高数竞赛辅导例题课件.ppt

1、高等数学习题选解高等数学习题选解、1.)1ln(lim,0)(lim,11nxnnnnnnxxx 求求满满足足已已知知数数列列解解.,0,02111成成立立有有时时当当 nnxxNnN)(.)()(123121 nnnxxxxxxxx123121.nnnxxxxxxxx11123121.1111 nnNNNNxxxxxxxxxxx.)(.2112312111 NnxxxxxxxNN 2.)1(111123121 nNnxxxxxxxnxNNn ;2.11123121 nxxxxxxxnxNNn0lim11123121.nxxxxxxxnNN.,02.2211123121 nxxxxxxxNNN

2、nN有有时时当当),max(21NNN 取取,时时当当Nn 22:nxn有有.0lim nxnn.01ln)1ln(lim)1ln(lim11 nnxnnnnxnn、2.432ln)21ln(lim122 nkknnnkCnnn求求解解,)1(0 nkkknnxCx;)1(:111 nkkknnkxCxn两两边边求求导导得得,)1(11 nkkknnkxCxnx.)1()1()1(:11221 knkknnnxkCxxnnxn两边求导得两边求导得;2)1(212112 nnnkknnnnkCx代入代入将将 432)1(2lim221nnnnnn.4143lim4122 nnnnn 2)1(24

3、3)21ln(lim432ln)21ln(lim212122nnnnnkknnnnnnnnkCnnn2)1(2432lim212 nnnnnnnnn、3.42cos1lim1221tdtttnnn 求求解解 tnttntttd tdtdtnnn1141141142cos141142cos1121222cos)1()1(tdnntttn2cos)1(114112cos4114111;)cos2(cos)1(12sin212411411dtnnttnn dtndtnnttnnnnnnnntttnn12sin21241141142cos11122lim)cos2(coslim)1(limlimdtn

4、ttnnnnnn 12sin2124142cos1limcoslimlim0;lim12sin21411dtnttnn dtdtnnttnttn 122112sin21110)(0)1(11 nnn0lim12sin211 dtnttnn.lim41142cos1122 dtntttnn、4.tan)cos1()sin(sin)tan(tan)1(lim2200 xxxxdtextx 求求解解 xxxxxxxxxxexxxxtan)cos1()sin(sin)tan(sinlimsintan)tan(sin)tan(tanlim2lim0004.1sintansec)sin(tanlimsin

5、tan)tan(sin)tan(tanlim20tan,sin0 xxxxxxxxxxxx 之间之间介于介于.1limtan)cos1()cos(sin1)tan(sinlimtan)cos1()sin(sin)tan(sinlim221221000 xxxxxxxxxxxxxxx.422tan)cos1()sin(sin)tan(tan)1(lim2200 xxxxdtextx xxxxxxxdtexxxxdtexxtxxtxtan)cos1()sin(sin)tan(tanlimtan)cos1()tan(tanlimtan)cos1()sin(sin)tan(tan)1(lim00000

6、2222 xxxxxxxdtexxtxtan)cos1()sin(sin)tan(sin)tan(sin)tan(tanlimlim02210022;tan)cos1()sin(sin)tan(sinlimsintan)tan(sin)tan(tanlim200 xxxxxxxxxx 、5.)1,0()(:,)1,0()(,)1,0()()(上上连连续续在在求求证证上上都都是是单单调调递递增增的的在在与与函函数数且且函函数数上上有有定定义义在在设设函函数数xfexfexfxfx 证明证明),1,0(0 x;10,000 xxxx时时当当);()(0)()(0 xfxfeexfxf );()()

7、()(0000 xfxfexfexfexxxx );()()(000 xfxfxfexx 夹逼准则夹逼准则);()(lim000 xfxfxx );()(lim000 xfxfxx 同理可得同理可得,)(0连续连续在在xxxf.)1,0()(上连续上连续在在xf、6.lim,arccos,:,)(),0()2(;),0()(,)1(:,cos)1(.coscos)(212122211221 nnnnnnnnnnnnnnnxxNnNNxfxxfnxCxCxCxf且且时时当当则则有有满满足足设设中中仅仅存存有有一一根根在在区区间间方方程程对对于于任任何何自自然然数数求求证证设设证明证明;)cos1

