文档介绍：第六节极限运算法则?一、极限运算法则?二、求极限方法举例?三、小结思考题一、极限的四则运算法则定理1则设,)(lim,)(limBxgAxf??BAxgxfxgxf?????)(lim)(lim)]()([lim)1(BAxgxfxgxf????)(lim)(lim)]()([lim)2(证明:仅证明结论(3),并考虑极限过程为由极限与无穷小的关系,要证明???BAxgxf)()(0lim0???xx其中???BAxgxf)()(或0xx?则若,0)3(?BBAxgxfxgxf??)(lim)(lim)()(limBAxgxf?)()(BABA??????)(??????BBAB,)(???????BBAB令,)(lim0Axfxx???,)(????Axf0lim0???xx其中,)(????Bxg,)(lim0Bxgxx??又0lim0???xx其中0)(limlim00??????????BBABxxxx只要证明0)(lim0?????ABxx?有界即可仅需证明)(1??BB证明:仅证明结论(3),并考虑极限过程为由极限与无穷小的关系,要证明???BAxgxf)()(0lim0???xx其中???BAxgxf)()(或0xx?,21)(2BBB????,2)(12BBB???故有界。,0,0???B?又,0???,00时当????xx,2B????????BBBB21??B21?,2||B???对于BAxgxf?)()(BABA??????)(??????BBAB,)(???????BBAB令0)(limlim00??????????BBABxxxx只要证明0)(lim0?????ABxx?有界即可仅需证明)(1??,)()(lim)3(;)]()(lim[)2(;)]()(lim[)1(,)(lim,)(lim??????????BBAxgxfBAxgxfBAxgxfBxgAxf其中则设推论1则为常数而存在如果,,)(,,)(limnxf推论2).(lim)](lim[xfcxcf?.)]([lim)](lim[nnxfxf?定理,)(lim,)(lim),()(bxaxxx???????而如果说明:(1)上述关于函数极限的四则运算法则对数列极限同样成立。ba?则有证明:令)()()(xxxf????,0)(?xf则由极限的保号性有,0)(lim?xf而由极限的四则运算性质有)(limxf)]()([limxx????)(lim)(limxx????ba??.,0baba????(2)、????xxxx求解)53(lim22???xxx?5lim3limlim2222??????xxxxx5limlim3)lim(2222??????xxxxx52322????,03??531lim232?????xxxx)53(lim1limlim22232????????3123??小结:则有设,)(.1110nnnaxaxaxf??????nnxxnxxxxaxaxaxf?????????110000limlim)(limnnnaxaxa??????10100).(0xf?nnxxnxxaxaxa????????110)lim()lim(00即当f (x) 是一个关于x的多项式时,有)()(lim00xfxfxx??其中设,)()()(.2xQxPxf?)(lim)(lim)(lim000xQxPxfxxxxxx????)()(00xQxP?).(0xf?,0)(0?xQ若nnnaxaxaxP??????110)(mmmbxbxbxQ??????110)(0)(0?xQ且则有注意:则上述商的运算法则不能用。解)32(lim21???xxx?,0?商的法则不能用)14(lim1??xx?又,03??1432lim21???????由无穷小与无穷大的关系,??????????xxxx