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1、精品文档高等数学练习题集pdf一.填空题a?1?x?1．设lim???tetdt,则a=_2._______.???x???x?ax2.=___112n??_____.lim?2?2???2?n??2n?n?n??n?n?1n?n?23.已知函数f???1?0
2、x
3、?1
4、x
5、?1,则f[f]___1____.4.lim=___limn??n??n?3n?n?n?3n?n?n?2____.5.limcotx?x?02016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档1??1??=______.?sinxx?解.limcosxx?sinxx?sinx1?co
6、sxsinx1??lim?lim?lim?x?0sinxx?0x?0x?06xxsinxx33x26n19906.已知limk?A,则A=______,k=_______.n??n?k所以k－1=1990,k=1991;111?A，A??kk1991二.选择题1.设f和?在内有定义,f为连续函数,且f?0,?有间断点,则?[f]必有间断点[?]2必有间断点f[?]必有间断点所以是答案.2.设函数f?x?tanx?esinx?必有间断点f,则f是偶函数无界函数周期函数单调函数解.是答案..2016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档极限lim?35
7、2n?1?的值是????2222?n??12?222?3n????01不存在解.,所以为答案.9554.设lim?8,则a的值为50x??11解.所以为答案..设lim均不对??,则?,?的数值为x???111?=,?=?=,?=均不对33?=1,?=解.为答案.6.设f?2x?3x?2,则当x?0时f是x的等价无穷小f是x的同阶但非等价无穷小f比x较低价无穷小f比x较高价无穷小解.所以为答案..设lim2016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档?a?6,则a的值为x?0x－11解.所以为答案..设limx?0atanx?bcln?d2?2，其
8、中a2?c2?0,则必有b=db=－4da=ca=－4c解.所以为答案.三.计算题1.求下列极限limx???x1x解.lim?limex???x???x1xlnx?elnx???xlim?ex???x?e?e1?e2016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档lim1?ex21?cos)xx??xx1解.令y?xlim?lim=ex??y?0xx1lny?e2cos2y?sinyy?0sin2y?cosylim?e2?1?tanx?lim??x?01?sinx???1?tanx?解.lim??x?01?sinx??x3x3tanx?sinx???
9、lim?1??x?01?sinx??22016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档x3taxn?sixn?tanx?sinx???lim??1??x?0?1?sinx???limsinxx31?sixntaxn?sixnn)x?limx?1lnarcsin2x?12解.当x?1时,ln~x?1,arcsin2x2?1~2x2?1.按照等价无穷小代换limx?1lnarcsin2x2?1?x???limx?1x?12016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档2x2?1?lim11?x?12x?122lim??12?c
10、otx?0x2?解.方法1：?1lim?2?cot2x?0x??sin2x?x2cos2?1cos2x??x?=lim?=lim???22??x?0x2sin2xsinx?x?0????xx?????1?cos2=lim?x?0?x4???2xcos2x?2cosxsinx?x????3?=lim?x?0?4x????2xcos2x?sin2x2x2cosxsinx?lim=lim3x?0x?04x4x?2cos2x?4xcosxsinx?2cos2x12016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档?=limx?012x22?2cos2x?2cos
11、2x114cosxsinx?4sin2x11???lim??=lim2x?0x?012x3224x32=lim?2sin2x111112???????x?024x3263233.求下列极限limnn??lnnxnn?1?1解.lim?lim令n?1?xlimx?0ln31?e?nxlimn??1?e?nx1?en??1?e?nx?nx解.limx?0?12016全新精品资料-全新公文范文-全程指导写作–独家原创36/36精品文档???0x?0??1x?0?n?a?b??,其中a>0,b>0l