8、(1cos)1(.coscos)()1(1221nnnnnnnnxxCxCxCxf )0()(;0)(,1)0(2122nnnnffff ,)(),0(212 nnnxfx满满足足介介值值定定理理 0sin)cos1()(1 xxnxfnn.)(仅存有一根仅存有一根单减单减 xfn;1)(arccoslim)1(1)(arccos)2(211111 ennnnnnnff211)(arccos;0 nnfNnN时时当当)()(arccos1nnnnxff nnfxn 1arccos单减单减;arccos21 nnx.limarccoslim221 nnnnx夹逼准则夹逼准则).(,1)()(xf

9、xfxxf求求函函数数已已知知 、7解解)1(,)()(1)()(222xxfxxfxxfxxfx 1)()(1)()(xfxxfxfxxf)2(),()(xfxxfx )()()2(),1(22xfxxxfx 22)()1(xxxfx 22222111)(xxxxxxxxf ;11)(222dxxxdxxxxf .arctan)1ln(21)(2cxxxxf ).0(),()(),1).(1)(1)(1()(,arcsinsin1sin11)(2842FxfgxFxxxxxgxxxxfn 求求设设、8解解xfxffx)0()(lim)0(0 ;1arcsinlimsin1sin10 xxxx

10、x);0()1()0()0()0(fgffgF )1).(1)(1)(1()()1(28442nxxxxxgx .1)()1(122 nxxgx;1,21,)(21211 xxxgnxxn 1)1()(lim)1(1xgxggx)2(lim21211111nxxxxn )1)(1()1(2112212limxxxxxnn2212)1()1(21121lim xxxxnn )1(2221211112100limxxxxnnn111112lim2 xxxxnn.24 1)12(lim222111100nnnxnnx .24)0()1()0(nnfgF .1)()(,),(:),()(,)(,),(

11、,)(2221 ffbaabdxxfaafbabaxfba使使得得内内至至少少有有一一点点在在求求证证且且有有内内可可导导在在上上连连续续在在设设、9证明证明 bababaxdxdxxfabdxxf)()()(2221;0)(badxxxf;0)()()(),(abccfdxxxfbacba使得使得积分中值定理积分中值定理;0)(ccf;)()(:xexxfxF 令令,),(,)(内可导内可导在在上连续上连续在在cacaxF;0)()(aeaafaF且且0)()(ceccfcF)()(cFaF,),(),(内内至至少少有有一一点点在在罗罗尔尔定定理理baca 0)(F使得使得.1)()(ff、

12、10.)(),0(:,)(0,1)0(,),0)(efexfxfxfx使使得得存存在在求求证证有有对对一一切切且且上上连连续续可可导导在在设设证明证明,),0()(,)()(:上上连连续续可可导导在在令令 xFexfxFx,01)0()0(fF且且,)(xexf 由于由于0lim)(lim xxxexf所以所以0)(lim xfx;0lim)(lim)(lim xxxxexfxF;0,0)(tan)(:22 tttFt令令),(0)(tanlim)(lim20022 tFttt,),0()(,0)(22内可导内可导在在上连续上连续在在 tt,0)()0(2 定理定理由由Rolle),0(2 ,

13、0)(使得使得0sec)(tan)(2 F,0)(tan F tan 取取;0)(),0(F,)()(xexfxF 0)(ef.)(ef.,),1,0(),1,0(:,0,0,1)1(,0)0(,)1,0(,1,0)()()(babaffxffbfa 使使得得求求证证内内可可导导在在上上连连续续在在设设、11证明证明;)(),1,0()1()0()1,0(kcfcfkfk 使使得得介介值值定定理理,)1,(),0(,1,0)(内可导内可导在在上连续上连续在在ccccxf)1)()()1(;)()0()(:),1,0()1,(),1,0(),0(cfcffcffcfcc 使得使得拉格朗日中值定理

14、拉格朗日中值定理;1;1;)(1)()()1()()()0()(ccccfkfcfffkffcf ;1)(1)(fkfk)1,0(1);1,0(babbaakk取取.1)()()()(bafbfaffbabbaa .2coscossin,0,为为等等价价无无穷穷小小量量与与时时使使得得当当求求kcxxxxxxkc、12 )()4(4(4sin2coscossin33!314141xoxxxxxxxxx解解);()(33383338xoxxoxxx .,3)0(1)(383)(38)(3382coscossin333 ckxxxkcxxoccxxokccxxxxxkk、13).1,0(,:arc

