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controller 接收上传文件的示例代码： ```java import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import java.io.*; import java.util.concurrent.CompletableFuture; @Controller public class FileUploadController { private static final String UPLOAD_DIRECTORY = "uploads"; @RequestMapping(value = "/upload", method = RequestMethod.POST) public @ResponseBody String handleFileUpload(@RequestParam("file") MultipartFile file, HttpServletRequest request) { String fileName = file.getOriginalFilename(); String filePath = request.getServletContext().getRealPath("/") + "\\" + UPLOAD_DIRECTORY + "\\" + fileName; File targetFile = new File(filePath); try { file.transferTo(targetFile); // 保存上传文件 } catch (IOException ex) { ex.printStackTrace(); return "上传文件失败"; } CompletableFuture future = CompletableFuture.runAsync(() -> { try { unzip(targetFile); // 解压缩上传文件 } catch (IOException e) { e.printStackTrace(); } }); future.thenRunAsync(() -> { try { storeImages(); // 存储解压后的图片 } catch (IOException e) { e.printStackTrace(); } }); return "上传成功"; } private static void unzip(File file) throws IOException { // 解压缩文件 } private static void storeImages() throws IOException { // 存储解压后的图片 } } ``` 1. 使用Spring MVC 提供的注解 `@RequestParam` 用来接收上传的压缩包文件，其中 `handleFileUpload` 方法返回的字符串即为上传成功或失败的提示信息。 2. 得到上传文件后，调用 `transferTo` 方法将上传的文件保存到服务器的指定文件夹中。 3. 使用 CompletableFuture 同步解压和存储文件操作，使用 `thenRunAsync` 方法将存储操作提交到另一个新线程中执行。 + 解压缩过程中，使用 `unzip` 方法解压缩文件 + 存储操作中，我们使用 `storeImages` 存储解压后的图片。需要注意的是，这里我们使用了两个异步操作，因此需要保证它们异步执行的顺序，确保解压缩完成后再进行存储操作。24:Tf4fd,## 2009年真题 **【2009年题44】**（13分）某计算机字长为 16 位，采用 16 位定长指令字结构，部分数据通路结构如下图所示，图中所有控制信号为 1 时表示有效、为 0 时表示无效。例如，控制信号 MDRE 为 1 表示允许数据从 DB 打入 MDR，MDRin 为 1 表示允许数据从内总线打入 MDR。假设 MAR 的输出一直处于使能状态。加法指令“ADD (R1), R0”的功能为 (R0) + ((R1)) → (R1)，即将 R0 中的数据与 R1 的内容所指主存单元的数据相加，并将结果送入 R1 的内容所指主存单元中保存。

下表给出了上述指令取指和译码阶段每个节拍（时钟周期）的功能和有效控制信号，请按表中描述方式用表格列出指令执行阶段每个节拍的功能和有效控制信号。

**解析：** 先分析题干中提供的示例：取指令的功能是根据 PC 的内容所指主存地址，取出指令代码，经过 MDR，最终送至 IR。这部分和后面的指令执行阶段的取操作数、存运算结果的方法是相通的。 C1: (PC)→MAR：在读写存储器前，必须先将地址（这里为(PC)）送至 MAR。 C2: M(MAR)→MDR, (PC)+1→PC：读写的数据必须经过 MDR，指令取出后 PC 自增 1。 C3: (MDR)→IR：然后将读到 MDR 中指令代码送至 IR 进行后续操作。 --- 1. 指令“ADD (R1), R0”的操作数一个在主存中，一个在寄存器中，运算结果在主存中。要读出 R1 的内容所指的主存单元，必须先将 R1 的内容送至 MAR，即(R1)→MAR。而读出的数据必须经过 MDR，即 M(MAR)→MDR。因此，将 R1 的内容所指主存单元的数据读出到 MDR 的节拍安排如下： C5: (R1)→MAR C6: M(MAR)→MDR 2. ALU 一端是寄存器 A，MDR 或 R0 中必须有一个先写入 A 中，如 MDR： C7: (MDR)→A 3. 然后执行加法操作，并将结果送入寄存器 AC： C8: (A) + (R0)→AC 4. 之后将加法结果写回到 R1 的内容所指主存单元，注意 MAR 中的内容没有改变： C9: (AC)→MDR C10: (MDR)→M(MAR) 有效控制信号的安排，还需注意特殊控制信号，如 PC+1、Add 等。于是得到参考答案如下：

本题答案不唯一，如果在 C6 执行 M(MAR)→MDR 的同时，完成(R0)→A（即选择将(R0)写入 A），并不会发生总线冲突，这种方案可节省 1 个节拍，见下表。

--- ## 2012年真题【2012年题44】某 16 位计算机中，带符号整数用补码表示，数据 Cache 和指令 Cache 分离。题 44 表给出了指令系统中部分指令格式，其中 Rs 和 Rd 表示寄存器，mem 表示存储单元地址，(x)表示寄存器 x 或存储单元 x 的内容。

该计算机采用 5 段流水方式执行指令，各流水段分别是取指（IF）、译码/读寄存器（ID）、执行/计算有效地址（EX）、访问存储器（M）和结果写回寄存器（WB），流水线采用“按序发射，按序完成”方式，没有采用转发技术处理数据相关，并且同一个寄存器的读和写操作不能在同一个时钟周期内进行。请回答下列问题： 1）若 int 型变量 x 的值为 -513，存放在寄存器 R1 中，则执行指令“SHR R1”后，R1 的内容是多少？（用十六进制表示） 2）若某个时间段中，有连续的 4 条指令进入流水线，在其执行过程中没有发生任何阻塞，则执行这 4 条指令所需的时钟周期数为多少？ 3）若高级语言程序中某赋值语句为 x = a + b，x、a 和 b 均为 int 型变量，它们的存储单元地址分别表示为[x]、[a]和[b]。该语句对应的指令序列及其在指令流水线中的执行过程如下图所示。 $I_1$ LOAD R1, [a] $I_2$ LOAD R2, [b] $I_3$ ADD R1, R2 $I_4$ STORE R2, [x]

