1第三节极限的运算法则定理则设,lim,limBvAu;limlim)lim()1(BAvuvu证略;limlim)lim()2(BAvuuv)0(limlimlim)3(BBAvuvu2vuvulimlim)lim(说明:1.有两层意思:(1)在limu和limv都存在的前提下,lim(u+v)也存在;(2)lim(u+v)的数值等于limu+limv.2.lim(u+v)存在,不能倒推出limu和limv都存在.3.若limu存在,而limv不存在,则lim(u+v)必不存在.4.可推广到有限多项.反证:uvuv)(若lim(u+v)存在,已知limu存在,由定理知limv存在,矛盾3推论1则为常数而存在如果,,limcu则是正整数而存在如果,,limnu推论2;limlim)lim()2(BAvuuv.lim)lim(uccu.)(limlimnnuu,lim00xxxx前已证.lim00nnxxxx所以例1)13(lim22xxx13limlim222xxxx1)2(3)2(2.14例2.531lim232xxxx求解531lim232xxxx)53(lim)1(lim2232xxxxx.3731235解例3.321lim221xxxx求.,,1分母的极限都是零分子时x.1后再求极限因子先约去不为零的无穷小x)1)(3()1)(1(lim321lim1221xxxxxxxxx31lim1xxx.21)00(型消零因子法6例4.147532lim2323xxxxx求解)(型332323147532lim147532limxxxxxxxxxx.72一般,为非负整数时有和当nmba,0,000.,,,0,,lim00110110mnmnmnbabxbxbaxaxannnmmmx当当当7例5.2arctanlim22xxxxx求解.,,,0,,lim00110110mnmnmnbabxbxbaxaxannnmmmx当当当xxxxxxxxxxxarctanlim2lim2arctanlim2222221.48共扼因子法解)1)(1()1)(1(lim1xxxxx原式)1(lim1xx.2.11lim1xxx求例69定理设函数y=f((x))是由()yfu000lim(),()xxxuxu但0ulim()ufuAu()x与复合而成,如果00lim[()]lim()uxuxfxfuA则0xx说明:可以为其他形式的极限。10证明0,010uu当时,0ulim()ufuA由于10根据极限的定义,0()uxu另外,定理的条件指出,所以()-fuA00xlim(),xxu又由于100对于上述的,00xx当时,01uu010uu0lim[()]Axxfx()-fuA11解令61tx,变量代换法,0x,1t11lim321ttt原式11lim21tttt.32.1111lim30xxx求例7326311111(),110t1,12t13xxtfttxtxt当时,当时,f(t)12例8).21(lim222nnnnn求解.是无穷多项之和时,n222221lim)21(limnnnnnnnn22)1(limnnnn.21先变形再求极限.