高等數(shù)學(xué)總復(fù)習(xí)1(2010.12)(西北工業(yè)大學(xué))

1、總復(fù)習(xí)(一) 一、主要內(nèi)容一、主要內(nèi)容 二、典型例題二、典型例題 作業(yè)集重點(diǎn)題 練習(xí)冊(cè)練習(xí)冊(cè)(上冊(cè)上冊(cè)) 5,20.)9( 6,17.)8( 5,15.)7( 6),1(5,12.)6( 4,11.)5( 8,7,10.)4( 4,8.)3( 5,7.)2( 3,3.)1( p p p p p p p p p 4,3,34.)19( 2);8(),4(1,33.)18( 7,32.)17( 4,31.)16( 5,30.)15( 5,28.)14( 10,27.)13( ),3(8,26.)12( 2,23.)11( 10,22.)10( p p p p p p p p p p )3(),1(

2、3,51.)29( 5,48.)28( )2()1(1,45.)27( 6,4,44.)26( )1(2,43.)25( 1,42.)24( 3,40.)23( 1210,39.)22( 5,37.)21( 3,36.)20( p p p p p p p p p p 、 、 )2(3,2),3(1,55.)32( 6),5(5,53.)31( )4(),3(4,52.)30( p p p 1,71.)42( 11,66.)41( 10,65.)40( p p p 5,64.)39( 8,61.)38( 6,60.)37( 5,4,59.)36( )2(3,2),4(1,58.)35( 86,5

3、7.)34( )3(5),2(4,56.)33( p p p p p p p 3,10.)9( 1210,8.)8( 9,7,7.)7( )1(5,6.)6( 3,2,5.)5( 10,4.)4( 86,3.)3( 5),4(),1(4,2.)2( 2,1.)1( p p p p p p p p p 作業(yè)集作業(yè)集(總冊(cè)總冊(cè)) 8,24.)20( 6,5,23.)19( 4,22.)18( 1210,21.)17( 97,20.)16( 6),1(5,4,19.)15( )2(3,2,1,18.)14( 3,2),10(1,17.)13( 10,13.)12( p p p p p p p p p

4、 96,12.)11( 5),4(),3(4,11.)10( p p 一、主要內(nèi)容一、主要內(nèi)容 1. 微分學(xué)基本概念微分學(xué)基本概念 函數(shù)、極限、無(wú)窮小、無(wú)窮大、無(wú)窮小的 比較(高階無(wú)窮小、同階無(wú)窮小、等價(jià)無(wú) 窮小)、連續(xù)、間斷點(diǎn)、導(dǎo)數(shù)、微分. 函數(shù)、極限、無(wú)窮小、無(wú)窮大、無(wú)窮小的 比較(高階無(wú)窮小、同階無(wú)窮小、等價(jià)無(wú) 窮小)、連續(xù)、間斷點(diǎn)、導(dǎo)數(shù)、微分. 2. 幾個(gè)重要關(guān)系幾個(gè)重要關(guān)系 收斂收斂 n x)1( 有界有界 n x = = )(lim)2( 0 xf xx 內(nèi)無(wú)界在某內(nèi)無(wú)界在某)()( 0 xUxf o ;)3(的關(guān)系函數(shù)極限與其子列極限 的關(guān)系函數(shù)極限與其子列極限 ;)4(的關(guān)系有

5、極限的變量與無(wú)窮小 的關(guān)系有極限的變量與無(wú)窮小 )()()(lim 0 xAxfAxf xx + += = = . 0)(lim 0 = = x xx 其中其中 ;)5(無(wú)窮大與無(wú)窮小的關(guān)系無(wú)窮大與無(wú)窮小的關(guān)系 幾個(gè)概念之間的關(guān)系幾個(gè)概念之間的關(guān)系)6( 可微可微可導(dǎo)連續(xù)可導(dǎo)連續(xù)極限存在極限存在 3. 求極限的方法求極限的方法 (1) 極限定義；極限定義； (2) 極限存在的充分極限存在的充分 必要條件；必要條件； (4) 極限運(yùn)算法則；極限運(yùn)算法則； (5) 極限存在準(zhǔn)則；極限存在準(zhǔn)則； (6) 兩個(gè)重要極限；兩個(gè)重要極限； (3) 有關(guān)無(wú)窮小的有關(guān)無(wú)窮小的 運(yùn)算；運(yùn)算； (7) 函數(shù)的連續(xù)

