极限的四则运算法则-课件PPT（演示稿）.ppt

文档介绍：第一章二、极限的四则运算法则三、复合函数的极限运算法则一、无穷小运算法则第五节机动目录上页下页返回结束极限运算法则时, 有??,,min21????一、:,0lim0???xx,0lim0???xx,0???,01???当100????xx时, 有2???,02???当200????xx时, 有2???取则当????00xx???????22??????)(lim0?????xx这说明当0xx?时,???:无限个无穷小之和不一定是无穷小!例如,?????????????????nnnnnn2221211lim?1?( P56 , 题4 (2) )解答见课件第二节例5机动目录上页下页返回结束类似可证: 有限个无穷小之和仍为无穷小. 定理2 .:设,),(10?xx????Mu?又设,0lim0???xx即,0???,02???当),(20?xx???时, 有M???取??,,min21????则当),(0?xx???时, 就有??u?u?????MM故,0lim0???uxx即?u是0xx?. ??解:1sin?x?01lim???xx利用定理2 ???xxxxxysin?说明:y = 0 是xxysin?、极限的四则运算法则,)(lim,)(limBxgAxf??则有??)]()(lim[xgxf)(lim)(limxgxf?证: 因,)(lim,)(limBxgAxf??则有??????BxgAxf)(,)((其中??,为无穷小) 于是)()()()(???????BAxgxf)()(??????BA由定理1 可知???也是无穷小,再利用极限与无穷小BA??的关系定理, .若机动目录上页下页返回结束推论:若,)(lim,)(limBxgAxf??且),()(xgxf??( P45 定理5 ))()()(xgxfx???:定理3 可推广到有限个函数相加、:令机动目录上页下页返回结束定理4. 若,)(lim,)(limBxgAxf??则有?)]()(lim[xgxf)(lim)(limxgxf提示:利用极限与无穷小关系定理及本节定理2 :定理4 .)(lim)](lim[xfCxfC?( C为常数)推论2 .nnxfxf])(lim[)](lim[?( n为正整数) 次多项式,)(10nnnxaxaaxP?????试证).()(lim00xPxPnnxx??证:??)(lim0xPnxx0axaxx0lim1?????nxxnxa0lim?)(0xPn?BA?机动目录上页下页返回结束为无穷小(详见P44)B2???B1)(1xg?)(0xx???定理5 .若,)(lim,)(limBxgAxf??且B≠0 , 则有?)()(limxgxf)(lim)(limxgxf证:因,)(lim,)(limBxgAxf??有,)(,)(??????BxgAxf其中??,设BAxgxf??)()(?BABA??????)(1???BB)(??AB?无穷小有界BA?因此?由极限与无穷小关系定理, 得BAxgxf?)()(lim)(lim)(limxgxf????BAxgxf)()(为无穷小,,lim,limByAxnnnn??????则有)(lim)1(nnnyx???nnnyx??lim)2(,00)3(时且当??BynBAyxnnn???limBA??BA?提示:因为数列是一种特殊的函数,故此定理可由定理3 , 4 , 5