兩個(gè)重要極限

1、11.6兩個(gè)重要極限兩個(gè)重要極限準(zhǔn)則準(zhǔn)則I如果數(shù)列xn、yn及zn滿足下列條件?(1)yn?xn?zn(n?1?2?3????)?(2)??aynn???limaznn???lim那么數(shù)列xn的極限存在?且?axnn???lim證明?因?yàn)??以根據(jù)數(shù)列極限的定義????0??Naynn???limaznn???lim1?0?當(dāng)n?N1時(shí)?有|yn?a|???又?N2?0?當(dāng)n?N2時(shí)?有|zn?a|???現(xiàn)取N?maxN1?N2?則當(dāng)n

2、?N時(shí)?有|yn?a|???|zn?a|???同時(shí)成立?即a???yn?a???a???zn?a???同時(shí)成立?又因yn?xn?zn?所以當(dāng)n?N時(shí)?有a???yn?xn?zn?a???即|xn?a|???這就證明了?axnn???lim簡(jiǎn)要證明?由條件(2)????0??N?0?當(dāng)n?N時(shí)?有|yn?a|??及|zn?a|???即有a???yn?a???a???zn?a???由條件(1)?有3?xxxcos1sin1??或?1sinc

3、os??xxx注意此不等式當(dāng)??x?0時(shí)也成立?而?根據(jù)準(zhǔn)則I??2?1coslim0??xx?1sinlim0??xxx簡(jiǎn)要證明簡(jiǎn)要證明?參看附圖?設(shè)圓心角?AOB?x()?20???x顯然BC?AB?AD?因此sinx?x?tanx?從而(此不等式當(dāng)x?0時(shí)也成立)?1sincos??xxx因?yàn)?根據(jù)準(zhǔn)則I???1coslim0??xx1sinlim0??xxx應(yīng)注意的問(wèn)題?在極限中?只要?(x)是無(wú)窮小?就有?)()(sinlim

4、xx??1)()(sinlim?xx??這是因?yàn)?令u??(x)?則u?0?于是?)()(sinlimxx??1sinlim0???uuu?(?(x)?0)?1sinlim0??xxx1)()(sinlim?xx??例1?求?xxxtanlim0?解??xxxtanlim0?xxxxcos1sinlim0???1cos1limsinlim00?????xxxxx例2?求?20cos1limxxx??解??20cos1limxxx??22