6、性；函數(shù)的連續(xù)性； (8) 導(dǎo)數(shù)定義；導(dǎo)數(shù)定義； (9) 利用微分中值公式；利用微分中值公式； (10) 洛必達(dá)法則；洛必達(dá)法則； (11) 定積分定義定積分定義. 二、典型例題二、典型例題 例例1 + = + + x x x axxeax 為何值時(shí)，問(wèn) 為何值時(shí)，問(wèn)a 的可去間斷點(diǎn)？ 是為何值時(shí)，處連續(xù)；在 的可去間斷點(diǎn)？ 是為何值時(shí)，處連續(xù)；在)(00)(xfxaxxf= = = 解解= )(lim)0( 0 xff xxx ax xarcsin )1ln( lim 3 0 + + ) 0 0 ( xx ax xarcsin lim 3 0 = = 2 2 0 1 1 1 3 lim x

7、ax x = = 2 3 )1( 2 2 1 6 lim 2 0 x x ax x = = a6 = = = + + + + )(lim)0( 0 xff x 4 sin 1 lim 2 0 x x axxe ax x + + + 4 1 lim 2 0 x x axxe ax x + = + = + + 2 2 0 1 lim4 x axxe ax x + = + = + + x axae ax x2 2 lim4 0 + = + = + + 2 2 lim4 2 0 + = + = + + ax x ea 42 2 + += = a 6)0(= =f 存在存在)(lim 0 xf x Q)

8、0()0( + + = = ff 得即得即, 426 2 +=+=aa. 2, 1 = = = =aa或或 處連續(xù)；在時(shí)，當(dāng)處連續(xù)；在時(shí)，當(dāng)0)(1= = =xxfa 處連續(xù)在而處連續(xù)在而0)(= =xxf)0()0()0(fff= + + , 6426 2 =+=+=aa即即 .)(0 , 6)0(12)(lim2 0 的可去間斷點(diǎn)是因而 時(shí)，當(dāng) 的可去間斷點(diǎn)是因而 時(shí)，當(dāng) xfx fxfa x = = = = = 例例2 討論討論 的連續(xù)性的連續(xù)性, = + + x x dtt x 0, 2= =x ), 2 , 1( sin lim)(limLQ= n x x xf nxnx . )()

9、, 2, 1( 無(wú)窮間斷點(diǎn) 的第二類是 無(wú)窮間斷點(diǎn) 的第二類是xfnnxL = = = 2 查分段點(diǎn)：查分段點(diǎn)：0= =x 1 sin lim)(lim)0( 00 = x x xff xx Q )(lim)0( 0 xff x + + + + = = 2 2 0 02 )1ln( lim x dtt x x + = + = + + x x x22 )21ln(2 lim 0 + + = = + 1= = 2)0(1)0()0(= + + fff .)(0的第一類可去間斷點(diǎn)是的第一類可去間斷點(diǎn)是xfx= = 再由初等函數(shù)的連續(xù)性可知，的連續(xù)范圍是再由初等函數(shù)的連續(xù)性可知，的連續(xù)范圍是)(xf

