兩個(gè)重要極限練習(xí)題

1、嚴(yán)謹(jǐn)規(guī)范求真鑄魂17兩個(gè)重要極限練習(xí)題兩個(gè)重要極限練習(xí)題教學(xué)過程：引入：考察極限xxxsinlim0?問題1：觀察當(dāng)x?0時(shí)函數(shù)的變化趨勢：x(弧度)0.500.100.050.040.030.02...xxsin0.95850.99830.99960.99970.99980.9999...當(dāng)x取正值趨近于0時(shí)，?1，即=1；xxsin??0limxxxsin當(dāng)x取負(fù)值趨近于0時(shí)，x?0x0sin(x)0于是)()sin(limsinl

2、im00xxxxxx????????綜上所述，得一1sinlim0??xxx的特點(diǎn)：1sinlim0??xxx(1)它是“”型，即若形式地應(yīng)用商求極限的法則，得到的結(jié)果是；0000(2)在分式中同時(shí)出現(xiàn)三角函數(shù)和x的冪推廣如果?(x)=0(a可以是有限數(shù)x0??或?)，ax?lim則==1ax?lim??????xx??sin????????xxx???sinlim0?例1求xxxtanlim0?解=xxxtanlim0?111cos1

3、limsinlimcos1sinlimcossinlim0000???????????xxxxxxxxxxxxx例2求xxx3sinlim0?解=xxx3sinlim0?3sinlim3)3(33sin3lim00????tttxxxtx令例3求20cos1limxxx??解=20cos1limxxx??2122sin22sin21lim)2(22sinlim2sin2lim0220220????????xxxxxxxxxxx例4求xx

4、xarcsinlim0?嚴(yán)謹(jǐn)規(guī)范求真鑄魂解令－=t，則x=－x2t2當(dāng)x??時(shí)t?0，于是==e–2xxx)21(lim???21020])1(lim[)1(lim???????tttttt例7求xxxx)23(lim????解令=1u，則x=2－xx??23u1當(dāng)x??時(shí)u?0，于是=xxxx)23(lim????])1()1[(lim)1(lim210120uuuuuuu?????????==e1])1(lim[])1(lim[20