第二天答案.pdf-2021-02-27-17-11-26-627.pdf

第二天第二天1.【答案】2.【解析】22121 cos222222001 cos22limlim2xxxxxxxxx2.【答案】2.【解析】2222lim sinlim211xxxxxxxx3.【答案】16.【解析】当0 x 时，111223336=xxxxxx4.【答案】C.【解析】题目考察无穷小量的性质和无穷小量的比较,采用洛必达法则计算如下：tantan00222311200001limlimtansec1tan1limlimlimlim,33xxx xxnnxxnnnnxxxxeeeexxxxxxxnxnxx洛必达tanxxee与3x同阶,故应选(C).5.【答案】B.【解析】当 0 x时,21211cosxxxa,65321lnxxxaxxa3111313,故从低阶到高级顺序：132,aaa，选 B.6.【答案】B.【解析】00ln1 22lim0,lim0,1xxxxxx故121001 cos22lim0,lim0,1,2.xxxxxx故故12，故选 B.7.【答案】B.【解析】1xex，ln(1)xx，1112xx，11 cos2xx，故答案选择 B.8.【答案】4.【解析】122241(1)1,sin,4axaxxxx:即即4a 9.【答案】14.【解析】401 cosln(1tan)limsinxxxxx2420022002201ln(1tan)ln(1tan)2limlim2sec11tansec1tanlimlim44sec2sectan1lim.44xxxxxxxxxxxxxxxxxxxxx10.【答案】A.【解析】sinf xxax与 2ln 1g xxbx是0 x 时的等价无穷小,则22000232000330()sinsinlimlimlim()ln(1)()sin1cossinlimlimlim36sinlim1,66xxxxxxxf xxaxxaxg xxbxxbxxaxaaxaaxbxbxbxaaxabaxb 等洛洛即36ab,故排除 B,C.另外,201coslim3xaaxbx存在,蕴含了1cos0aax0 x,故1,a 排除 D.所以本题选 A.由此极限为 2，故9092 312aabab .6.【答案】1.【解析】22222222222222()+()=limlim+2=1limlim+xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx7.【答案】sin1.【解析】0lim(sin1sin)=sin1xxx 8.【答案】极限不存在.【解析】0000limlim11,limlim11,xxxxf xxf xx 左右极限不等，所以在0 x 处极限不存在.9.【答案】26.【解析】221111313+131lim=lim223+12212=limlim62 212212xxxxxxxxxxxxxxxxxxxxxx 10.【答案】D.【解析】11211111limlim11xxxxxexex+11211111limlim1=+1xxxxxexex，11211111limlim1=01xxxxxexex，左右极限不等，所以极限不存在，左极限为 0，不是.