高等數(shù)學中求極限的方法小結(jié)

1、16高等數(shù)學中求極限的方法小結(jié)2.求極限的常用方法2.12.1利用等價無窮小求極限利用等價無窮小求極限這種方法的理論基礎(chǔ)主要包括：(1)有限個無窮小的和、差、積仍是無窮小.(2)有界函數(shù)與無窮小的乘積是無窮小.(3)非零無窮小與無窮大互為倒數(shù).(4)等價無窮小代換(當求兩個無窮小之比的極限時，分子與分母都可用等價無窮小代替).[3]設(shè)、且；則：與是等價無窮小的充分必要條件為：???~~???limlim????????0()?????常

2、用等價無窮小：當變量時，0x?21sin~tan~arcsin~arctan~1~ln(1)~1cos~2xxxxxxxxxexxxxx???11~(1)1~xxxxx???????例1求01coslimarctanxxxx??解，2101cos~arctan~2xxxxx???時故，原式220112lim2xxx???例2求1230(1)1limcos1xxx????解因此:12223110(1)1~1cos~32xxxxx?????

3、時原式202123lim132xxx????例3求30131limtanxx???解，故:原式=0x?時3111~tan~3xxxx??0113lim3xxx??18（此為強行代入以定型）.22224431100(00)(00)0000000000?????????可能是比高階的無窮小，倘若不這樣，或??00???00?422(00)(00)0000000?????或.43(00)(00)0000000?????解22222222400

4、01cossincos(sincos)(sincos)lim()limlimsinsinxxxxxxxxxxxxxxxxxx?????????，33000sincossincossincoslimlim2limxxxxxxxxxxxxxxx????????由洛必達法則的.22222001cossin4sin422limlim333xxxxxxx??????有：上式=例7求.201limxxexx???解.22000(1)1limlim1

5、lim1()21xxxxxxeeexxxxx????????????????例8求.332132lim1xxxxxx??????解原式.（二次使用洛必達法則）.22113363limlim321622xxxxxxx?????????例9求.02limsinxxxeexxx?????解原式.0002limlimlim21cossincosxxxxxxxxxeeeeeexxx???????????????例1010求.22143lim21x