则这 4 条指令执行过程中，$I_3$ 的 ID 段和 $I_4$ 的 IF 段被阻塞的原因各是什么？ 4）若高级语言程序中某赋值语句为 x = x*2 + a，x 和 a 均为 unsigned int 类型变量，它们的存储单元地址分别表示为[x]、[a]，则执行这条语句至少需要多少个时钟周期？要求模仿题 44 图画出这条语句对应的指令序列及其在流水线中的执行过程示意图。 **解析：** 1）x 的机器码为 $[x]_补 = \text{1111 1101 1111 1111B}$，即指令执行前 (R1) = FDFFH，右移 1 位后位 1111 1110 1111 1111B，即指令执行后 (R1) = FEFFH。（2 分） 2）每个时钟周期只能有一条指令进入流水线，从第 5 个时钟周期开始，每个时钟周期都会有一条指令执行完毕，故至少需要 4 + (5 - 1) = 8 个时钟周期。（2 分） 3）$I_3$ 的 ID 段被阻塞的原因：因为 $I_3$ 与 $I_1$ 和 $I_2$ 都存在数据相关，需等到 $I_1$ 和 $I_2$ 将结果写回寄存器后，$I_3$ 才能读寄存器内容，所以 $I_3$ 的 ID 段被阻塞（1 分）。$I_4$ 的 IF 段被阻塞的原因：因为 $I_4$ 的前一条指令 $I_3$ 在 ID 段被阻塞，所以 $I_4$ 的 IF 段被阻塞（1 分）。注意：要求“按序发射，按序完成”，故 2）中下一条指令的 IF 必须和上一条指令的 ID 并行，以免因上一条指令发生冲突而导致下一条指令先执行完。 4）因 2*x 操作有左移和加法两种实现方法，故 x=x*2+a 对应的指令序列为 I1 LOAD R1, [x] I2 LOAD R2, [a] I3 SHL R1 //或者 ADD R1, R1 I4 ADD R1, R2 I5 STORE R2, [x] 这 5 条指令在流水线中执行过程如下表所示。故执行 x=x*2+a 语句最少需要 17 个时钟周期。 --- ### 表与图的 img markup 代码块 **表格 3（解答流水线示意图）markup：** ```html

``` ## 2014年 **【2014年题44】**（12分）某程序中有如下循环代码段 P：“for (int i = 0; i < N; i++) sum+=A[i];”。假设编译时变量 sum 和 i 分别分配在寄存器 R1 和 R2 中。常量 N 在寄存器 R6 中，数组 A 的首地址在寄存器 R3 中。程序段 P 起始地址为 0804 8100H，对应的汇编代码和机器代码如下表所示。

执行上述代码的计算机 M 采用 32 位定长指令字，其中分支指令 bne 采用如下格式：

OP 为操作码；Rs 和 Rd 为寄存器编号；OFFSET 为偏移量，用补码表示。请回答下列问题，并说明理由。 1）M 的存储器编址单位是什么？ 2）已知 sll 指令实现左移功能，数组 A 中每个元素占多少位？ 3）表中 bne 指令的 OFFSET 字段的值是多少？已知 bne 指令采用相对寻址方式，当前 PC 内容为 bne 指令地址，通过分析表中指令地址和 bne 指令内容，推断出 bne 指令的转移目标地址计算公式。 4）若 M 采用如下“按序发射、按序完成”的 5 级指令流水线：IF（取值）、ID（译码及取数）、EXE（执行）、MEM（访存）、WB（写回寄存器），且硬件不采取任何转发措施，分支指令的执行均引起 3 个时钟周期的阻塞，则 P 中哪些指令的执行会由于数据相关而发生流水线阻塞？哪条指令的执行会发生控制冒险？为什么指令 1 的执行不会因为与指令 5 的数据相关而发生阻塞？ **解析：** 该题涉及指令系统、存储管理以及 CPU 三个部分内容，考生应注意各章节内容之间的联系，才能更好地把握当前考试的趋势。 1）已知计算机 M 采用 32 位定长指令字，即一条指令占 4B，观察表中各指令的地址可知，每条指令的地址差为 4 个地址单位，即 4 个地址单位代表 4B，一个地址单位就代表了 1B，所以该计算机是按字节编址的。（2 分） 2）在二进制中某数左移两位相当于乘以 4，由该条件可知，数组间的数据间隔为 4 个地址单位，而计算机按字节编址，所以数组 A 中每个元素占 4B。（2 分） 3）由表可知，bne 指令的机器代码为 1446FFFAH，根据题目给出的指令格式，后 2B 的内容为 OFFSET 字段，所以该指令的 OFFSET 字段为 FFFAH，用补码表示，值为 -6（1 分）。当系统执行到 bne 指令时，PC 自动加 4，PC 的内容就为 08048118H，而跳转的目标是 08048100H，两者相差了 18H，即 24 个单位的地址间隔，所以偏移址的一位即是真实跳转地址的 -24/-6 = 4 位（1 分）。可知 bne 指令的转移目标地址计算公式为 (PC) + 4 + OFFSET×4（1 分）。 4）由于数据相关而发生阻塞的指令为第 2、3、4、6 条，因为第 2、3、4、6 条指令都与各自前一条指令发生数据相关。（3 分）第 6 条指令会发生控制冒险。（1 分）当前循环的第五条指令与下次循环的第一条指令虽然有数据相关，但由于第 6 条指令后有 3 个时钟周期的阻塞，因而消除了该数据相关。（1 分） --- ## 2015年 **【2015年题43】**（13 分）某 16 位计算机的主存按字节编码，存取单位为 16 位；采用 16 位定长指令字格式；CPU 采用单总线结构，主要部分如下图所示。图中 R0～R3 为通用寄存器；T 为暂存器；SR 为移位寄存器，可实现直送（mov）、左移一位（left）和右移一位（right）3 种操作，控制信号为 SRop，SR 的输出由信号 SRout 控制；ALU 可实现直送 A（mova）、A 加 B（add）、A 减 B（sub）、A 与 B（and）、A 或 B（or）、非 A（not）和 A 加 1（inc）7 种操作，控制信号为 ALUop。

请回答下列问题。 1）图中哪些寄存器是程序员可见的？为何要设置暂存器 T？ 2）控制信号 ALUop 和 SRop 的位数至少各是多少？ 3）控制信号 SRout 所控制部件的名称或作用是什么？ 4）端点①～⑨中，哪些端点须连接到控制部件的输出端？ 5）为完善单总线数据通路，需要在端点①～⑨中相应的端点之间添加必要的连线。写出连线的起点和终点，以正确表示数据的流动方向。 6）为什么二路选择器 MUX 的一个输入端是 2？ **解析：** 1）程序员可见寄存器为通用寄存器（R0～R3）和 PC。因为采用了单总线结构，因此，若无暂存器 T，则 ALU 的 A、B 端口会同时获得两个相同的数据，使数据通路不能正常工作。【评分说明】回答通用寄存器（R0～R3），给分；回答 PC，给分；部分正确，酌情给分。设置暂存器 T 的原因若回答用于暂时存放端口 A 的数据，则给分；其他答案，酌情给分。 2）ALU 共有 7 种操作，故其操作控制信号 ALUop 至少需要 3 位；移位寄存器有 3 种操作，其操作控制信号 SRop 至少需要 2 位。 3）信号 SRout 所控制的部件是一个三态门，用于控制移位器与总线之间数据通路的连接与断开。【评分说明】只要回答出三态门或者控制连接/断开，即给分。 4）端口①、②、③、⑤、⑧须连接到控制部件输出端。【评分说明】答案包含④、⑥、⑦、⑨中任意一个，不给分；答案不全酌情给分。 5）连线 1，⑥$\rightarrow$⑨；连线 2，⑦$\rightarrow$④。【评分说明】回答除上述连线以外的其他连线，酌情给分。 6）因为每条指令的长度为 16 位，按字节编址，所以每条指令占用 2 个内存单元，顺序执行时，下条指令地址为 $(\text{PC})+2$。MUX 的一个输入端为 2，可便于执行 $(\text{PC})+2$ 操作。 ## 心得笔记 1. 2009年题，加法运算有一个暂存器A，而加法运算有两个操作数，一个是R0保存的数据，而对于 R1，其中保存的是地址，要用其值从主存中取数据，所以要走 MAR ——M—— MDR，但是哪个操作数放到暂存器 A ，谁不放 A，答案是要将无需访存、立即可用的通用寄存器数据（R0）先送入暂存器 A ，这样效率最高。另外要注意编排时不要造成总线冲突。图中有系统总线和片内总线。系统总线和片内总线倒是相互不会冲突，可以并行的25:T1736,## 题目 **48. Directions:** Write an essay based on the chart below. In your essay, you should 1) describe and interpret the chart, and 2) give your comments. Write your answer in about 150 words on the ANSWER SHEET. (15 points)