10、), 2, 1, 0(RxnnxxI = = = =L 類似題類似題 1.)()(sin 1 ln )(有，則設(shè)函數(shù)有，則設(shè)函數(shù)xfx x x xf = = 2008考研考研 (A) 1個(gè)可去間斷點(diǎn)，個(gè)可去間斷點(diǎn)，1個(gè)跳躍間斷點(diǎn)；個(gè)跳躍間斷點(diǎn)； (B) 1個(gè)可去間斷點(diǎn)，個(gè)可去間斷點(diǎn)，1個(gè)無(wú)窮間斷點(diǎn)；個(gè)無(wú)窮間斷點(diǎn)； (C) 2跳躍間斷點(diǎn)；跳躍間斷點(diǎn)； (D) 2個(gè)無(wú)窮間斷點(diǎn)個(gè)無(wú)窮間斷點(diǎn). 解解f (x) 無(wú)定義的點(diǎn)無(wú)定義的點(diǎn): x = 0, x = 1. xxxf xx sin)(lnlim)(lim 00 = = )0( x x x sin 1 ln lim 0 = = )( x x x x

11、2 0 sin cos 1 lim = = x x x x sin sin lim 0 = 001= = = = A 1 1 lim1)(sin 1 = = x x .)(0的可去間斷點(diǎn)是的可去間斷點(diǎn)是xfx = = x x x xff xx sin 1 ln lim)(lim)1( 11 = x x x xfsin 1 ln )( = = x x x = = 1 ln lim1)(sin 1 ) 0 0 ( 1sin = = x x x xff xx sin 1 ln lim)(lim)1( 11 = + + + 1 ln lim1)(sin 1 = = + + x x x 1 1 lim1

12、)(sin 1 x x + + = =1sin= = ),1()1( + + ff.)(1的跳躍間斷點(diǎn)是的跳躍間斷點(diǎn)是xfx = = 2.上的第一類間斷在函數(shù)上的第一類間斷在函數(shù), e)(e tane)(e )( 1 1 + = + = x x x x xf 點(diǎn)是點(diǎn)是 x= ( ). . 0)(A. 1)(B . 2 )( C. 2 )(D A 解解)(lim)0( 0 xff x = = e)(e tane)(e lim 1 1 0 + = + = x x x x x 1 = = )(lim)0( 0 xff x + + + + = = e)(e tane)(e lim 1 1 0 + =

13、+ = + + x x x x x 1= = 3.,)(lim),()(axfxf x = =+ + 內(nèi)有定義，且在設(shè)內(nèi)有定義，且在設(shè) = = 00 0), 1 ( )( x x f xg 是間斷點(diǎn)，處的連續(xù)性，若在討論是間斷點(diǎn)，處的連續(xù)性，若在討論00)(= = =xxxg 請(qǐng)指出其類型.請(qǐng)指出其類型. 解解 a x fxg xx = ) 1 (lim)(lim 00 Q ;0)(0處的連續(xù)在時(shí)，當(dāng)處的連續(xù)在時(shí)，當(dāng)= = =xxga .)(00的第一類可去間斷點(diǎn)是時(shí)，當(dāng)?shù)牡谝活惪扇ラg斷點(diǎn)是時(shí)，當(dāng)xgxa= = 例例3, 0,31lim 0,)1ln( )( + + + + = = xxx x

14、bxa xfn nn n 設(shè)設(shè) .0)(,處可導(dǎo)在，使試確定常數(shù)處可導(dǎo)在，使試確定常數(shù)= =xxfba 解解 nnn n x+ 31limQ + = + = 30,) 3 (1) 3 1 (lim3x x n nn n )1 3 0(+ x xx x n nn n )1 3 0( + + = = xx xx xbxa xf .0)(處可導(dǎo)，必連續(xù)在由于處可導(dǎo)，必連續(xù)在由于= =xxf 處連續(xù)在而處連續(xù)在而0)(= =xxf )0()0()0(fff= + + )(lim)0( 0 xff x = =b= = )0(03lim)(lim)0( 00 fxxff xx = + + + )1ln(l