15、sin)1ln(11 xxxxx证明证明证明证明).1ln()1(arcsin1)(:2xxxxxf 令令1)1ln(1arcsin)(21 xxxfxx).1,0(,0)1ln(arcsin)(21 xxxxfxx0)0()()(fxfxf单减单减,0)1ln()1(arcsin12 xxxx).1,0(,arcsin)1ln(11 xxxxx、14证明证明).(:2baababeeabee 证明证明;ba 不妨假设不妨假设.,);()(:2baxaxeexfaxeeax 令令;)()(22axxeeexaxexf ,(,0)()()(2222baxaxaxexfxxxxeeeex ,(,0

16、)()()(baxafxfxf ,(,0)()()(baxafxfxf 0)(0)(2 abeebfabeeab).(2baababeeabee 、15.)(,0:),()(,1)0(,1)0(,),0)(xexfxxfxfffxf 时时求证求证上二阶可导上二阶可导在在设设证明证明,1)0()0();,0),()(:fFxxfexFx令令);()()()()(xfxfexfexfexFxxx );()()(:xfxfexGx 令令;01)0()0()0()0(fffG )()()()()(xfxfexfxfexGxx )(0)()(xGxfxfex0)0()(,0 GxGx时时0)()(xfx

17、fex0)()()(xfxfexFx)(xF1)0()(,0 FxFx时时.)(1)(xxexfxfe 、16.,)(:,1)0()2(;,),()()(:)1(,0)()()(,0)(,)()0(212221221RxexffRxxfxfxfRxxfxfxfxfxfxfxx 证明证明若若证明证明且且二阶可导二阶可导若若证明证明0;);(ln:)1()()()()()()(22 xfxfxfxfxfxfyyxfy令令是凹函数是凹函数y).()()()(ln222122)(ln)(ln212121xxxxxfxffxfxff ;)()0()0()()2(22110 xxyxyyxyMaclaul

18、in 公式公式;)0()0(ln)(ln2)()()()(21)0()0(22xfxxfxfxfxfxfxfff .)()0(xfexf ;arccosarcsin xdxx求求、17解解 xxxdxxxxdxxarccosarcsinarccosarcsinarccosarcsin dxxxxxxxxx)(arccosarcsin221arcsin1arccos 21)arcsinarccos(arccosarcsinxdxxxxx )arcsin(arccos11)arcsin(arccosarccosarcsin22xxdxxxxxxx dxxxxxxxxxx)(11)arcsin(ar

19、ccosarccosarcsin22111122.21)arcsin(arccosarccosarcsin2cxxxxxxx .)(1lim,)(0 xxdxxfxxxxf求求极极限限设设函函数数解解)(11)1(xfxxxxxf .1)(为周期函数为周期函数以以xf 10101010)()(dxxxdxdxxxdxxf.2110 xdx;)()()(11000dxxfdxxfdxxfxxxxxx ;1111xxx ;)(1)(1)(111000dxxfxdxxfxdxxfxxxx dxxfxxdxxfxxxx 100)(1lim)(11lim;211lim21 xxxdxxfxxdxxfxx

20、xx 1010)(1lim)(1lim;211lim21 xxx夹逼准则夹逼准则.21)(1lim0 xxdxxfx、18.cossinsin203 dxxxx求求解解 dtttttttdtdxxxxtx22220302222303cossincos)cos()sin()()(sincossinsin ;cossincos203 dxxxx 2203303cossincossin21cossinsin dxxxxxdxxxx 2022cossin)coscossin)(sincos(sin21 dxxxxxxxxx 20)cossin1(21 dxxx.414sin41221202 x、19.

