第六节极限运算法则本节要建立极限的四则运算法则和复合函数的极限运算法则.在下面的讨论中，记号“”下面没有表明自变量的变化过程，是指对和以及单则极限均成立.但?x?limxx?0在论证时，只证明了的情形.x?x0分布图示★极限运算法则★例1★例2★例3-4例6★例5★★例★例79★例811例★★例10★复合函数的极限运算法则★例12★例13★★内容小结课堂练习1-6★习题内容要点一、极限的四则运算：定理1推论1推论2二、复合函数的极限运算法则：定理2例题选讲极限的四则运算2?3x?lim(x5).例1(E01)求x?222225?limlim?3x?lim5?(limxlim(x)?3x?5)?limx?lim3x?解35?3?2??2x?2x?2x?2x?2x?2x?2x?2nn?1设注：??a?ax,?axf(x)?则有n10nn?1nn?1?x??aax?aa)?(limx)??a(limxa?).f(x?)f(xlim??010n0n100x?xx?xxx?0002?x92lim.求例2(E02)2?7xx?253?x2?9)xlim(2229?22x?9?393?x解mil??.?22222235??75x?x?27??3)?lim(5x7?2x3?x3?xP(x)Q(x)?0,,f(?x)则有且：设注0)x(Q．limP(x)P(x)x?x0?flimf(x)?(?x).00Q()limQ(xx)x?x00xx?0Q(x)?0时，则商的法则不能应用当.04x?1lim.例3(E03)求2?2xx?31x?2lim(x?2x?3)?0,x????0)lim(4,31?解又商的法则不能用.x?11?x2?2x?3x04x?1?lim??0.lim??.由无穷大与无穷小的关系，得24x?13x?2x?31x?1?x2?1x例4(E04)求.lim2x?2x?31x?0解时，分子和分母的极限都是零先约去不为零的无穷小因子后再求1?x?1x).型(0极限.2)1x?x?1)((?1xx?11（消去零因子法）?limlim?lim?.2)?1?3)(x(xx?323x?2x?1x?1x?1?x32?x5?32x.lim计算例5(E05)32?1?47xx?x??3x?x?去除分子分母，分出无穷型时，分子和分母的极限都是无穷大(解).先用?.小，再求极限35?2?23253x?2x?3xx.lim?lim?(无穷小因子分出法)14237?1x7x?4???xx??7?3xxa?0,b?0,mn为非负整数时，有注：当和00a?0,当n?m?b1?mmaax?ax????00m10,当n?lim?m.?1?nnb??bxb?x??x??n01?,当n?m??无穷小因子分出法：以分母中自变量的最高次幂除分子和分母，以分出无穷小，然后再求极限的方法.2331??56xx8x?.lim例6计算23x??x??x?,?此类极限也不能直接用极限运算法则，可把分子时，分子分母均趋于解.分母同除以绝对值最大的项，再用极限运算法则165?8??32332?1?8x5?6xx32xxx.lim?lim?233x?2???x?x?3x1n2??.?lim?????求(E06)例7222nnn???n?.先变形再求极限解本题考虑无穷多个无穷小之和1)1n(n?1?2n1?2?n111?????2.lim???1?lim?lim?lim???????222222n2nnnnn???????nn???n?n?34)?)(?x1?xx)(1(1.lim计算例83)x(1?1x?，故不能应用极限运算法则，而要先对函数做必要的变形，因分因分母的极限为0解.子中含有根式，通常用根式有理化，然后约去分子分母中的公因子34)x)(1??x)(1?x(1lim3)x(1?1x?)1)(?xx)(1?x1(?lim?33223441?x34)x1x??x)(?x??xx)(1?x(1?)(111lim??.243223441x?34)xx?1x(1?)(?x?x)(1?x?).lim(sinx?1?sinx例9计算???x???xsin?xsinx1与的极限均不存在，但不能认为它们差的极限也解时，不存在，要先用三角公式变形：x?1?xx?1?xcosx??(sinlimx1sin)2?limsin22???x???xx??1x1.0?lim2sincos?2)x2(x?1???x?.最后这一步用了“有界量与无穷小的乘积为无穷小”的结论计算下列极限：10例23!sinnnxtan;1()lim.(2)lim11n???n0x?e2?x231n,lim?lim?0!sinn是有界量，由“有界量与无穷小而解(1)由于11?n1??n?n??n323n23sinn!n?0lim.之积为无穷小”知n?1?n?11111,,2?e?2e?0,0??,x?0limtanxx为有界又即因为(2)从而1120x?e2?xe2?xtanx?0.lim量，所以10x?e?2xx?1,x?0??2?)(x,f),(xx),limf(x),limflimf(?3x?x1已知求例11?,x?0?????xx?0x??3x?1?limf(x),因为先求解x?0limf(x)?lim(x?1)??1??x?0x?02?3x?1x??1limf(x)?lim3?1x??0x?0x?limf(x)??1.此外，易求得所以x?0131??21x3?x?32xxxlim?(?limx)limf,?0131x?????x?x???x??13xlimf(x)?lim(x?1)???.x???x???2??1?xlimln..求(E07)例12??2(x?1)1x???221?1x??x1x?1,u???,u1?x令则当时，解一2(x?1)2(x?1)2.?limlnu?0故原式1u?22????11?xxx??1??.?0ln?lim?ln1limln?lnlim解二??????)1x2(2(x?1)?2??1???1x1xx????2,c)?2lim(5x?ax?bx?ba,.之值已知例13求???x22)??axc?bx(5x?axx?bx?c)(52lim(5x?)?bx?cax?lim因解2???xx???cbx?5x?ax?c?)xb?(25?a2c(25?a)x??bxx,?2?lim?limcb2??x???x?cx5?ax??bx?5?a?2xx0?a?25??b.?20ba?25,,故解得?2??a5??课堂练习:1.求极限11?x?3sinx;)(1lime.limx(2)3x?20?x??x?f(x)g(x)f(x)?g(x)是否有极限？为什那么在某个过程中，2.若有极限，无极限，么？