本篇笔记文档名 👇
本文档为考研数学二复习笔记,内容涵盖函数、极限、连续、导数、微分、积分等核心知识点,以选择题、填空题、解答题形式呈现典型例题。通过系统梳理函数极限计算、导数应用、积分计算等题型,帮助备考学生巩固知识体系,提升解题能力,适配考研数学二考试要求,适合备考学生系统复习使用。
文档的预览图如下,需要完整PDF文件的同学,文末有文档编码,保存后即可直接打印使用。
📑27数学-【竖版】27李良700题做题本高数
第2页选择题:1. 设函数f(x)在x=0处可导,则lim(x→0) [f(1-cosx)]/[x²] = ( ),选项含0、1/2 f’(0)等;2. 设f(x)在x=0处可导,f(0)=0,则lim(x→0) [f(x)]/x = f’(0);3. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;4. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;5. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;6. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;7. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;8. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;9. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;10. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;11. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;12. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;13. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;14. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;15. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1。第3页填空题:1. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;2. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;3. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;4. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;5. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;6. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;7. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;8. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;9. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;10. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;11. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;12. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;13. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;14. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;15. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1。第4页选择题:1. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;2. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;3. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;4. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;5. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;6. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;7. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;8. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;9. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;10. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;11. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;12. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;13. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;14. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;15. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1。第5页填空题:1. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;2. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;3. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;4. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;5. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;6. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;7. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;8. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;9. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;10. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;11. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;12. 设f(x)在x=0处可导,f(0)=0,f’(0)=1,则lim(x→0) [f(x)]/x = 1;13. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1;14. 设f(x)在x=0处可导,f(0)=0,f’(0)=2,则lim(x→0) [f(x)]/x = 2;15. 设f(x)在x=0处可导,f’(0)=1,则lim(x→0) [f(x)-f(0)]/x = 1。
声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。