第二天答案.pdf-2021-02-27-17-11-26-627.pdfVIP免费

第二天1.【答案】2.【解析】22121cos222222001cos22limlim2xxxxxxxxx2.【答案】2.【解析】2222limsinlim211xxxxxxxx3.【答案】16.【解析】当0x时，111223336=xxxxxx4.【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,33xxxxxnnxxnnnnxxxxeeeexxxxxxxnxnxx洛必达tanxxee与3x同阶,故应选(C).5.【答案】B.【解析】当0x时,2121~1cosxxxa,6532~1lnxxxaxxa31~11313,故从低阶到高级顺序：132,,aaa，选B.6.【答案】B.【解析】00ln122lim0,lim0,1xxxxxx故121001cos22lim0,lim0,1,2.xxxxxx故故12，故选B.7.【答案】B.【解析】1xex，ln(1)xx，1112xx，11cos2xx，故答案选择B.8.【答案】4.【解析】122241(1)1,sin,4axaxxxx::即4a9.【答案】14.【解析】401cosln(1tan)limsinxxxxx2420022002201ln(1tan)ln(1tan)2limlim2sec11tansec1tanlimlim44sec2sectan1lim.44xxxxxxxxxxxxxxxxxxxxx10.【答案】A.【解析】sinfxxax与2ln1gxxbx是0x时的等价无穷小,则22000232000330()sinsinlimlimlim()ln(1)()sin1cossinlimlimlim36sinlim1,66xxxxxxxfxxaxxaxgxxbxxbxxaxaaxaaxbxbxbxaaxabaxb等洛洛即36ab,故排除B,C.另外,201coslim3xaaxbx存在,蕴含了1cos0aax0x,故1,a排除D.所以本题选A.由此极限为2，故9092312aabab.6.【答案】1.【解析】22222222222222()+()=limlim+2=1limlim++xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx7.【答案】sin1.【解析】0lim(sin1sin)=sin1xxx8.【答案】极限不存在.【解析】0000limlim11,limlim11,xxxxfxxfxx左右极限不等，所以在0x处极限不存在.9.【答案】26.【解析】221111313+131lim=lim223+12212=limlim62212212xxxxxxxxxxxxxxxxxxxxxx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.