微積分基礎(chǔ)(國(guó)家開(kāi)放大學(xué))---第1章---第2節(jié)---極限的概念和計(jì)算.

1、121A2AnA123(n邊)3注意注意 1 數(shù)列對(duì)應(yīng)著數(shù)軸上一個(gè)點(diǎn)列數(shù)列對(duì)應(yīng)著數(shù)軸上一個(gè)點(diǎn)列.可看作一可看作一 動(dòng)點(diǎn)在數(shù)軸上依次取動(dòng)點(diǎn)在數(shù)軸上依次取.,21nxxx1x2x3x4xnx2 數(shù)列可看作定義在整數(shù)列可看作定義在整數(shù)集上的函數(shù)數(shù)集上的函數(shù)).(nfxn 4,81,41,21,21n,34,21, 2,11nnn, 2lg, 1lg,lgn,1, 1 , 1, 1,11n(1)(2)(3)(4)0814121034202lg3lgnlg011,n1021n,n111nnn,nnlg,n不趨于一確定值215limnnxA63(2) 1,8, 27 , 64,n1234(1),23451

2、nn11111(3),23451n78lim( )( )()xf xAf xA x 或1limxx01limcosxx1900lim( )( )()xxf xAf xA xx或3lim 26xx0limcos1xx21lim(23)0 xxx10 0limxxfxA 0limxxfxA Axfxlim Axfxlim Axfxx0lim .limAxfx 0limxxfx 0limxxfxA limxf x Axfxlim111(1)lim(1)xx211(2)lim1xxx111(3)lim ( )( )11xxxf xf xx其中121lim(1)2xx211lim1xxx1lim(1)2

3、xx1lim( )xf x1lim(1)2xx12xx1sinlim00 xxxsinlim13xx21lim0yxOxxxy210014.lim0不存在不存在驗(yàn)證驗(yàn)證xxxyx11 o00limlimxxxxxx左右極限存在但不相等左右極限存在但不相等,.)(lim0不存在不存在xfx證證0lim( 1)1x 00limlimxxxxxx0lim11x15例例5 5).(lim,0, 10,1)(02xfxxxxxfx 求求設(shè)設(shè)yox1xy 112 xy解解兩個(gè)單側(cè)極限為兩個(gè)單側(cè)極限為是函數(shù)的分段點(diǎn)是函數(shù)的分段點(diǎn),0 x)1(lim)(lim00 xxfxx , 1 )1(lim)(lim2

4、00 xxfxx, 1 左右極限存在且相等左右極限存在且相等,. 1)(lim0 xfx故故161718111lim10nnnn 19sinlim?xxx1lim0,xxsin1limlimsin0 xxxxxx01lim sin?xxx2021定理定理. 0,)()(lim)3(;)()(lim)2(;)()(lim)1(,)(lim,)(lim BBAxgxfBAxgxfBAxgxfBxgAxf其中其中則則設(shè)設(shè)22推論推論1 1).(lim)(lim,)(limxfcxcfcxf 則則為常數(shù)為常數(shù)而而存在存在如果如果常數(shù)因子可以提到極限記號(hào)外面常數(shù)因子可以提到極限記號(hào)外面.)(lim)(l

5、im,)(limnnxfxfnxf 則則是正整數(shù)是正整數(shù)而而存在存在如果如果推論推論2 22314lim22xxx22lim xxxx4lim21lim2x22limxxxx2lim41lim2x2224113101( ),nnnf xa xa xa設(shè)則有nnxxnxxxxaxaxaxf 110)lim()lim()(lim000nnnaxaxa 10100).(0 xf 24121lim220 xxxx ,limAxf ,limBxg 0lim Bxg xgxflimBA xgxflimlim1lim20 xx112lim20 xxx10121lim220 xxxx) 12(lim) 1(l

6、im2020 xxxxx1110( )( ),()0,( )P xf xQ xQ x設(shè) 有理分式且則有)(lim)(lim)(lim000 xQxPxfxxxxxx )()(00 xQxP ).(0 xf ., 0)(0則商的法則不能應(yīng)用則商的法則不能應(yīng)用若若 xQ25解解)32(lim21 xxx, 0 商的法則不能用商的法則不能用)14(lim1 xx又又, 03 1432lim21 xxxx. 030 由無(wú)窮小與無(wú)窮大的關(guān)系由無(wú)窮小與無(wú)窮大的關(guān)系,得得.3214lim21 xxxx求求.3214lim21 xxxx2693lim23xxx)3(lim3xx0)9(lim23xx000未定

