第六天答案.pdf-2021-02-27-17-11-26-672.pdf

第六天第六天1.【答案】A.【解析】01limln 1001lim 11xxxxxeex，A 正确，B 错误.1limln 111lim 1xxxxxeex,C 错.1limln 111lim 1xxxxxeex,D 错误.2.【答案】2e【解析】12111100002cos lnln 11limlimlimlimcot lncot2sinsinsin01lim=1xxxxxxxxxxxxxxxxxxxxeeeeex 等价替换3.【答案】3e【解析】22222222000cosln 1 3tan3tanlimlimlimcotln 1 3tancot32sinsin0=(1 3tan)limxxxxxxxxxxxxeeeex 等价替换4.【答案】1e【解析】22244444sin2 ln tantan1tan11lim tan2 ln tanlimlimlim1 tanlim1 tancos2cos21 tan1 tantan2x14lim tan=xxxxxxxxxxxxxxxxxxxeeeeee5.【答案】13()abc【解析】000011131lim1limlnlimlimlnlnln333330=()lim3xxxxxxxxxxxxxxxxabcabcabcxxxaa bb ccxxxxxabceeeeabc 洛必达6.【答案】e【解析】111limlimln+lim=xxxxxex exxx exxxeeee 洛必达7.【答案】13e【解析】22223200001111arctan1arctanarctan11+lim1limlnlimlim330arctanlim=xxxxxxx xxxxxxxxxxxeeeeex 洛必达8.【答案】C【解析】22222lim1limlnlimlimlim()()xxxxxx ax ba b xabxxxxxxx ax bx ax bx ax bx ax ba bxxeeeeexa xb9.【答案】1e【解析】ln1ln2111 ln1 1 lnln11limlimlimln1111lnlimln11lnlnlim(1)=xxxxxxxxxxxexxxxxexxxxxxxeeeee 洛必达等价替换10.【答案】12e【解析】22201lncoslncoslimln(1)ln(1)ln(1)00lim(cos)=limxxxxxxxxxee而2200lncosln(1 cos1)limlimln(1)ln(1)xxxxxx20cos1limxxx(等价无穷小替换ln(1)xx)220112lim2xxx(等价无穷小替换211 cos2xx)故原式=.121ee