21、sin1224 xdxeeIxx求求解解 222244sin1)()(sin1 tdteetdteeItttttx;sin11sin11222244 xdxetdtext 222244sin11sin12 xdxexdxeeIxxx 222220444sin2sinsin11 xdxxdxxdxeexx 202sin432 xdx 2021432 dx;83221432 .163 I、20)0(,)1)(1(02 xxdx求求解解 022102)1)(1()1)(1(ttdttxxdxtx 02222)1)(1(tttdttt;)1)(1()1)(1(0202 xxdxxttdtt 02020

22、2)1)(1()1)(1()1)(1(2 xxdxxxxdxxxdx 02021)1)(1(1xdxdxxxx ;2arctan0 x.4)1)(1(02 xxdx、21.)()(sinlim,0)(020dxxfdxxfnxxfn 试试证证上上连连续续在在设设.22;sin)()(sin)(sin111110dxnxfdxxfnxdxxfnxnknknkknknknknknkk 积积分分中中值值定定理理证明证明 dttdxnxkkntnxnknk )1(11sinsin dttn 01sin.sin201nndtt nknkknnkffdxxfnx12210)()()(sin nknknnk

23、nknnffdxxfnx12120)(lim)(lim)(sinlim .)(02dxxf ).()0()2(sin)(:,2,0)(220为为正正整整数数其其中中证证明明上上单单调调增增加加且且连连续续可可微微在在设设nffnxdxxfxfn 证明证明 nxdxfnnxdxxfcos)(1sin)(2020 )(cos1)0()2(20 xdfnxnnff nxdxxfnnffcos)(1)0()2(20 dxnxxfnnffnxdxxfcos)(1)0()2(sin)(2020 ;0)()(xfxf单单增增.0)0()2(ff dxxfnnff 20)(1)0()2(dxxfnnff 20

24、)(1)0()2().0()2(2ffn .23、24.2ln)(:,0)1(,arctanarctan)()(,10,1,0)(2110 dxxffyxyfxfyxxf求求证证又又时时且且当当上上可可积积在在设设证明证明 dxxfdxxf1010)()(dxfxf 10)1()(dxx101arctanarctan dxxdxx)arctan(arctan410104 )arctan()arctan(410104xxdxx .2ln)1ln(021102211012 xdxxx.1)(cos,)(sinmax),(:112112 dxxfxdxxfxxxf有有对对任任意意连连续续函函数数证证

25、明明、25证明证明 dxxfxxfxxdxxfxdxxfxx）（)(cos)(sin)(cos)(sin2211112112 dxxfxxfxx1122)(cos)(sin;2)1(11111 dxxdxx.1)(cos,)(sinmax112112 dxxfxdxxfxx.)()(max,0)()(),(,:2)(4dxxfxfbfafxfbabaabbxa 则则有有如如果果上上的的连连续续可可微微函函数数对对于于证证明明证明证明,)(max:xfMbxa 令令:),(由由拉拉格格朗朗日日中中值值定定理理得得bax )2()()()()()()1()()()()()(2211bxMbxfxf

26、bxfbfxfaxMaxfxfaxfafxf :),(),(),(),(21stbabxbaxa dxxfdxxfdxxfbbabaaba22)()()(dxbxMdxaxMbbabaa22 dxxbMdxaxMbbabaa22)()(bbaMbaaMbxax222222)()(.)()(max)()()()(2412422222222abxfabMbabxaMabbaMbaM .)()(max2)(4dxxfxfbaabbxa 、26.)(,)(,0,sin41221时时液液面面上上升升的的速速率率求求液液面面高高度度为为的的速速率率注注入入液液体体如如果果以以轴轴旋旋转转所所得得的的旋旋转

27、转面面绕绕线线一一容容器器的的内内表表面面是是由由曲曲msmyyyyx 解解.,0;:2102 ydyxVy液液体体的的体体积积为为dyxdtddtdVy 02 dtdydyxdydy02;)(2dtdyyx ;)(1)()(222yxyxyxdtdVdtdy )4(124 xdtdyy ).()224(1)4sin(4(122sm 、27、28.),(,)0,0(),(:,)0,0(),()0,0(),(0)tan(),()0,0(2222yxdyyxfyxyxyxyxfyxyx并求并求处可微处可微在在证明证明设设 证明证明;1lim)0tan(limlim)0,0(22tan02110)0