## 范文 The bar chart vividly depicts public attitudes toward children's outdoor activities. As illustrated, satisfying curiosity and fostering observation skills rank as the top two priorities, commanding nearly identical shares at 54.60% and 54.50% respectively. Meanwhile, building physical fitness accounts for 37.20%, followed by enhancing parent-child interaction at 33.20%. This survey highlights an evolving mindset in modern parenting: prioritizing cognitive exploration and holistic well-being over rigid academic routines. In a fast-paced society where young parents face relentless competition, children are often confined to classrooms and digital devices, triggering unnecessary anxiety. However, true personal growth does not stem merely from textbook memorization or exam scores, but from direct contact with nature, where youngsters can develop keen perception and stay emotionally healthy. In my view, this trend will prove highly beneficial for long-term childhood development. Parents and educators should take the initiative to encourage kids to step out of their indoor comfort zones. Ultimately, embracing the outdoors enriches children's minds and equips them with lasting vitality for the future. ## 全文参考翻译这幅柱状图生动展现了公众对儿童户外活动所持的态度。如图所示，满足好奇心与培养观察力位列前两位首要诉求，所占比例几乎持平，分别为 54.60% 和 54.50%。与此同时，强身健体占比 37.20%，紧随其后的是增强亲子互动，占比 33.20%。这项调查凸显了当代育儿理念中正在发生的一种观念转变：相比死板刻板的学业日常，家长们更看重认知探索与身心的全面发展。在一个年轻父母面临激烈竞争的快节奏社会中，孩子们往往被局限在教室内和电子屏幕前，这常常引发不必要的焦虑。然而，真正的个人成长并不单单来源于死记硬背课本或应付考试分数，而是源自与大自然的直接接触——在自然中，孩子们可以培养敏锐的感知力，并保持身心健康。在我看来，这一趋势对儿童的长远发展大有裨益。父母与教育工作者应当主动采取行动，鼓励孩子们走出室内的“舒适区”。归根结底，拥抱户外能够丰富孩子们的内心世界，并赋予他们面向未来的持久生命力。 ## 写作进阶生词与高分表达词表 | 英文 | 音标 | 释义 | |---|---|---| | foster | /ˈfɒstə(r)/ | v. 培养，促进 | | priority | /praɪˈɒrəti/ | n. 首要事务，优先考虑的事 | | command | /kəˈmɑːnd/ | v. 拥有，占据（比例/注意等） | | identical | /aɪˈdentɪkl/ | adj. 完全相同的，极为相似的 | | interaction | /ˌɪntərˈækʃn/ | n. 互动，相互影响 | | evolving | /ɪˈvɒlvɪŋ/ | adj. 逐渐发展的，演变中的 | | mindset | /ˈmaɪndset/ | n. 思维模式，观念 | | parenting | /ˈpeərəntɪŋ/ | n. 育儿，父母对子女的教养 | | cognitive | /ˈkɒɡnətɪv/ | adj. 认知的，感知的 | | exploration | /ˌekspləˈreɪʃn/ | n. 探索，探究 | | holistic | /həʊˈlɪstɪk/ | adj. 全面的，整体的 | | well‑being | /ˌwel ˈbiːɪŋ/ | n. 健康，幸福，良好状态 | | relentless | /rɪˈlentləs/ | adj. 持续强烈的，残酷无情的 | | confine | /kənˈfaɪn/ | v. 限制，使局限于 | | trigger | /ˈtrɪɡə(r)/ | v. 引起，触发 | | memorization | /ˌmeməraɪˈzeɪʃn/ | n. 记忆，死记硬背 | | perception | /pəˈsepʃn/ | n. 感知力，洞察力 | | initiative | /ɪˈnɪʃətɪv/ | n. 主动权，积极性（take the initiative 主动） | | enrich | /ɪnˈrɪtʃ/ | v. 丰富，充实 | | vitality | /vaɪˈtæləti/ | n. 活力，生命力 | 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575 的成绩通过了英语六级考试的人，我的英语底子很好。这次考研我要考的英语科目是英一，我做了真题，发现阅读部分不怎么错，无需复习，需要提升一下的就是翻译和写作。其中，作文英一占30分，英二占25分。 ## 英语写作提升计划我觉得资料上给出的范文都很不地道，于是主要依靠 gemeni 来生成范文。然后把范文当作课文来学习、背诵，围绕课文，首先整理出生词表，通过熟悉词汇和短语、做到能顺畅通读。为了辅助背诵，我用 NaturalReader 生成语音，然后录屏，并转成音频来跟读、听、背诵。每天上下班的时候听。从范文中，我总结出，在作文中能用到的通用性较强的词汇有 genuine enrich 等。后续 gemini 生成范文时，应让其有意地想办法复用一些之前已经出现过的词汇，或者相近的表达，方便我加深印象，真正在考试中能应用上这些金句、词汇。 ### 怎样让 AI 根据范文总结生词表 ``` 翻译这篇文章。另外，我的词汇量在 6000 左右，帮我根据这篇文章，总结出对我写作有帮助的生词表，按如下 markdown 格式给我。 | 英文 | 音标 | 释义 | | ------------------------ | -------------------------------- | ------------ | | depict | /dɪˈpɪkt/ | v. 描绘，刻画 | | dismiss | /dɪsˈmɪs/ | v. 不予理会，轻视 | | remark | /rɪˈmɑːk/ | v. 评论，说道 | | beneficial | /ˌbenɪˈfɪʃl/ | adj. 有益的 | | highlight | /ˈhaɪlaɪt/ | v. 凸显，强调 | | specialization | /ˌspeʃəlaɪˈzeɪʃn/ | n. 专业化，专攻 | | intellectual | /ˌɪntəˈlektʃuəl/ | adj. 求知的；智力的 | | short‑sighted | /ˌʃɔːt ˈsaɪtɪd/ | adj. 