15、im 0 bxa x + + = = 由由 得得. 0= =b 處可導(dǎo)在又處可導(dǎo)在又0)(= =xxfQ)0()0( + + = = ff 0 )0()( lim)0( 0 = x fxf f x x xa x 0)1ln( lim 0 = = . a = = , 3= = a. 3 = =a .0)(03處可導(dǎo)在時(shí)，即當(dāng)處可導(dǎo)在時(shí)，即當(dāng)= = = = =xxfba 0 )0()( lim)0( 0 = + + + + x fxf f x 3 03 lim 0 = = = = + + x x x 例例4, 0, 0, cos)( )( = = = = xa x x xxg xf設(shè)設(shè)有其中有其中

16、)(xg (0)1.g= = 確定 的值，使在處連續(xù)；確定 的值，使在處連續(xù)；(1)( )0af xx = = 在成立的情形下，求在成立的情形下，求(2)(1)( ).fx 二階導(dǎo)數(shù)，二階導(dǎo)數(shù)， 在處連續(xù)在處連續(xù)1.( )0f xx = = afxf x = = = )0()(lim 0 解解 )(lim 0 xf x Q x xxg x cos)( lim 0 = = ) 0 0 ( )0( g = = )0(ga = = 1 sin)( lim 0 xxg x + + = = 時(shí)，當(dāng)時(shí)，當(dāng)0.2 x cos)( )( = = x xxg xf 2 cos)(cos)( x xxgxxgx

17、= = 2 cos)(sin)( x xxgxxgx + + = = 時(shí)，當(dāng)時(shí)，當(dāng)0= =x 0 )0()( lim)0( 0 = x fxf f xx g x xxg x )0( cos)( lim 0 = = 2 0 )0(cos)( lim x xgxxg x = = ) 0 0 ( x gxxg x 2 )0(sin)( lim 0 + + = = sin 0 )0()( lim 2 1 0 x x x gxg x + + = = 1)0( 2 1 + =+ =g =+ + = =+ + = 0,1)0( 2 1 0, cos)(sin)( )( 2 xg x x xxgxxgx xf

18、 例例5設(shè)在上有定義，在區(qū)間上設(shè)在上有定義，在區(qū)間上( )(,)0, 2f x + + 2 ( )(4)f xx x= = 若對(duì)于任意 都滿足：若對(duì)于任意 都滿足：x ( )(2)f xk f x= =+ + 其中 為常數(shù)其中 為常數(shù).k問(wèn)： 為何值時(shí)，在處可導(dǎo)？問(wèn)： 為何值時(shí)，在處可導(dǎo)？( )0kf xx = = 解解1( ) 2 0)f x o o 求在， 的表達(dá)式.求在， 的表達(dá)式. ( )(2)f xk f x= =+ + 2 (2)(2)4k xx=+=+ (2)(4)kx xx=+=+ ( 022 )x + (20 )x (2004年考研年考研) 于是當(dāng)時(shí)，有于是當(dāng)時(shí)，有 2,2x

19、 ， 2 (2)(4),20 ( ) (4)02 kx xxx f x x xx + + + = = 2( )0f xx = = o o 討論在處的可導(dǎo)性討論在處的可導(dǎo)性 0 ( )(0) (0)lim 0 x f xf f x = 0 (2)(4)0 lim x kx xx x + + = + = (0)0.f= =由題設(shè)知由題設(shè)知 8k= = 0 ( )(0) (0)lim 0 x f xf f x + + + + = 2 0 (4)0 lim x x x x + + = =4= = 在處可導(dǎo)在處可導(dǎo)( )0(0)(0)f xxff + = =Q=Q 即即 1 84, 2 kk= = =

20、= 當(dāng)時(shí)，在處可導(dǎo)當(dāng)時(shí)，在處可導(dǎo) 1 ( )0. 2 kf xx= = = 例例6 ),(, )( lim ,)()()( 0 1 0 xAA x xf dtxtfxxf x = = = = 為常數(shù)，求其中 且連續(xù)，設(shè) 為常數(shù)，求其中 且連續(xù)，設(shè) 解解)(lim)0( 0 xff x = =x x xf x = )( lim 0 00 = = = = A 00)0()0( 1 0 1 0 = dtdtf .0)(處的連續(xù)性在并討論處的連續(xù)性在并討論= = xx = = 1 0 )()(dtxtfx xtu = = )0()( 0 x x du uf x ).0( )( 0 = x x duuf

