14两个重要极限-课件PPT（演示稿）.ppt

文档介绍：一. 极限存在的准则ESC§. 无穷小量等价代换§. ,那么 ??)(lim)(lim)(limxfA)(xg)(xh准则Ⅰ如果函数 , ,在同一变化过程中满足)()()(xhxfxg??,)(xf准则Ⅱ如果数列 单调有界,则一定存在.??nxnnxlim??. ()1sinlim0??xxx证因为 ,)sin(??????x作单位园O,设圆心角,延长交过点的切线于于,则面积<扇形面积<??OBADAOB?AOBAOD?ESC一. ??.因为,当20???x时,0sin?x所以,在()式中除以,得xsin210?xxxxcossin1??,xxxcos1sin1??.三项都为正数,取它们的倒数,得()1sincos??xxx即 . ()ESC一. , ,于是有222112sin21cosxxx????1sincos2112???? ,由准则Ⅰ可得1)211(lim20???xx1sinlim0??. ?.例1求xxxxxxxcos1sinlimtanlim00????解.()1cos1limsinlim00?????xxxxxESC一. 例2求 .)0(sinlim0??kxkxx解即令 .则当 时, .于是kxt?0?x0?kxkkxkxxkxxx????sinlimsinlim00.()kk???1ttktsinlim0???ESC一. 例3求 .)0,0(sinsinlim0???babxaxxbxaxxsinsinlim0?xbxxaxxsinsinlim0??00sinsinlimlimxxbxbbxaxaax???. ()ba?ESC一. .)2tan7sin3tan(lim0xxxx??)2tan7sin2tan3tan(lim)2tan7sin3tan(lim00xxxxxxxxx?????解227232tan7sinlim2tan3tanlim00????????. 例5求 .20cos1limxxx??2022020]22sin[lim212sin2limcos1limxxxxxxxxx?????=解21121]22sinlim[21220?????xxx.