目光短浅的 | | utilitarianism | /juːˌtɪlɪˈteəriənɪzəm/ | n. 功利主义 | | vocational | /vəʊˈkeɪʃənl/ | adj. 职业的 | | genuine | /ˈdʒenjuɪn/ | adj. 真正的 | | broaden | /ˈbrɔːdn/ | v. 拓宽（眼界） | | perspective | /pəˈspektɪv/ | n. 视角，眼界 | | restrict | /rɪˈstrɪkt/ | v. 限制 | | cross‑disciplinary | /ˌkrɒs ˈdɪsəplɪnəri/ | adj. 跨学科的 | ``` 另一项训练是：跟豆包对练口语，会说了自然会写。练到出口成章，至少不会担心写不出内容。基本策略是：通过口头表达的能力，把框架理出来，再打磨文字表达。在训练过程中，我用以下提示词，让豆包为我纠正对话中自己的不太地道的表达，给出准确的英文表达。我相信从自身的表达过的内容入手，进行纠正，这对于快速提升口语能力是最有效的。 ``` Please polish everything I’ve said in our earlier conversation to sound natural for a native speaker. I know you can understand my awkward expressions, but I want to improve. Please analyze and show me how to rephrase them like a native. Stick only to what I’ve asked for. ``` | Your original sentence | Native rephrasing | Key notes | | ---- | ---- | ---- | | I just can't answer I just can't understand the question | I can’t quite understand the question. | Remove repetition; "can’t quite understand" is natural spoken English | | What research what what ancient what i i didn't you're clearly | Wait, what ancient figure? I didn’t catch that clearly. | "I didn’t catch that" = I failed to hear/understand | | But if I know nothing about ancient Chinese scientist, what can I do? Oh, no. This question is difficult to me. | But what if I know nothing about ancient Chinese scientists? Oh no, this question is hard for me. | Use plural "scientists"; "what if" for hypotheticals; collocation: hard for me | | for all things i said in previous conversation, polish it as native skpear, I know you can understand my poor expression , but i need to improve it , so analyse and teach me how to replace it with native speaker expression. put nothing more than i required | Please polish everything I’ve said in our earlier conversation to sound natural for a native speaker. I know you can understand my awkward expressions, but I want to improve. Please analyze and show me how to rephrase them like a native. Stick only to what I’ve asked for. | skpear → speaker; poor → awkward; put nothing more than I required → Stick only to what I’ve asked for | launch your career / pursue work | Your original sentence | Native rephrasing | |--------------------------------------------------------------------------|-------------------------------------------------------------------------------| | Now show me 2023. OK, do it. | Show me the 2023 question. Okay, go ahead. | | Is the question, does question include sport Or it is your creation? | Does the original question mention sports, or did you add that part yourself? | | Repeat. | Repeat that. / Can you repeat it? | | What is the meaning of careful for paperwork? | What does “careful with paperwork” mean? | | Give me another sample. | Can you give me another sample? | | Can you show me the 2020s 4 English I writing part A. | Can you show me the 2024 English I Writing Part A question? | | Question. | Show me the question. | | You, you chat with me in English and you do it. You give an example. OK. | Chat with me in English and write a sample for me. Okay. | | Do it as a native speaker. | Write it the way a native speaker would. | --- ## 在哪里看真题 zhenti.burningvocabulary.cn/kaoyan/2026/01 ## 如何用 gemini 写考研英语作文范文 ``` Solve this ,show 3 samples which are handy for exam, sounds like writing by native speaker, but be sure it is not hard to compose during exam The cartoon vividly depicts two contrasting attitudes toward work under the title "Habit." On the left, a diligent girl