主讲老师:高昆轮新浪微博:@考研数学高昆轮启航教育官方微博:@启航教育启航教育官方b站:启航教育官方《张宇考研数学1000題》刷题训练营Day2极限专题lnvvuue=、原则:求(复杂)极限务必先化简,如,通分、约分、有理化、换元、提取公因子、非零因子先计算等.()():lim211xIxxxxx→+=+−+−+−解11lim211xxxxxx→+=−+++++()()2lim211xxxxxxxx→+−+=+++++()()()2lim2112xxxxxxxx→+−=+++++++0.=11111111111:lim1nnnnnnnaaIbb−++→−++−=−解11lnln1lim.11lnln1naannbbnn→−+==−+1111111lim1nnnnnab−+→−+−=−()()()()()()()lim,limlimlim,.hxfxgxhxfxgxhx=若存在则即“加减中存在极限的部分可以先计算”技巧:()20:limln10,xaax→+=解注意到(存在)()()2200ln1limlimln1xxaxaIaaxxx→→+=−++于是()20ln1lim0xaxaxx→−+=+()222012lim.2xaxax→==()()222001ln1:limlimxxxxeexIxx→→+−+=+解()2ln1220limxxxeeex+→−=+()2ln122201limxxxeeex+−→−=+()2202ln12limxxxeex→+−=+()2220ln12limxxxeex→+−=+220.ee=−+=一、用好泰勒()()240ln1ln11.lim____.xxxxx→+−+=()2442405512lim.12xxxoxxIx→−−++==−所以()()()2233011:11,lim0.xxxeBxCxAxeBxCxAxoxx→++−−++−−==分析即()()2332301111126:limxxxxoxBxCxAxx→++++++−−解()()23330111262limxBBAxBCxCxoxx→+−+++++++=0=1110,0,0.262BBABCC+−=++=++=于是二、背好等价2:01cos,2xxx→−解利用时,()201322limxxIxx→=直接有2102lim0,3xxbx+→==()212,1,,.3bD+===于是即此时选32032032200322001cos2cos3:lim____.1cos2cos3lim1cos2cos2cos2cos3limlim1cos21cos3limlcoimcoss2cos22xxxxxxxxxxxxxxxxxxxxxxxxx→→→→→→−=−=−−=+−−−=++思考2:01cos,2xxx→−解利用时,三、明察洛必达()200sin2:lim4xxtdttx→+分析()2000sin24:lim11xxxtdttItdt→+=+−解20sin24lim11xxxx→+=+−012lim2.122xxx→==()()202000sin21sin24limlim.4xxxxtdtttdtxxt→→+==+四、凑拉格朗日()()()()()()22222211:ln1ln1sin11sinsin,fxxxxxxx=+−+=+−+=−解()222211sin,01,sin,xxxfxxx++→→−其中介于与之间于是在时,有此时()()sinsinxxxx=−+34112,4.63xxxn==于...