2015版微積分疑難分析講座第一

微積分疑難分析第一關(guān)于極限與連續(xù)的若干 實實題本概念義本理論 能函函數(shù)是微積分一、怎樣理解數(shù)列極限的“一、怎樣理解數(shù)列極限的“,N”與函數(shù)極限的描要多小有多小，則這個常數(shù)就是變量的極描數(shù)列的極限當n無限增大時數(shù)列{xn}無限趨近于alimxlimxna0,NN,n:nN,|xna|的任意性描述了極限的無限過程的給定性描述一個確定的有限過程-N定義用無限多個有限過程來描述無限變化過程lim 0,NN,n:nN |xna|0,NN,n:nN,axna任給一個a的鄰域(a,a)，總存在正整數(shù)N，使得從xN1項起，數(shù)列所有的點全部落入(a,a中.nx:1,1,1,1,1,1,,(1)nnlimxn0,NN,n:nN |xna|0,NN,n:nN,axna使得從xN1項起，數(shù)列對應的點全部落入(a,a中.limx2k1lim k k{xn的子列 }都有l(wèi)im 函數(shù)極限的概念函數(shù)極限的概念fx)在Ux)有定義limfx) x x:0|xx0| f(x)A|0,0,xU°(y

,),A f(x)A0yf(AAAo

x0

x0 fx)在Ux)有定義mfx x0,0,x:0|xx0|,|f(x)f(x00)f(x00)fxA無窮xx0fx)Axx0 xxUx),都有l(wèi)imfx

n limlimf(x)Ax,xxn0n(0),都有l(wèi)imfn)偶函有界狄利偶函有界D(x)

x為有理數(shù)周期函 周期函x0(,

limD(x)x取有理數(shù)xnx0:取無理數(shù)xnx0

limD(xn)limD(xn)xnlimDxnlimDxnlimDx) x9xlim|f(x)|0limf(x)xxlim|f(x)|0limf(x)xx xlim|sgnx|1,但limsgnx不存在x xxxlimf(x)Alim|f(x)||Ax0,0,x:0|xx0||f(x)||A||f(x)A|lim|f(x)||A limf(x)Ax x函數(shù)極限定義的擴充函數(shù)極限定義的擴充fx)在Ux)有定義mfx x0,0,x:0|xx0|,|f(x)xx,xx,xx x,x,x

f(x)limf(x)M0,X0,x:xX,f(x)M局部有界ffxlimflimgBAx保序性 四則運算法則xxlimf(x),limg(x)lim(fg)limflimx x x x[例1]設fx)在x0的某個鄰域連續(xù)且f(0)0,

f(

2,則在x0處f(x) x01cos(C)取得極大值

(B)可導f(0)(D)取得極小[分析]

f(

2U

f(

0(局部保號性x01cos 1cos1cosx0f(x)0f(0),xUfx0取得極小值四則運算法則要注意 f(x)g四則運算法則要注意lim[g(x)(x)]0,則 f(x) x(A)且等于零;(C)必不lim[g(x)(x)]x

x(B)不一定等于零;D)不一定.limg(x)x

lim(x)0limg(x x

limx 準則知 f(limf,limglim(fg)limflimx x x x x[例2]x:(x) f(x)g(x，lim[g(x)(x)]0,則 f(x) x(A)(C)必不

x(B)不一定等于零;D)不一定.[分析

f(x)g(x)lim[g(x)(x)] limf(x)1 f(x)g(x)lim[gx(x limfxlimx不

0, 不xx(1ex

x1ex

x1exlim1 1lim 0x 1ex

x x1elim1 0

1,

x

1ex

xx(1exlimxarctan limarctan1不

limx

arctan x

利用左利用夾[例3]設limsin6xxf(x)0,則lim6f(x) (A) (B) (C) (D)0limsin6xxf(x)sin6xlimxf(

lim6xlimf(x)lim6f( xxlimf(x),limg(x)lim(fg)limflimxxxxsin6xxf( 6xxf( 6f(0 x0 x3

x0 x3

x0 x2[例3]設limsin6xxf(x)0,則lim6f(x)

