文档介绍：xxxsinlim0?极限xxx)11(lim???极限§1-4? 0 sin?sintancosxxx? logex x?以e为底的指数函数y=ex的反函数y = logex,叫做自然对数,在工程技术中经常被运用,常简记为y = ,它的前八位数是: e = 281 8??? cos0=1 | sin | 1 x? | cos | 1 x?( )n n nab a b?n m n ma a a????mnm na a?,以0为极限的变量称为在这个变化过程中的无穷小量,常用字母性质无穷小量与有界变量的乘积仍为无穷小量., ,a b g等表示。 lim lim lim(1) ( ( ) ( )) ( ) ( )f x g x f x g x? ? ?2) lim[ ( ) ( )] lim ( ) lim ( )(f x g x f x g x? ? ?( ) lim ( )lim( ) lim ( ).f x f xg x g x? lim ( )(3) 0g x?若,(4) lim[ ( )] lim ( )cf x c f x?(5) lim[ ( )] [lim ( )]k kf x f x?X 1 …. ??xxxxxsin0sinlim ?xxx??X -1 - - - - …. ?第一个重要极限OxBACD0sinlim ??+证明证 sin tanx x x? ?即sin ( sin 0),x x?各式同除以因为得,cos1sin1xxx??.1sincos??xxx即0sinlim ?? ?+0 0tan sin 1lim lim( )cosx xx xx x x? ?? ?0 0sin 1lim limcosx xxx x? ?? ?解1 1 1? ??0sin 1lim( )cosxxx x?? ?这个结果可以作为公式使用1tanlim0??xxx0tanlimxxx?例1求例2 5 , 0 , 0x t x t? ??令当时有0sin, 5limttt??所以原式注:在运算熟练后可不必代换,直接计算:0sin5 limxxx?求,?推广:设为某过程中的无穷小量sinlim 1???某过程0sin5 limxxx?解:05sin5lim5xxx??0sin55lim5xxx??0sin55lim 5 1 55xxx?? ???0sin5limxxx?5 1 5? ??练习1. 求下列极限:00sin3 1 limsin52 lim3xxxxxx??()()0 0sin3 3sin3 lim lim3x xx xx x? ??解:0sin33lim3xxx??3 1 3? ??0 0sin5 sin5 5 lim lim( )( )3 5 3x xx