focuses on her studies surrounded by piles of books, believing that finishing her work as soon as possible leaves her with peace of mind. On the right, a relaxed boy leans back in his seat, telling himself that he never starts until the deadline approaches. This image highlights the fundamental difference between prompt action and procrastination. The girl understands the value of time management, taking the initiative early to avoid unnecessary panic. Conversely, the boy relies entirely on last-minute pressure, a common habit that often leads to sloppy work and high anxiety. From my firsthand experience, while putting things off feels easier in the short term, it creates greater trouble down the road. We should learn to break large goals into daily steps and act before we are forced to. Through self-discipline, we can truly stay calm, productive, and in control of our own future. The cartoon depicts a candid conversation between a father and his son, who is dressed in a traditional opera costume. The boy hesitantly mentions that his classmates find learning opera unappealing. In response, his father calmly reassures him: "Don't you love it yourself? That is enough." The illustrator vividly captures the tension between individual passion and peer pressure. In a fast-paced society where trends dictate what is deemed "cool" or worthwhile, young people often feel pressured to conform to mainstream preferences. However, true fulfillment does not stem from external validation or collective approval; it comes from intrinsic motivation. When we abandon our unique interests simply to fit in, we risk losing our individuality and authentic joy. In my view, the father’s advice offers a valuable lesson in resilience and self-belief. Instead of letting peer opinions dilute our ambitions, we should have the courage to follow our genuine passions. Parents and educators should likewise encourage children to march to the beat of their own drum, for uniqueness is what ultimately enriches life. The cartoon depicts two students standing in front of a bulletin board promoting campus lectures. Looking at the poster, the girl on the left dismisses it, remarking that since the topic falls outside their major, attending would be useless. In contrast, the student on the right suggests that going would definitely prove beneficial. This simple contrast highlights two distinct attitudes toward education: narrow specialization versus intellectual curiosity. The first student falls into the trap of short-sighted utilitarianism, treating college simply as a vocational training ground. However, genuine higher education is meant to broaden one’s perspective rather than restrict it to a single discipline. Real-world challenges rarely fit neatly into one academic field, and cross-disciplinary knowledge is essential for creative problem-solving and long-term adaptability. In my view, college students ought to step out of their academic comfort zones. While mastering one’s major remains fundamental, exploring interdisciplinary lectures fosters critical thinking and well-rounded competence. True personal growth begins when we stop asking what is merely "useful" for tomorrow's exam and start pursuing what enriches our minds for the future. The picture and chart vividly depict the rapid growth of urban green spaces. On the left, residents are happily exercising in a newly built neighborhood park, expressing their satisfaction. Beside it, the bar chart illustrates a steady rise in the city's parks, increasing