文档介绍：,)()(lim)3(;)]()(lim[)2(;)]()(lim[)1(,)(lim,)(lim??????????BBAxgxfBAxgxfBAxgxfBxgAxf其中则设证.)(lim,)(limBxgAxf???.0,0.)(,)(???????????其中BxgAxf由无穷小运算法则,得极限的四则运算法则一、?)()]()([BAxgxf???????成立.)1(?)()]()([BAxgxf???ABBA????))((?????????)(?(2)成立.?BAxgxf?)()(BABA??????)(??????????AB?,0,0??B??又,0???,00时当????xx,2B????????BBBB21??B21?推论1).(lim)](lim[,,)(limxfcxcfcxf?则为常数而存在如果常数因子可以提到极限记号外面..)]([lim)](lim[,,)(limnnxfxfnxf?则是正整数而存在如果推论2,21)(2BBB????,2)(12BBB???故有界,(3)成立.?定理1给出了极限的四则运算法则,它可以推广到??A??B或以及(3)中的某些情形:(1)当 时,而 时,??A??B???)]()(lim[xgxf(2)当 时,而 时,??A0?B???)]()(lim[xgxf(3)当 时,而 时,??A??B??)()(limxgxf(4)当 时,而 时,??A??B0)()(lim?xgxf(5)当 时,而 时,0?B0?A??)()(limxgxf关于数列极限也有类似的四则运算法则定理2(复合函数的极限运算法则)设函数 是由函数 与)]([xfy??)(ufy?)(xu??复合而成,)]([xfy??)(lim0uxxx???Aufuu??)(lim,0且存在 ,当 时,00??),(000?xUx?有0)(ux??则Aufxfuuxx????)(lim)]([lim00?证按函数极限的定义,需要证:对任意的0??0??????00xx????Axf)]([,存在,当由于,对任意 ,存在Aufuu??)(lim00??0??当 时,????00uu???Auf)(又由于,)(lim00uxxx???对上面得到的 ,0??存在01??????00xx,当 时,????0)(ux由条件当 时,),(000?xUx?0)(ux??取},min{10????,则当时,????00uu????0)(ux0)(0??ux??????0)(0ux成立,从而??????AufAxf)()]([.)(lim)]([lim00ufxfuuxx????二、极限的不等性证明:令0)()()(???xgxfxF根据保号性定理,有0)]()(lim[)(lim???xgxfxF从而,0)(lim)(lim?????BAxgxfBxgAxf????则有且若)()(,)(lim,)(????xxxx求解)53(lim22???xxx?5lim3limlim2222??????xxxxx5limlim3)lim(2222??????xxxxx52322????,03??531lim232?????xxxx)53(lim1limlim22232????????3123??三、求极限方法举例