高数第一章无穷小比较

第七节无穷小的比较一、无穷小的阶二、等价无穷小第一章1引例当x?0时,3x,x,sinx都是无穷小,2x但lim?0,x?03x2sinx1lim?,x?03x3sinxlim2??,x?0x可见无穷小趋于0的速度是多样的.2一、无穷小的阶定义1设?,?是自变量同一变化过程中的无穷小,?若lim?0,则称?是比?高阶的无穷小,记作???o(?)?若lim??,则称?是比?低阶的无穷小;??若lim?C?0,则称?是?的同阶无穷小;?23例如,当x?0时x?o(6x)11?cosx如：lim??2x?02x故时是关于x2的同阶无穷小。2x2sin2lim2xx?04()23二、等价无穷小定义2设?,?是自变量同一变化过程中的无穷小,?若lim?1,则称?是?的等价无穷小,记作?~??或?~?例如,当x?0时sinx～x;tanx～x；arcsinx～x；1?cosx1又如，lim?2x?02x故时211?cosx～2xarctanx～x.4例1.证明:当证:时,～a?b?(a?b)(annn?1?an?2b???bn?1)n1?1x?1～xn5例2.求解:原式说明:当时,有ln(1?x)~x6例3.求解:令x?loga(1?t),1t原式?lim?limt?01t?0loga(1?t)loga(1?t)t1t?a?1,则x?limt?0loga(1?t)1t说明:当时,有e?1~xa?1~xlnax7x高等第八讲主讲教师:王升瑞8定理1.设且存在,则?lim?此定理称为等价无穷小的替代定理。????????证:lim?lim??????????????????lim?limlimlim???????tan2x2x2lim?例如,x?0sin5x?limx?05x59(1??1.例4.求limx?0cosx?1解:123x)例5求解10?1例6求I?limn??ln?n?b??lnn解：利用数列极限与函数极限的关系axaene?1limx???ln?x?b??lnxaae?1x?lim???limx????b?x???bbln?1???x?x?a?I??bax11e?e例7求I?lim.x?0x?sinx解xsinxI?limex?0sinxe?1?1x?sinxx?sinx一般，若f(x)g(x)则f(x)?g(x)e?1e?eaa?eI?lim?elimf(x)?g(x)f(x)?g(x)12例8求I?limn(a?an??21n1n?1)解I?limnan??021n?1(a21n(n?1)?1)n?alnalim?lnan??n(n?1)a1n(n?1)1?1~lnan(n?1)13limu(x)?0,limv(x)??,则有说明:若x?xx?x00x?x0lim?1?u(x)?v(x)?e例9.求x?x0limv(x)u(x)(?lim(1?2x))x?03x解:原式2x?3sinx3?2xx14定理2.证:～～????o(?)?lim?1?????lim(?1)?0,即lim?0??????o(?),即????o(?)例如,x?0时,～tanx～x,故tanx?x?o(x)15x?0时,极限运算法则设对同一变化过程,?,?为无穷小,由等价无穷小的性质,可得简化某些极限运算的下述规则.(1)和差取大规则:若?=o(?),则???~?证明例9?????1lim＝lim（1?）??????~?1x?lim3?lim3x?0x?3xx?03xsinx例1016(2)和差代替规则:若?~??,?~??且?与?不等价,?????????lim,则???~?????,且lim??但?~?时此结论未必成立.tan2x?sinx2x?xlim例如,?lim1?21?x?1x?0x?0x2例11求解原式tanx?sinxlim.3x?0xx?x原式?lim3x?0x21x?2x?limx?0x317(3)因式代替规则:若?~?,且?(x)极限存在或有界,则lim??(x)?lim??(x)11例如,limarcsinx?sin?limx?sin?0?x?0xxx?0例12I?lim解xln?1?2x??tan3x1?x?x?1232x?011?u?1~u2xln?1?2x??tan3xI?2lim23x?0x?x2x?x~x2322x?3x?10?2lim2x?0x22ln?1?2x?~2xtan3x~3x2218ln(1?x?x)?ln(1?x?x)例13limx?0secx?cosx解:22ln[(1?x)?x]原式=lim2x?01?cosxcosxln(1?x?x)?lim2x?0sinxx?lim2x?0sinx1922224x?0limcosx?12例14求1xx???lime?0x解:f0?????2?e?sinx???lim???2?1?14?x?x?0??1?ex????1?2?exsinx??lim????1x?x?0?4x?1?(e)?f0???0?1?1原式=1(2000考研)20内容小结1.无穷小的比较设?,?对同一自变量的变化过程为无穷小,且??0?是?的高阶无穷小?是?的低阶无穷小?是?的同阶无穷小?是?的等价无穷小?是?的k阶无穷小21常用等价无穷小:～～～～～ln(1?x)e?1xa?1x2.等价无穷小替换定理(1)和差取大规则:若?=o(?),则???~?22(2)和差代替规则:若?~??,?~??且?与?不等价,?????????lim,则???~?????,且lim??但?~?时此结论未必成立.(3)因式代替规则:若?~?,且?(x)极限存在或有界,则lim??(x)?lim??(x)23作业P631口答；2单号；3;4；5；6.24xx2（14）lim[]x??(x?a)(x?b)22010考研(a?b)x?abx?lim[1?]x??(x?a)(x?b)x[(a?b)x?ab]limx??(x?a)(x?b)a?b?e?e252（16）解因为当当令时，，则时，所以266.当时，若均是比x高阶的无穷小，求的取值范围。解(2014考研)所以及27