極限問(wèn)題

1、極限問(wèn)題極限時(shí)數(shù)學(xué)分析的精髓，既是重點(diǎn)也是難點(diǎn)，既需要一些常規(guī)的代數(shù)變化和一些經(jīng)驗(yàn)，公式（如stolz定理，洛必達(dá)法則），還要求在解決這類型問(wèn)題是能夠靈活變形，適當(dāng)?shù)膽?yīng)用等價(jià)量，無(wú)窮小量和泰勒公式。(一)一些著名的經(jīng)驗(yàn)極限（1）；（2）（3）1sinlim0??xxxexxx???10)1(lim1)1(lim0???xxInx(4)(5)(6)11lim0???xexx??????xxx1)1(lim0121cos1lim20???x

2、xx(7)(8)(9)121sintanlim30???xxxx161sinlim30???xxxx0!lim???nann(10)；(11)(0為任意實(shí)數(shù))0lim?????nnInn0lim?????nnInn??(二)求極限的一些方法第一種類型（直接代入）32)12)(1()1)(1(lim121lim1221???????????xxxxxxxxxnmxxxxxxxxnnmmxnmx??????????????????11lim

3、11lim212111??第二種類型（進(jìn)行一些簡(jiǎn)單的代換變型）（n代表一個(gè)自然數(shù)）21)()()(limaxaxnaaxnnnax??????解：設(shè)x=ay，則當(dāng)xa時(shí)y0代入??22320210212)1(])2)(1(!31)1(!2[lim)(lim)()()(lim??????????????????????????nnnnynnnynnnaxannyyannnannyynaayaaxaxnaax?第三種類型（利用stolz公式

4、）Stolz定理1.已知13lim)0(321sin01???????nnnnxnxnxx求證??證：事實(shí)上，因?yàn)?(311)(132???????nnxnxnnn1.12121lim1)sin11()1tan1(lim21sin21sin1121tan211tan10x1sin1tan1lim.553)6(2lim2arctan3tan2sin)sin(sinlim62arctan3tan22sin)(sinsin0x2arctan3

5、tan2sin)sin(sinlim.4000000??????????????????????????????????????????????xxxexxxxxxxxexxxxxxxxxxxxxxxxxxxxxxxxxxxxxx于是原式時(shí)有當(dāng)于是有時(shí)有當(dāng)?shù)谖宸N類型（利用洛必達(dá)法則注意所需條件）1coscos2sin212sincoslimsinsincoslimcos1coslimsinlim.26124sincos)sin(sin2

6、sin)cos(sin23cos)sin(sinlim12cossin)sin(sincos)cos(sinlim4sincos)sin(sinlimsincos)(sincoslim.1sinsinsin3sin0sin2sin0sin0sin0302203040???????????????????????????????????xxexexexeexxexeexxeexxeexxxxxxxxxxxxxxxxxxxxxxxxxxxx