28、,0()0,(0 xxxxxxxfxfxxxf;1lim)0tan(limlim)0,0(22tan02110)0,0(),0(0 yyyyyyyfyfyyyf )tan(221)0,0()0,0()0,0(),(222222yxyxyxyxyxyxyfxffyxfyx )sincostan(2sincos1cossin rrrrrrxry)0(0)1)(sin(cos22tan rrr ,)0,0(),(处处可可微微在在yxf.),()0,0(dydxyxdy 、29).2,()2,(,)2,(,)2,(),(),(,),(2xxfxxfxxxfxxxfyxfyxfyxfxyxxxyyxx

29、与与试试求求且且的的二二阶阶偏偏导导数数皆皆连连续续已已知知函函数数解解xxxfxxfxxxfyx2)2,(2)2,()2,(2 两边求全导两边求全导2)2,(4)2,(2)2,(2)2,(xxfxxfxxfxxfyyyxxyxx两边求全导两边求全导);1(2)2,(4)2,(5)2,()2,()2,()2,(xxfxxfxyxxxxfxxfxxfxxfyyxxyxxy);2(1)2,(2)2,()2,(xxfxxfxxxfxyxxx两边求全导两边求全导.21)2,(;0)2,()2(),1(2145112521452142 xxfxxfxyxx.1),(:2),(42222上的最大值和最小值

30、上的最大值和最小值在椭圆域在椭圆域求函数求函数 yxyxDyxyxf解解,2)2(),(22xyxyxfxx .2)2(),(22yyxyxfyy ,000),(0),(yxyxfyxfyx);1(2:42222 yxyxL 令令 014020222221yxLyyLxxLyx 01;20yxyx,3)0,1(,2)2,0(,2)0,0(fff.2,3:mM比较得比较得、30、31.)sin(1lim2060dyxydxttxtt 计计算算解解 dxxydytdyxydxtytttxtt020602060)sin(1lim)sin(1lim交换积分次序交换积分次序dxxttttHL 0250.

31、)sin(61lim00 dututttxtu1sin61lim20250duuttt 20260sin61lim 540.36sin2lim00ttttHL.181sinlim181440 ttt、32.)cos(sin,1:2222 DdxdyyxyxD求求设设解解;)cos(sin)cos(sin2222 DxyDDdxdyxydxdyyx对对称称关关于于 )cos(sin)cos(sin(21)cos(sin222222DDDdxdyxydxdyyxdxdyyx dxdyxyyxD)cossincossin(212222.Ddxdy、33.)(1lim,2:,0)0(,0)(22250

32、222 dVzyxfttzzyxfuuft求求可可导导在在设设.arccos0,20,202:2222trtrtzzyx 解解 drddrrfdVzyxfsin)()(22222 drrfrddrrfrdttrttr)()1(2sin)(22202arccos02022202 ;)()(2202312022 tttdrrfrdrrfr 5202312022022250)()(2lim)(1limtdrrfrdrrfrdVzyxftttttt 6202320220)()(2limtdrrfrdrrfrtttt 5232220220.6)4(82)4(44)(2lim00ttfttfttdrrfr

33、ttHL 5202206)(2limtdrrfrtt ).0(15324)0()4(lim153215)4(8lim2204220.00ftftfttftttHL .,.)4,3()2,1(),(2所所做做的的功功对对质质点点求求力力角角小小于于轴轴的的正正向向夹夹其其与与与与原原点点的的连连线线其其方方向向垂垂直直于于质质点点的的距距离离与与原原点点的的大大小小等等于于质质点点作作用用的的过过程程中中受受变变力力运运动动从从如如图图红红线线所所示示为为直直径径的的半半圆圆周周沿沿以以设设质质点点PFyPPFFBAABP 解解.443:sin23cos22:yxLAB为为直直径径的的半半圆圆周

34、周).,(,),(xyFLyxP LxdyydxW 443)sin23)(cos22()cos22)(sin23(d).1(2)cos2)(cos22()sin2)(sin23(443 d、34、35.)(:,1:),(2)(22222222eyfDxfyxyfxfdxdyyxeyxDyxf 证证明明且且偏偏导导数数有有二二阶阶连连续续在在区区域域设设函函数数证证明明;)sincos()(2010cossin drrrdrdxdyyxyfxfrxryyfDxf ):()sincos(22220正向正向 ryxldxdydrrlyfxfyfxf dxdyedxdyrrrDyxDyfxfGreen

