文档介绍：目录上页下页返回结束第一章二、极限的四则运算法则一、无穷小运算法则第六节极限运算法则目录上页下页返回结束一、: ???2 2 21 1 ( ) ?nnn n n??? ?????????????个+22 2 21 1 ( ) ?nnn n n??? ?????????????个+32 2 21 1 ( ) ?nnn n n??? ?????????????个+1limnn??22limnnn??32limnnn??0?1???目录上页下页返回结束定理2 .. ??解:1sin?x?01lim???xx利用定理2 ???xxx说明:y = 0 是xxysin??目录上页下页返回结束二、,)()(lim)3(;)]()(lim[)2(;)]()(lim[)1(,)(lim,)(lim??????????BBAxgxfBAxgxfBAxgxfBxgAxf其中则设推论1 .)(lim)](lim[xfCxfC?( C为常数)推论2 .nnxfxf])(lim[)](lim[?( n为正整数)目录上页下页返回结束思考:,)(lim,)(lim不存在存在若xgxf)]()(lim[xgxf?是否存在? 为什么?答: )()]()([)(xfxgxfxg???利用极限四则运算法则可知)(limxg存在, ( ) , lim ( ) ,f x g x若不存在不存在lim[ ( ) ( )]f x g x?,是否一定不存在? 问lim[ ( ) ( )]f x g x,)(lim,)(lim不存在存在若xgxflim[ ( ) ( )]f x g x是否一定不存在? : ,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 次多项式,)(10nnnxaxaaxP?????试证).()(lim00xPxPnnxx??证:??)(lim0xPnxx0axaxx0lim1?????nxxnxa0lim?)(0xPn?( )( ) ,( )P xF xQ x?其中)(,)(xQxP都是多项式,,0)(0?xQ试证: 00lim ( ) ( ) .x xF x F x??证: 0lim ( )x xF x??)(lim)(lim00xQxPxxxx??)()(00xQxP?0( )F x???322 1lim5 3xxx x??? ?322 1lim5 3xxx x???? ?322 12 10 3??? ?例?求解?思考:若,0)(0?xQ怎么求函数极限?3222lim( 1)lim( 5 3)xxxx x???? ?x = 3 时分母为0 !31lim3????????xxxx)3)(3()1)(3(lim3??????xxxxx62?31?目录上页下页返回结束例5 .????xxxx解:x = 1 时,3245lim21????xxxx0?31241512?????????????4532lim21xxxx分母= 0 , 分子≠0 ,但因目录上页下页返回结束结论: ( ) ( ) .x xF x F x??( )( ) ,( )P xF xQ x?,0)(0?xQ若则0( ) 0 ,Q x?若0lim ( ).x xF x?求0( ) 0 ,P x?0( ) 0 ,P x?去公因子再求0lim ( )x xF x??? ( ) ( ) .x xF x F x??( ),F x则