21、 x ).()1(x 求求 = = ) )( ()(0 0 x duuf xx x 時(shí)，當(dāng)時(shí)，當(dāng) . )()( 2 0 x duufxxf x = = 0 )0()( lim)0( 0 0 = = = = x x x x 時(shí)，當(dāng)時(shí)，當(dāng) x x duuf x x 0 )( lim 0 0 = = 2 0 0 )( lim x duuf x x = = x xf x2 )( lim 0 = = ) 0 0 ( 2 A = = = = = = . 0, )()( 0, 2 )( 2 0 x x duufxxf x A x x .0)()2(處的連續(xù)性在討論處的連續(xù)性在討論= = xx )0()(li

22、m ? 0 = x x . )()( lim)(lim 2 0 00 x duufxxf x x xx = = Q )( )( lim 2 0 0 x duuf x xf x x = 2 0 00 )( lim )( lim x duuf x xf x xx = 22 AA A=)0( = = .0)(處的連續(xù)在處的連續(xù)在= = xx 例例7 求下列極限：求下列極限： . dtan dsin lim)1( sin 0 tan 0 0 x x x tt tt 解解= =原式原式 ) 0 0 ( xx xx xcos)tan(sin sec)sin(tan lim 2 0 )tan(sin )si

23、n(tan cos 1 lim 3 0 x x x x = )tan(sin )sin(tan lim 0 x x x = = ) 0 0 ( )tan(sin )sin(tan lim 0 x x x = = xxxtan)sin(tan0時(shí)，當(dāng)時(shí)，當(dāng)Q xxsin)tan(sin 1 sin tan lim )tan(sin )sin(tan lim 00 = x x x x xx 11 = = =原式從而原式從而 (2)( () ) x x x x 1 e1 1 lim 0 + + )( 解解=原式 =原式 )e1( e1 lim 2 0 x x x x xx + + ) 0 0 ( .

24、1e 0 u u u 時(shí)，當(dāng) 時(shí)，當(dāng) 2 2 0 e1 lim x xx x x + = + = x x x x 2 e21 lim 0 + = + = ) 0 0 ( 2 e2 lim 0 x x + = + =. 2 3 = = 注注 下列做法是錯(cuò)誤的：下列做法是錯(cuò)誤的： 2 2 0 e1 lim x xx x x + 2 2 0 lim x xxx x + = + = 1= =原式 =原式 (3) .d 1 1 )( )(2)2()2( lim 2 0 3 2 0 t t xf r xfrxfrxf I x r + + = = + + + = = 其中 ， 其中 ， 解解 r rxfrx

25、f I r2 )2()2(2)2( lim 0 + + + + = = ) 0 0 ( r rxfrxf r )2()2( lim 0 + + = = 1 )2(2)2(2 lim 0 rxfrxf r + + + = = ).(4x f = = x x xf2 )(1 1 )( 32 + = + = t t xf x d 1 1 )( 2 0 3 + + = =Q 6 1 2 x x + + = 26 6 6 )1( 51 2) 1 (2)( x x x x xf + = + = + = + = )(4xfI = = 26 6 )1( )51(8 x x + = + = (4)設(shè)，求設(shè)，求

26、21 01lim (123). n n xxxnx +L+L 解解 21 123 n xxnx +L+L 23 ( )()()() n xxxx=+L=+L 23 () n xxxx =+L=+L 23 (1)1 n xxxx = =+L+L 1 1 (1) 1 n x x + + = = 1 2 (1)(1)(1) ( 1) (1) nn nxxx x + + + = + = 21 123 n xxnx +L+L 1 2 (1)(1)(1) ( 1) (1) nn nxxx x + + + = + = 1 2 (1)1 1 (1) nn nxx x x + + + =+ + =+ 當(dāng)時(shí)， 可