(C) (D)恒等變[解1]lim6fx)lim6xxf恒等變 lim6xsin6xsin6xxf( lim6xsin6xlimsin6xxf(xx x

x

x1(6lim66cos6x02lim

3

[例3]設limsin6xxf(x)0,則lim6f(x)

(C)

(D)[解

0limsin6xxf(恒等變 恒等變limsin6x6x6xxf( limsin6x6xlim6xxf(x0 x0 lim6f(x)lim6xsin6

[例3]設limsin6xxf(x)0,則lim6f(x)

(C) (D)[解3]sin6xxf(x)0 f(x)x2(x)sin6lim6f(x)

x lim(x)lim6xsin6x0lim6xsin6xx0 x0 x0 函函數(shù)、極限與無窮小的關(guān)系或無窮小的概[例3]設limsin6xxf(x)0,則lim6f(x)

(C) (D)[解4]sin6xxfx)ox3

o(x3 sin6f(x) 高階無小的定6o(x3)高階無小的定

6f(x)x2

x26xsin6xo(x3 6xsin6 0x0 x3 x0 x3[例3]設limsin6xxf(x)0,則lim6f(x) [解

(C) (D)特殊函數(shù)檢驗令sin6xxfx)0,f(x特殊函數(shù)檢驗xlim6 f(x)lim

6sin6xx x x xlim6xsin6x [例4]x

cos3cos變cos3cos變量[解66cos

u,則sin2x1u12，當x0,uu3u原式

limu2

u 1

u

u12u1(u1)(u11u10u u1

u10u lim[lim[xx2ln(11xx令xt=2fgegln12fgegln5]x

3

等價無窮小等價無窮小代原式=limx

xln2cosex1~ex1~x(xxxln

2cos

cosx1ln =lim

=lim x x

x2=lim

cosx3

=1lim

1x2

x x 3x x ln(1ln(1x)~x(x1cosx~12(x[例x

sinxsin(sinx)x3

需要注意limxsinxlim1cosxlim x

x

3

x

3 sinxsinx~x(xsin(sinx)~sinx(x必必須證明limsinxsin(sinx)1,否則xsinsinxsin(sinx)xsinx(xlimsinxsin(sinx)limxsinxlim1cosxx

x

x

3x2 利用兩個重要極限求limsin利用兩個重要極限求 limlimsin[f(x)]1(x:f(x)f( 1lim1

lim1xx 11x x 11

lim[1f(x)]f(x)

(x:f(x)lim[1lim[1f(x)]g(x)xlimf(xg(x(1(x:f(x) g(x)xlimf(x)g(x lim[1f(x)1]g(x)=elim[f(x)1]g(xx (x:f(x) g(x)[例7]求