7、型93lim23xxx)3)(3(3lim3xxxx31lim3xx)03(x61方法：約去零因子法不可利用法則3 ,limAxf ,limBxg xgxflimBA xgxflimlim27357243lim2323xxxxx)243(lim231xxx)357(lim231xxx357243lim2323xxxxx33357243limxxxxx7312lim2xxx212limxxx112lim2xxx012lim2xxx28為非負(fù)整數(shù)時(shí)有為非負(fù)整數(shù)時(shí)有和和當(dāng)當(dāng)nmba, 0, 000 , 0,lim00110110mnmnmnbabxbxbaxaxannnmmmx當(dāng)當(dāng)當(dāng)當(dāng)當(dāng)當(dāng)無(wú)窮小分出

8、法無(wú)窮小分出法 以自變量的最高次冪除分子以自變量的最高次冪除分子,分分母母,以分出無(wú)窮小以分出無(wú)窮小,然后再求極限然后再求極限.29).21(lim222nnnnn 求求解解是無(wú)限多個(gè)無(wú)窮小之和是無(wú)限多個(gè)無(wú)窮小之和時(shí)時(shí), n222221lim)21(limnnnnnnnn 2)1(21limnnnn )11(21limnn .21 先變形再求極限先變形再求極限.302121limnnn221limnnnn221limnnnn21)1112(lim21xxx112lim21xxx11lim21xxx11lim1xx21)1(limxxxxxxxxxx1)1)(1(limxxx11lim031xx

9、xcoslimxxlimxxcoslim不存在xxxcos1limxxxcoslim01limxx1cosx有界函數(shù)根據(jù)無(wú)窮小乘以有界函數(shù)仍是無(wú)窮小x1是當(dāng)x時(shí)的無(wú)窮小量所以xxxcos1limxxxcoslim0無(wú)窮小有界函數(shù)無(wú)窮小32從條件, 0)11(lim2baxxxx求常數(shù).,babaxxx1121)()()1 (2xbaxbaxa01a0ba1a1b33341sinlim0 xxxAC)20(, xxAOBO 圓心角圓心角設(shè)單位圓設(shè)單位圓,tan,sinACxABxBDx 弧弧于是有于是有xoBD.ACO ，得，得作單位圓的切線作單位圓的切線,xOAB的圓心角為的圓心角為扇形扇形,

10、BDOAB的高為的高為 ,tansinxxx , 1sincos xxx即即0limcos1,xx0lim11,x0sinlim1.xxx0()00sinsin()sinlimlimlim1.()xxuxxuxxu0sinlim1.xxx351sinlim0 xxx注100型(含三角函數(shù))注2x為弧度注31sinlim0在形式上完全一致，且 xf注41sinlim0 xxx如：220sinlim1xxx111()01sinlim sinlim1xxxxxx0sinlim(0)xkxk kxsin()lim1xaxaxa36xxx5sinlim0555sinlim05xxx5xtgxx0limx

11、xxsinlim0 xcos1xxxsinlim0 xxcos1lim0120cos1limxxx2202sin2limxxx2022sinlim21xxx211sinlim0 xxxsinlim0 xxxxxx1sinlim00 xxx1sinlim11sinlim0 xxx37xxx11lime11nn1lim 1nnn1lim 1nnne590457182818284. 2e11111111nxnnxn1lim 1xxx1(1)11lim 1lim 11xuxuxuxu (1)lim1uuuu(1)1lim 1uuu11lim 11uueuu 1lim 1xxex38xxx11lime注11注211lim在形式上完全一致,且可以 xfe無(wú)窮大量無(wú)窮小量1lim如：exxx101lim39101lim 1lim 1xxxxexex211limxxx2111limxxx21exxx11lim111limxxx1 exxx1021lim221021limxxx221021limxxx2ennn111lim11111limnnn1111limnnn1111limnne232limxxxx3 51lim 13xxx 3511lim1133xxxxe4041424344454647484950515253545556575859606162636465666