=}e^{\lim\limits_{t\to0}\frac{\tan^2(2t)+\cos t-1}{t^2}}
+\\&=e^{\lim\limits_{t\to0}\frac{4t^2}{t^2}-\frac{\frac{1}{2}t^2}{t^2}}(四则运算和等价无穷小)
+\\&=e^{\frac{7}{2}}
+\end{aligned}$$
 
-**答案：** $e^{7/2}$
+**答案：** $e^{\frac{7}{2}}$
 
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