27、以證明： 當(dāng)時(shí)， 可以證明： 01 lim (1)0 n n x nx + += = 21 lim (123) n n xxnx +L+L 1 2 (1)1 lim 1 (1) nn n nxx x x + + + =+ + =+ 2 1 . (1)x = = 例例8填空題填空題 ._ 0. 1 tan = = n xeex nxx 則 是同階無(wú)窮小，與時(shí)，設(shè) 則 是同階無(wú)窮小，與時(shí)，設(shè) 解解 n xx xx ee c = = tan 0 lim n xx x xx e e 1 lim tan 0 = = n xx x x xx e e 1 limlim tan 00 = = n xx xx

28、= = tan lim 0 )0( c 3 1 2 0 1 2 0 tan lim 11sec lim = = = n x n xx x nnx x ._ )1(sin sin)1ln()cos1(0. 2 2 2 = + = + n exx xxxxx xn n 的無(wú)窮小，則正整數(shù) 高階是比高階無(wú)窮小，而 是時(shí)，設(shè)當(dāng) 的無(wú)窮小，則正整數(shù) 高階是比高階無(wú)窮小，而 是時(shí)，設(shè)當(dāng) 解解時(shí)，當(dāng)時(shí)，當(dāng)0 x , 22 )1ln()cos1( 4 2 2 2 x x x xx=+=+ ,sin 1+ +nn xxx 2 1 2 xe x 依題 設(shè)， 依題 設(shè)， 2 lim sin )1ln()cos1(

29、lim0 3 0 2 0 n x n x x xx xx = + = + = 2 3 n得得 . 2= = n ._lim ,1 2 3 . 3 1 0 1 = += = += + + n n n n n n n na dxxxa則極限設(shè)則極限設(shè) 解解 )1(1 2 3 1 0 nn n n n xdx n a+=+= + + 1 0 2 3 )1( 1 + + +=+= n n n x n 1) 1 (1 12 3 + + +=+= n n n n = = n n nalim 1 ) 1 1( 1 1lim 2 3 + + + + n n n . 1)1( 2 3 1 +=+= e 1)1(

30、 2 3 1 + e 4._ )(2 lim, 0 )21ln()( lim 0 2 0 = = + + = = + + + x xf x xxxf xx 則已知?jiǎng)t已知 解解(方法方法1) 2 00 )(2 lim )(2 lim x xxfx x xf xx + = + = + + 22 0 2)21ln()21ln()( lim x xx x xxxf x + + + + + = = 2 0 2)21ln( lim x xx x + + = = 0 0 x x x2 2 21 2 lim 0 + = + = 2)2(= = = = 2 (方法方法2)，知由，知由0 )21ln()( lim

31、 2 0 = = + + + x xxxf x )()21ln()( 2 xoxxxf=+=+ x xxo xf )21ln()( )( 2 + = + = = + = + x xf x )(2 lim 0 故故 2 2 0 )21ln()(2 lim x xxox x + 2 0 )21ln(2 lim x xx x + + = = 2 2 21 2 2 lim 0 = + = + = x x x 錯(cuò)錯(cuò)解解 2 0 )21ln()( lim0 x xxxf x + + + = = 得得 由等價(jià)無(wú)窮小代換，由等價(jià)無(wú)窮小代換，)0(2)21ln(+xxx 2 0 2)( lim x xxxf x