1e2x…enx1x0

exe2x…enxnx0lim

exe2x…enxex0 1e

x

(ex1)(e2x1)…(enx1xlim

ex

e2x

…

enx1en

x

x

x 1(12…n

1n(n1

ne

e

limfx)A(常數(shù)gxlimfx)A(常數(shù)gx0fxg(f(x)A(g(

g(x)f(x)Ag(x)g(x)limlimf(x)B0,f(x)0g(x)g(x.[例8]試確定abc,使x

ax2bxcx2x22x

lim(ax2bxcxx22xaxx22x

)0x22xco(x2)x22xcx22xaxx22x x x2x22xb2有理化b 有理化 a316x22a316x22xalim 0

limn

nnn化數(shù)列極為函數(shù)極nnn化數(shù)列極為函數(shù)極

1 tat

b c

at btctlim

1lim x

t0 atbtct lim1

btct3t

limt0 3t0 1

at1bt1ct1 3t0

lnalnbln3 e 3aax1xlna(xynxnzn,limynlimznAlimxn [例10]lim

…

n

n

n2n

n2nn

n2nn

n2n

…

n2n

xn

n2n

n

…

n2n1n(n1) 1n(n2n2 2n2n

… 1n

n2 n2

n2n [11]a0i1,2,…,m),limnanan…an a

nnnnanan…12mlimnm

nann a,b,c0,limnanbncnmax n1xnx2n1xnx22n求fx)的顯式表達式

(xn1xnn1xn x22

0x

1xy2

y2

2y

,2

x yx 13]lin

1n2nn

[證]1

n11n2n

定 n定 n14]limnnn

1nn

nnn1…n nn12 2(n)? a1a2a1,a2,,ana1a2 定14]limnn定n[證 1nn

n nn n n1…n

nnn 1 1(n a1,a2,,an

a a

a1a2nn[例14]證明n

n用二項式定lim(nn1)n(n用二項式定lim(nn1)nn[證nn1

1)n(n1)nnnn

1)2…

nnn(n1)n2

1)2

1)2

(nnn2n0 1 0n2n單調(diào)增加有上界的數(shù)列必有極限存{xn}:xnxn1,M,xnMlimxnn單調(diào)減少有下界的數(shù)列必有極限存{xn}:xnxn1,m,xnmlimxnn[例15]設x12,xn

2x3n 1,3n 遞歸數(shù)列證明數(shù)列xn的極限,并遞歸數(shù)列

令lim

aa

2a31a

單調(diào)有單調(diào)有界2x3

1 1n]n

2 nn

3 3 xnx xnxnx21n3xnxnx2

1有下界1,xnababc3abc(a,b,c3[例15]設x12,xn

2x3n 1,3n 單調(diào)有界證明xn的極限,并單調(diào)有界已證已證xn}有下界1,即xnxn1

2x3 1 n nn3n

3xn1xnxn}單調(diào)減少令

故數(shù)列極限2a3limxna,則由遞推公式得an 2x3 3x31

a 1 1. 3 3 16]利用不等式

ln(11)1nN )證明：n 數(shù)列x: 111lnn(n )收斂

xn1

ln(n1)lnnn1 ln(11 0 }單調(diào)遞減n ln(11)ln(11)…ln(11)ln ln2ln3…lnn1lnn

34…n1ln

歐拉常ln(n1lnn0xn有下歐拉常 1 lim1 … lnnC …n Euler1707~1783數(shù)Euler1707~1783 數(shù)學界的莎士數(shù)學界的莎士用單調(diào)有界準則證明數(shù)列極限存在的基本 xnxn10( 1(1)等abab(ab0), sinxxx0)等;或兔群繁殖問題兔群繁殖問題 Fn1Fn

1,證明

1

n n兔對成長兔對成年兔對Fn1121324355687FnFn1Fn(n2,n1234567891123581232538512ananFn Fn(n2,3,n123456789122

單增,

單減,1

n n 1 ananFn Fn1an2. Fn

1 1 n n

已已知a1, 111na(n2,liman存在，并求極限值a1,1n1a1,1n1a1(n2,證明極liman存在，并求極限用數(shù)學歸納法證有界a11,a221a1,a21an則由 11知1 aan

由數(shù)學歸納法知an既有上界又有下界a1a3,a2a4

Fn1, Fn1,nn1na設a2n1a2n1, 1

2n1

1

a2n1

a2n2 1

1

a2n

2n,故由單調(diào)有界定理 lima2n1 ,lim n Fn1,nFn1na1令lima2n1A,lima2n an1

兩邊取極

A11 B11 55AB1 lim lim 155 n黃金分黃金分割lim

Fn

15 5

1910--聰明在于學習聰明在于學習天才在于積 fx)x0連續(xù)fx)在Ux0)有定義，且limfxfx0x0,0,x:|xx0 |f(x)f(x0)|f(x00)f(x00)f(x0limf(x0x)f(x0 (xx0limy yf(x0x)f(x0x0, x為無理數(shù)Dx) 0, x為無理數(shù) limDx,(處處不連續(xù)xx0為第二類間斷點只在一點連續(xù)，而在其他點都不連續(xù)的函數(shù)0,x為無理fxxDx0,x為無理