from 406 in 2020 to 670 in 2022.This combination highlights a shift in urban life: prioritizing well-being over sheer material speed. In a fast-paced society where work pressure often triggers anxiety, accessible green spaces offer citizens essential peace of mind. True fulfillment does not stem solely from economic gains; it comes from having an environment where one can stay calm, healthy, and in control of daily life. These pocket parks provide a welcome escape from concrete blocks, allowing people to slow down.In my view, this trend proves highly beneficial for sustainable growth. Local authorities should continue to take the initiative in creating communal areas, encouraging people to step out of their indoor comfort zones. Ultimately, building more green spaces enriches our minds and revitalizes community life. 想办法复用上面链接中一些之前已经出现过的词汇，或者相近的表达方式，方便我加深印象，真正在考试中能应用上这些金句、词汇，但不能强求，要保证文章质量 ``` 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1. 数列的极限 **定义** 如果对于任意给定的 $\varepsilon > 0$，总存在正整数 $N$，当 $n > N$ 时，恒有 $|x_n - a| < \varepsilon$ 成立，则称常数 $a$ 为数列 $\{x_n\}$ 当 $n$ 趋于无穷时的**极限**，记为 $\lim_{n \to \infty} x_n = a$。 **【注】** ： 1. $\varepsilon$ 用来刻画 $x_n$ 与 $a$ 的接近程度，$N$ 用来刻画 $n \to \infty$ 这个极限过程。 2. $\lim_{n \to \infty} x_n = a$ 的几何意义是：对于 $a$ 点的任何 $\varepsilon$ 邻域即开区间 $(a-\varepsilon, a+\varepsilon)$，一定存在 $N$，当 $n > N$ 即第 $N$ 项以后的点 $x_n$ 都落在开区间 $(a-\varepsilon, a+\varepsilon)$ 内，而只有有限个（最多有 $N$ 个）在这区间之外。 3. 数列 $\{x_n\}$ 的极限是否存在，如果存在其极限值等于多少，与数列的前有限项无关。 4. $\lim_{n \to \infty} x_n = a \Leftrightarrow \lim_{k \to \infty} x_{2k-1} = \lim_{k \to \infty} x_{2k} = a$。 --- ## 2. 函数的极限 (1) **自变量趋于无穷大时函数的极限**。 **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $X > 0$，当 $x > X$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为 $f(x)$ 当 $x \to +\infty$ 时的**极限**，记为 $\lim_{x \to +\infty} f(x) = A$。 **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $X > 0$，当 $x < -X$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为 $f(x)$ 当 $x \to -\infty$ 时的**极限**，记为 $\lim_{x \to -\infty} f(x) = A$。 **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $X > 0$，当 $|x| > X$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为 $f(x)$ 当 $x \to \infty$ 时的**极限**，记为 $\lim_{x \to \infty} f(x) = A$。 **【注】** 这里的 $x \to \infty$ 是指 $|x| \to +\infty$，而数列极限中的 $n \to \infty$ 是指 $n \to +\infty$。 **定理** 极限 $\lim_{x \to \infty} f(x)$ 存在的充要条件是极限 $\lim_{x \to -\infty} f(x)$ 及 $\lim_{x \to +\infty} f(x)$ 存在并且相等。 (2) **自变量趋于有限值时函数的极限**。 **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $\delta > 0$，当 $0 < |x - x_0| < \delta$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为函数 $f(x)$ 当 $x \to x_0$ 时的**极限**，记为 $\lim_{x \to x_0} f(x) = A$。 **【注】** (1) $\varepsilon$ 用来刻画 $f(x)$ 与 $A$ 的接近程度，$\delta$ 用来刻画 $x \to x_0$ 这个极限过程。 (2) 几何意义：对任意给定的 $\varepsilon > 0$，总存在 $\mathring{U}(x_0, \delta)$，当 $x \in \mathring{U}(x_0, \delta)$ 时，曲线 $y = f(x)$ 夹在两直线 $y = A - \varepsilon$ 和 $y = A + \varepsilon$ 之间。 (3) 这里 $x \to x_0$，但 $x \neq x_0$。极限 $\lim_{x \to x_0} f(x)$ 是否存在、极限值（如果存在）等于多少，与 $f(x)$ 在 $x = x_0$ 处有没有定义、函数值（如果有定义）等于多少无关，只与 $x = x_0$ 的去心邻域 $\mathring{U}(x_0, \delta)$ 函数值有关。而要使 $\lim_{x \to x_0} f(x)$ 存在，$f(x)$ 必须在 $x = x_0$ 的某去心邻域 $\mathring{U}(x_0, \delta)$ 处处有定义。 **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $\delta > 0$，当 $x_0 - \delta < x < x_0$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为函数 $f(x)$ 当 $x \to x_0^-$ 时的**左极限**，记为 $$ \lim_{x \to x_0^-} f(x) = A, \text{或 } f(x_0^-) = A, \text{或 } f(x_0 - 0) = A. $$ **定义** 若对任意给定的 $\varepsilon > 0$，总存在 $\delta > 0$，当 $x_0 < x < x_0 + \delta$ 时，恒有 $|f(x) - A| < \varepsilon$，则称常数 $A$ 为函数 $f(x)$ 当 $x \to x_0^+$ 时的**右极限**，记为 $$ \lim_{x \to x_0^+} f(x) = A, \text{或 } f(x_0^+) = A, \text{或 } f(x_0 + 0) = A. $$ **定理** 极限 $\lim_{x \to x_0} f(x)$ 存在的充要条件是左极限 $\lim_{x \to x_0^-} f(x)$ 及右极限 $\lim_{x \to x_0^+} f(x)$ 存在并且相等。 **【注】** 需要分左、右极限求极限的问题，常见的有以下三种： (1) 求分段函数在分界点处的极限，而在该分界点两侧函数表达式不同（这里也包括带有绝对值的函数，如 $\lim_{x \to 0} \frac{|x|}{x}$）。 (2) $e^\infty$ 型极限（如 $\lim_{x \to 0} e^{\frac{1}{x}}, \lim_{x \to \infty} e^x, \lim_{x \to \infty} e^{-x}$）。 $\lim_{x \to 0^-} e^{\frac{1}{x}} = 0, \lim_{x \to 0^+} e^{\frac{1}{x}} = +\infty$，则 $\lim_{x \to 0} e^{\frac{1}{x}}$ 不存在。 $\lim_{x \to -\infty} e^x = 0, \lim_{x \to +\infty} e^x = +\infty$，则 $\lim_{x \to \infty} e^x$ 不存在。 $\lim_{x \to -\infty} e^{-x} = +\infty, \lim_{x \to +\infty} e^{-x} = 0$，则 $\lim_{x \to \infty} e^{-x}$ 不存在。 **【注】** $e^\infty \neq \infty, e^{+\infty} = +\infty, e^{-\infty} = 0$。 (3) $\arctan \infty$ 型极限（如 $\lim_{x \to 0} \arctan \frac{1}{x}, \lim_{x \to \infty} \arctan x$）。 $\lim_{x \to 0^-} \arctan \frac{1}{x} = -\frac{\pi}{2}, \lim_{x \to 0^+} \arctan \frac{1}{x} = \frac{\pi}{2}$，则 $\lim_{x \to 0} \arctan \frac{1}{x}$ 不存在。 $\lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}, \lim_{x \to +\infty} \arctan x = \frac{\pi}{2}$，则 $\lim_{x \to \infty} \arctan x$ 不存在。 **【注】** $\arctan \infty \neq \frac{\pi}{2}, \arctan(+\infty) = \frac{\pi}{2}, \arctan(-\infty) = -\frac{\pi}{2}$。 --- ## 相关例题 **【例 1】** （2006，数三）$\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{(-1)^n} =$ \_\_\_\_\_\_\_\_. **解【方法 1】** 当 $n$ 为奇数时，$x_n = \left(\frac{n+1}{n}\right)^{-1} = \frac{n}{n+1}$。 $$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{n}{n+1} = 1. $$ 当 $n$ 为偶数时，$x_n = \frac{n+1}{n}$。 $$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{n+1}{n} = 1. $$ 则 $\lim_{n \to \infty} x_n = \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{(-1)^n} = 1$。 **【方法 2】** 由于 $\left(\frac{n+1}{n}\right)^{-1} \leqslant \left(\frac{n+1}{n}\right)^{(-1)^n} \leqslant \frac{n+1}{n}$，由夹逼准则知 $\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{(-1)^n} = 1$。 --- **【例 2】** 试证明： (1) 若 $\lim_{n \to \infty} x_n = a$，则 $\lim_{n \to \infty} |x_n| = |a|$，但反之不成立。 (2) $\lim_{n \to \infty} x_n = 0$ 的充分必要条件是 $\lim_{n \to \infty} |x_n| = 0$。 **证** (1) 由于 $\lim_{n \to \infty} x_n = a$ 及数列极限定义知，$\forall \varepsilon > 0, \exists N > 0$，当 $n > N$ 时， $$ |x_n - a| < \varepsilon. $$ 又 $||x_n| - |a|| \leqslant |x_n - a|$，则 $\forall \varepsilon > 0, \exists N > 0$，当 $n > N$ 时， $$ ||x_n| - |a|| < \varepsilon. $$ 故 $\lim_{n \to \infty} |x_n| = |a|$。反之不成立，例如 $x_n = (-1)^n$，则 $\lim_{n \to \infty} |x_n| = 1 = |1|$，但 $\lim_{n \to \infty} x_n = \lim_{n \to \infty} (-1)^n$ 不存在。 (2) 由 (1) 可知，若 $\lim_{n \to \infty} x_n = 0$，则 $\lim_{n \to \infty} |x_n| = |0| = 0$。所以，只需证明：若 $\lim_{n \to \infty} |x_n| = 0$，则 $\lim_{n \to \infty} x_n = 0$。由 $\lim_{n \to \infty} |x_n| = 0$ 可知，$\forall \varepsilon > 0, \exists N > 0$，当 $n > N$ 时， $$ ||x_n| - 0| < \varepsilon, $$ 即 $|x_n - 0| < \varepsilon$。故原题得证。 **【注】** 函数极限有对应的结论。 **【例 3】** （1992，数一、二、三）当 $x \to 1$ 时，函数 $\frac{x^2-1}{x-1} e^{\frac{1}{x-1}}$ 的极限 (A) 等于 2。 (B) 等于 0。 (C) 为 $\infty$。 (D) 不存在但不为 $\infty$。 **解** 本题中出现 $e^\infty$，所以要分左、右极限 $$ \lim_{x \to 1^-} \frac{x^2-1}{x-1} e^{\frac{1}{x-1}} = \lim_{x \to 1^-} \frac{x+1}{1} e^{\frac{1}{x-1}} = 2 \times 0 = 0, $$ $$ \lim_{x \to 1^+} \frac{x^2-1}{x-1} e^{\frac{1}{x-1}} = \lim_{x \to 1^+} \frac{x+1}{1} e^{\frac{1}{x-1}} = +\infty, $$ 所以 $\lim_{x \to 1} \frac{x^2-1}{x-1} e^{\frac{1}{x-1}}$ 不存在，但不是 $\infty$，应选 (D)。 --- **【例 4】** （2021，数三）已知 $\lim_{x \to 0} \left[ a \arctan \frac{1}{x} + (1+|x|)^{\frac{1}{x}} \right]$ 存在，求 $a$ 的值。 **解** 由于 $\lim_{x \to 0} \left[ a \arctan \frac{1}{x} + (1+|x|)^{\frac{1}{x}} \right]$ 存在，且 $$ \lim_{x \to 0^-} \left[ a \arctan \frac{1}{x} + (1+|x|)^{\frac{1}{x}} \right] = \lim_{x \to 0^-} a \arctan \frac{1}{x} + \lim_{x \to 0^-} (1-x)^{\frac{1}{x}} = -\frac{\pi}{2}a + \frac{1}{e}, $$ $$ \lim_{x \to 0^+} \left[ a \arctan \frac{1}{x} + (1+|x|)^{\frac{1}{x}} \right] = \lim_{x \to 0^+} a \arctan \frac{1}{x} + \lim_{x \to 0^+} (1+x)^{\frac{1}{x}} = \frac{\pi}{2}a + e, $$ 所以 $-\frac{\pi}{2}a + \frac{1}{e} = \frac{\pi}{2}a + e$，故 $a = \frac{1-e^2}{\pi e}$。【例 13】 (1999, 数二) “对任意给定的 $\varepsilon \in (0,1)$，总存在正整数 $N$，当 $n \geqslant N$ 时，恒有 $|x_n - a| \leqslant 2\varepsilon$” 是数列 $\{x_n\}$ 收敛于 $a$ 的 (A) 充分条件但非必要条件。 (B) 必要条件但非充分条件。 (C) 充分必要条件。 (D) 既非必要也非充分条件。解由数列极限定义知，如果数列 $\{x_n\}$ 收敛于 $a$，则对于任意给定的 $\varepsilon > 0$，总存在正整数 $N$，当 $n > N$ 时，恒有 $|x_n - a| < \varepsilon$ 成立。由于 $\varepsilon < 2\varepsilon$，故必有 $|x_n - a| \leqslant 2\varepsilon$。如果对任意给定的 $\varepsilon \in (0,1)$，总存在正整数 $N$，当 $n \geqslant N$ 时，恒有 $|x_n - a| \leqslant 2\varepsilon$。由于 $\varepsilon$ 的任意性，则对于任意给定的 $\varepsilon_1 > 0$，取 $\varepsilon = \frac{\varepsilon_1}{3}$，当 $n \geqslant N$ 时，恒有 $|x_n - a| \leqslant 2\varepsilon = \frac{2\varepsilon_1}{3} < \varepsilon_1$。由数列极限定义知，数列 $\{x_n\}$ 收敛于 $a$，故应选 (C)。 --- 【例 14】 (2015, 数三) 设 $\{x_n\}$ 是数列，下列命题中不正确的是 (A) 若 $\lim_{n \to \infty} x_n = a$，则 $\lim_{n \to \infty} x_{2n} = \lim_{n \to \infty} x_{2n+1} = a$。 (B) 若 $\lim_{n \to \infty} x_{2n} = \lim_{n \to \infty} x_{2n+1} = a$，则 $\lim_{n \to \infty} x_n = a$。 (C) 若 $\lim_{n \to \infty} x_n = a$，则 $\lim_{n \to \infty} x_{3n} = \lim_{n \to \infty} x_{3n+1} = a$。 (D) 若 $\lim_{n \to \infty} x_{3n} = \lim_{n \to \infty} x_{3n+1} = a$，则 $\lim_{n \to \infty} x_n = a$。解 (D) 选项的命题不正确，例如 $x_{3n} = x_{3n+1} = 1$，而 $x_{3n+2} = 0$。显然 $\lim_{n \to \infty} x_{3n} = \lim_{n \to \infty} x_{3n+1} = 1$，而 $\lim_{n \to \infty} x_{3n+2} = 0$，则 $\lim_{n \to \infty} x_n$ 不存在。 --- 【例 15】 (2022, 数一，数二) 已知数列 $\{x_n\}$，其中 $-\frac{\pi}{2} \leqslant x_n \leqslant \frac{\pi}{2}$，则 (A) 当 $\lim_{n \to \infty} \cos(\sin x_n)$ 存在时，$\lim_{n \to \infty} x_n$ 存在。 (B) 当 $\lim_{n \to \infty} \sin(\cos x_n)$ 存在时，$\lim_{n \to \infty} x_n$ 存在。 (C) 当 $\lim_{n \to \infty} \cos(\sin x_n)$ 存在时，$\lim_{n \to \infty} \sin x_n$ 存在，但 $\lim_{n \to \infty} x_n$ 不一定存在。 (D) 当 $\lim_{n \to \infty} \sin(\cos x_n)$ 存在时，$\lim_{n \to \infty} \cos x_n$ 存在，但 $\lim_{n \to \infty} x_n$ 不一定存在。解【方法 1】直接法由于 $\sin x$ 在 $[-\frac{\pi}{2}, \frac{\pi}{2}]$ 上单调、连续，则当 $-\frac{\pi}{2} \leqslant x_n \leqslant \frac{\pi}{2}$ 时，若 $\lim_{n \to \infty} \sin x_n$ 存在，必有 $\lim_{n \to \infty} x_n$ 存在。但 $\lim_{n \to \infty} \cos \frac{(-1)^n \pi}{2} = 0$，而 $\lim_{n \to \infty} \frac{(-1)^n \pi}{2}$ 不存在。$\lim_{n \to \infty} \cos x_n$ 存在时，$\lim_{n \to \infty} x_n$ 不一定存在。由此可知选 (D)。【方法 2】排除法取 $x_n = (-1)^n \frac{\pi}{2}$，则排除 (A)(B)(C)，故应选 (D)。 --- 【例 16】 (1993, 数三) 当 $x \to 0$ 时，变量 $\frac{1}{x^2} \sin \frac{1}{x}$ 是 (A) 无穷小。 (B) 无穷大。 (C) 有界的，但不是无穷小。 (D) 无界的，但不是无穷大。解对任意给定的 $M > 0$ 及 $\delta > 0$，总存在 $x_n = \frac{1}{2n\pi + \frac{\pi}{2}}, y_n = \frac{1}{2n\pi}$，使得 $0 < x_n < \delta, 0 < y_n < \delta$，此时 $\left| \frac{1}{x_n^2} \sin \frac{1}{x_n} \right| = \left( 2n\pi + \frac{\pi}{2} \right)^2 > M, \frac{1}{y_n^2} \sin \frac{1}{y_n} = 0$，则当 $x \to 0$ 时，变量 $\frac{1}{x^2} \sin \frac{1}{x}$ 是无界的，但不是无穷大。选 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