32、 + + = = x xf x 2)( lim 0 + + = = 0. )(2 lim 0 = = + + x xf x 例例9 ：的某鄰域內(nèi)滿足關(guān)系式 的連續(xù)函數(shù)，它在是周期為已知 ：的某鄰域內(nèi)滿足關(guān)系式 的連續(xù)函數(shù)，它在是周期為已知 0 5)( = =x xf )(8)sin1(3)sin1(xoxxfxf+ += = + + .)6(, 6( )(1)( 0)( 處的切線方程在點(diǎn) 處可導(dǎo)，求曲線在且 高階的無(wú)窮小，時(shí)比是當(dāng)其中 處的切線方程在點(diǎn) 處可導(dǎo)，求曲線在且 高階的無(wú)窮小，時(shí)比是當(dāng)其中 f xfyxxf xxxo = = 解解的連續(xù)性，及由的連續(xù)性，及由)(xf )(8)sin1

33、 (3)sin1 (xoxxfxf+ += = + + )sin1(3)sin1(lim 0 xfxf x + + 得得 0)(8lim 0 = =+ += = xox x , 0) 1(3) 1(= = ff即即 . 0) 1(= =f sin )( sin 8 lim sin )sin1(3)sin1( lim 0 0 x x x xo x x x xfxf x x +=+= + + 又又 8= = x xfxf xsin )sin1(3)sin1( lim 0 + + 而而 t tftf t xt )1(3)1( lim 0 sin + + = = = = )1()1( 3 )1()1(

34、 lim 0t ftf t ftf t + + + = = )0) 1(= =fQ ) 1(3) 1(ff += =) 1( 4f = = ,8)1(4= = f.2)1(= = f ),()5(xfxf= =+ +由于由于 )1()6(ff= =得得 , 1= =x所以令所以令 0= = 1 )()1( = = = = x xff又又 11 )5()5( = = = + + + + = = xx xxf )6(1)6(ff = = = = 2)1()6(= = = = ff 故所求切線方程為：故所求切線方程為： ).6(2 = =xy 例例10 :, 1)(lim , 0)(), 0()(

35、且滿足 內(nèi)可導(dǎo)，在已知 = 且滿足 內(nèi)可導(dǎo)，在已知 = + + + xf xfxf x , )( )( lim 11 0 xh e xf hxxf h = = + + ).(xf求求 解解 h xf hxxf h 1 )( )( lim 0 + + Q )( )( ln 1 0 lim xf hxxf h h e + + = = hx xfhxxf x h )(ln)(ln lim 0 + + = = )(ln = =xfx )( )( ln 1 00 lim )( )( lim 1 xf hxxf h hh e xf hxxf h + + = + = + )(ln = = xfx e , 1

36、 )(ln x ee xfx = = 由已知條件得由已知條件得 )( )( ln 1 lim 0 xf hxxf hh + + 而而 = =dx x xf 2 1 )(ln 1 1 c x +=+= x cexf 1 )( =即=即 得由得由, 1)(lim= = + xf x 1= =c .)( 1 x exf = = 即即故故, 1 )(ln x xfx= = 2 1 )(ln x xf= = 例例11 ).0( )1ln( sin lim , 3 0 = + = + cc dt t t xax cba x b x 的值，使確定常數(shù)的值，使確定常數(shù) 解解 + + x b x dt t t

37、)1ln( lim 3 0 Q )sin( sin )1ln( lim 3 0 xax xax dt t t x b x + = + = 00 1 = c 0 )1ln( 0 3 = + = + b dt t t 利用定積分的保號(hào)性，可以斷定：利用定積分的保號(hào)性，可以斷定：.0= =b + + = = xx dt t t xax c 0 3 0)1ln( sin lim于是于是) 0 0 ( x x xa x)1ln( cos lim 3 0+ + = = )1ln( )cos( lim 3 0 x xxa x+ = + = ) 0 0 ( 3 0 )cos( lim x xxa x = = 2 0 cos lim x xa x = = 2 2 00 cos lim)cos(limx x xa xa xx = = 00 = = = = c , 01 = = a.1= =a 2 0 cos1 lim x x c x = = 從而從而. 2 1 2 sin lim 0 = x x x 例例12 ,)(li