limf(x)0fx處處不連續(xù)的函數(shù)復合成處處連續(xù)的函數(shù)D[D(x)]18]求fx

ln|x1

的間斷點,并判斷其類型[

ln|x1|沒有定義的x11是間斷點分母為零的點|x1|1,即x20,x32. x1ln|x1 0: x0ln|x1 x0ln(1

x0xx20x32為無窮間 2:x32為無窮間 x2ln|x119]試確定ab的值,使fx)

ex(xa)(x有無窮間斷點x0與可去間斷點xe 若x0是無窮間斷x0(xa)(x

,lim(xa)(x1)0

0a0,bx

ex

1

e 若x1xlim(exb)0b

x(x1)aa0,b七、怎樣理解與運用閉區(qū)間上連續(xù)函數(shù)的性質(zhì)七、怎樣理解與運用閉區(qū)間上連續(xù)函數(shù)的性質(zhì)有界性定理：fxC[abfx)[ab]上有界最值定fxC[abf(x)必在[a,b]上取得最大值與最小值介值定理：fxC[ab],f(af(bf(a),f推論：fxC[a,b],其最大、小值分別為M、(mM(a,b)f(零點定理(根的存在定理fxC[a,b],f(af(b0(a,b),f(零點定理(根的存在定理零點定理(根的存在定理的推廣fxC(a limfx),limfx)limfxlimfxx x x x(a,b), 需證x1,x2(a, f(x1)f(x2)

不妨limfxAxa由局部保號性知,x1(a,a1), f(x1)0,limf(x)B0, f(x2)[x1x2(a fxC[x1x2fx1fx2零點定理(根的存在定理零點定理(根的存在定理fxC(a limfxlimfx)limfxlimfx xa (a,b), [證

不妨limfx)A0,limfx)Bx xf( ax作輔助函數(shù)F(x)

xaxblimF(x)limf(x)AF(a)xa limF(x)limf(x)BF(b)x xF(x)C[a,b],F(a)F(b)ffC(a,b)且limf( limf(x)x xf在(ab)f在()fC(, limf(limf(x)21]2xtanx在(0,)內(nèi)至少有一實根.2[分析 令f(x)2xtan

f(0)[證 令f(x)2tanx f(x)C(0, f(x) 2tanx210, x x f(x) 2tanx x(fx0,由推廣的零點定理 (0, ) )使f()0, [例22]設f(x)C[a,b]，acd (p,q0,常數(shù)證明：(a,b) pf(c)qf(d)(pq)f[證

f(x)C[a,b]fmaxM,fminmf(c)M mf(d)Mpmpf(c)pM qmqf(d)介值定(pq)mpf(c)qf(d)(pq)Mmpf(c)qf(d)介值定(p(a,b)使f() pf(c)qf(d)(p[例22]設f(x)C[a,b]，acd (p,q0,常數(shù)證明：(a,b) pf(c)qf(d)(pq)f[分析]需證方pf(c)qf(dpq)fx0有根2Fxpf(cqf(dpqfF(c)pf(c)qf(d)(pq)f(c)q[f(d) f(c)]F(d)pf(c)qf(d)(pq)f(d)p[f(c)f(d零點定理F(c)F(d)pq[f(c)f(d)]2零點定理若f(cf(d)，則F(cF(d0c或若f(cf(d則F(cF(d0(cdF([例23]證明

x2x1在(0,內(nèi)必有[分析

惟一實根xn,并求limxn(n2,3,nx2x1在(0,1內(nèi)必有惟一實根2x3 x2x1在(0,1內(nèi)必有惟一實根3xnxn xn

xx

x1在(0,1內(nèi)必有惟一實根xn1x1在(0,1內(nèi)必有惟一實根數(shù)列x1,x2,,xn1,xn xn[例23]證明

x2x1在(0,內(nèi)必有n惟一實根xn,并求limxn(n2,3,n

令

(x)

xn

nx2xfn(x)在[0,1]連續(xù) fn(0)1 fn(1)n1fn(x)在(0,1)內(nèi)有零點，即方程有根.x1x2,x1,x2(0,1),fn(x2)fn(x1(xn xn)(xn xn1)(x2x2)( x