35、ryxD)(:222222222)(公式公式);1(22020rreded 10)1()(2rdredxdyyxryfDxf.21021212ere 、36.21,222222轴轴正正方方满满足足右右手手法法则则其其方方向向与与为为其其中中求求曲曲线线积积分分zzzyxzyxzdzdyxy 解解.;:21:432221222222 所所为为圆圆盘盘yxzzzyxzyx.1cos;0cos,0cos:法法向向量量方方向向余余弦弦 zdzdyyxzdzdyxyzdzdyxyxyxy)()()()(dSzyxdSzxyxyzyxxyzyxStokes)()(01000100公式公式.0)1(4321

36、4321)()(xyxydSdS zdzdyyxzdzdyyxzdzdyxyzdzdyxyABxyBAxy)()()()()()(、37).,(,2),()()()(2),(),(,)0(1222222yxfdxdyeyxzfdzdxexydydzezxyxyxfyxfzzyxzzz求求满满足足连连续续函函数数的的外外侧侧为为设设 解解;2),()()(:22dxdyeyxzfdzdxexydydzezxazzz 令令;)(2),(2ayxyxf .:.1,0:22 所围区域所围区域与与下侧下侧yxz;2),();();(:22zzzeyxzfRezyQezxP 令令 RdxdyQdzdxdy

37、dzPRdxdyQdzdxdydzPRdxdyQdzdxdydzPa dxdydVeyxfezezyxzzzGauss1222222),(公式公式 dxdydVeyxfezezayxzzzGauss1222222),(公式公式;)(2),(2ayxyxf 2),(22 dVyxfz 2)(2222dVayxz 24)(2222dVaxyzyx 24)(232222adVxydVzyx;20sin23256321040202 aadrrdd )32(5183256 aaa.)(2),()32(5182 yxyxf、38.,.),3,2(3,.,5,2,1:1111321的的敛敛散散性性判判别别级

38、级数数记记已已知知数数列列 nnannnnnxxnaaaaaaan解解,0,0,01,02,0111221 nnnaaaaaaa假设假设已知已知.0;0)(21111 nnnnnnnnnaaaaaaaaa则则.单单调调增增加加由由数数学学归归纳纳法法na11111233 nnnnnnnaaaaaaa;32113211nnaaxxnn ;)()(.)(013211322232132 nnnnnxxxx.,)(11132收收敛敛比比较较判判别别法法收收敛敛 nnnnx、39).(,1,1:)2()1(,2,1011127210 xSxxaxnaaaaannnnnn和和函函数数时时并并求求当当收收敛

39、敛幂幂级级数数时时当当证证明明设设 解解,1limlim,)1(212111111 nnnaannnaannnnnnnaa.,110收收敛敛幂幂级级数数时时当当收收敛敛半半径径 nnnxaxR),3()1()2()1(11111 naanaannnnnn 32111211)()()(nnnnnnnnnnnnnaaa 273122342111)1()1().()()(nannnnnnn);3()1()1(67 nnn;)1()1(21)(36722732210nnnnnnxnxxxaxaxaaxS nnnxnxf 3)1()1()(:令令;)()(13104xxnnxxdxxf 2434)1(3

40、41)()(xxxxxxf .1),1()(331221)1(12 xxxxSx、40.,2cos1求求此此微微分分方方程程并并求求通通解解解解为为次次微微分分方方程程有有一一个个设设四四阶阶实实常常系系数数线线性性齐齐xxeyx 解解,212cos1ixxeyx 特特征征方方程程有有二二重重特特征征根根特特解解0)21()21(22 ii 故特征方程为故特征方程为02520144)52(23422 化简化简.0252044:)4(yyyyy微分方程为微分方程为.2sin)(2cos)(:4321xxCCxxCCeyx 通解为通解为、41.,)(221222222的的表表达达式式试试求求函函数数且且满满足足具具有有连连续续二二阶阶偏偏导导数数设设uyxuyxuuxuxyuxu 解解22:yxt 令令,22dtdutxdtduyxxxtdtduxu )()(22211dtduxtxdtduxttdtdutxux txdtudtxdtdutxtdtdutx222)(11;)(2222321dtudtxdtdutxt ;)(222232221dtudtydtdutytyu 同理同理;222tudtud 代入原方程代入原方程.2sincos:221 ttCtCu通解通解.2sincos)(2222222122 yxyxCyxCyxu