2015考研高數(shù)基礎(chǔ)模塊精講講義

型-不定

考研高數(shù)基礎(chǔ)模塊精講講義-0 、、0 窮 法 Taylor分u(x)vx)evxlimu②ee~ (③lnln(1)~ (④xsinxtanx x ex2cosx0xln(1(ex21)(1解：I

x21 2例2.limtanxsin Q:Ilimxxx0 ·解：Ilimtanx1-cosx · 例3.lim1[2cosx)xx0 (2cosx)x解：I

xln2coslim

ln2cos l1cosxlim 1limcosx13x0 6例4.lim(2x)x

(1x)xlim2x

xln(1xlim 2

ln(1x 2 恒等lim(1)12 1解：Ilim{[1

sinx1)]

e2x1( ( 例2.lim(cos1 1limx2(cos

c1 x limcostet0te1tanxx01sin解：I

)tanxsinx1)Δ]x31 0,nP(x)lim xm

,n ,nx2xsin x 0即11 0·1 1 1 xo sin2sin2x sin2x解：Ix0x2sin2

(lim ·limsinxxsinx-· 2limsinx 2limcosx13x0 3x24xx24x方法一：I

4x x24x10414x14x14x14x

2(1410)12 x1lim(14t10t2)2t ∵(1x)a1~ax(x1∴(14t10t2)21~2t5t2~∴I0型：0e0ln即例1lim(

)tan解：Iex01limtanxlne2 1limtanxln 2x0

lim2x0 1lime2e0解：I2 limsin2xln x02 e0? 定?定積分定例1.lim(1x)(1x2(1x2n1x2n1

(|x|11In1 1例2.limcosxcosxcosx(x 2cosx2cosx2cos

2n2n

sin sinn2nsinxsinlim 2xsinx

nnsinn2例3.lim(11 n (2n1)(2n解：Ilim1n

2n

)2例

解： n2

∴In2例5.limn2ni1n3 1 解 i 3n

i1n

nnn

n(n1)(2n6·∴l(xiāng)im左1lim右·∴I3

3 型三左、右極限?分段(?若f(x)中含ab

(x例1.f(x)ex2解：f(20)

limfxf(20)limf(x)sin 12例 x0|x 12解:f(00)11f(00)1(1)f(0)f(00)∴l(xiāng)imf(x) 1 2 x?rctanx2

xlim[x] lim[x]xm

(m準(zhǔn)側(cè)I定理Case1.{a無(wú)上界liman 如

n

n{a}且 1 n nn無(wú)下界liman 注

n?{an且liman,則ann?{an且liman,則ann方法二：an1an例1.0a16,an1 an(6an),證：liman存n

an6an3an1an

an(6an)2an(3an an(6an)an1an{an}liman2 2 例2.a1 2,a2 2 2,a3 ?liman存在,2 2 解?an1a2設(shè)ak1

2 2ak1 2ak,即ak2akn,an1an{an}現(xiàn)證ana1 2設(shè)akak1 2akn有anliman?令liman2由a22211設(shè)ak1ak求證an2a11則1ak11akak21故{an下證{an}

0,證liman存在證a20a1ak證

12a

5設(shè)a1 5 52則ak

15{an}單調(diào)減少且有下界nliman1、連?f(x)在xa處連續(xù)若limf(x)f(a)得f(x)在xa注：f(x)在x0處連續(xù)f(a0)f(a0f?f(x)在[ab]上連續(xù)

f(x)在(ab)f(a)f(a0),f(b)f(b得f(x)在(ab)上連續(xù)，記f(xC[a間斷若limf(x

f(a),f(x)在xa第一類間斷點(diǎn)：f(a0),f(a0)f(a0)f(a0)(f(a)af(a0)f(a0)a為跳躍間x第二類間斷點(diǎn)：f(a0),f(a0)至少一個(gè)不存在例1.f(x) x1e1 解x0,x x0時(shí) 0e1x1~x1e1x

1x1

1limf(x) xx0f(10)f(10)f(10)f(1x1為跳躍間例2.f(x) sin解：xk(k)為間斷點(diǎn)limf(x)x0f(0)x為第二類同理：x[k且k0]為第二類間斷點(diǎn)二、f(x)C[a,b]性質(zhì)1.(最值)若f(xC[ab]則m.有界）若f(xC[ab]，則K0,f(x)MYmYm0ab3.(零點(diǎn))若f(xC[ab]且f(af(b0則（a,b,使f(4.(介法)若f(x)C[a,b],[m,n], [a,batleastonepo

f(介值：thevalue （位于m和n之間值f(x)至少可取到一次b]出現(xiàn)函數(shù)值之和用介值定理例1.f(xC[0,2].f(0)2f(1)3f(2)12證：C[0,2],使f(c)證：f(xC[0,2m6mf(0)2f(1)3f(2)m2C[0,2]使f(c例2.f(xC[0,2].2f(0)f(1)ff(xC[1,2f(x)在[1,2]上有mf(1)f(2)2c[1,2使f(cf(1)f(2)2

f(1)f(2)2ff(0)f②f(x)C[a,b].[a,b],介值定例3.f(xC[abp0q(pq)mpf(a)qf(b)(p2.mpf(a)qf(b)p[a,b],pb

f(x)dxf()(bb2.mf(x)Mm(ba)af(x)dxM(ba)m bf(x)dx ( bf(x)dxfbba[a,

baPartI

00

題型

例1.f(xC(ab)在(ab)內(nèi)可導(dǎo)f(af(b0,f(afab2證：ab)，使f()1.f(afab0x(aab 使f(x1)f(a)f(b)f(ab)f(b)ax b)使f(x 2.f(x1)f(x2x1x2(ab)使f()例2.f(xC[0,3]在(0,3)內(nèi)可導(dǎo)f(0)f(1)f(23,f(3證：0,3)使f()：1f(xC[0,2f(x)在[0,2]上有mMmf(0)f(1f(2)M3即m1C[0,2]使f(c2.f(c)f(3)c,3(0,3)使f(例3.f(x)三階可導(dǎo)，f(1)0,F(xx3f2.F(x)3x2f(x)x3fF(0)F(1) F(2)3.F(x)6xf(x)3x2f(x)3x2f(x)x3f

(x 12證：f(0)1,f()2

1,f(1)2

0左低右高(a非極值點(diǎn) f(a)0左高右低(a非極值點(diǎn))f()

f(a)

f(a)為極值f(a)型二含ζ方法一：還原法[lnf(x)]ff例1f(xC[0,1]在(0,1)內(nèi)可導(dǎo)，f(1)0證：0,1)使f(3f()分析xf(x)3f(x)f(x)3f [lnf(x)](lnx3)[lnx3f(x)]證：令(xx3f例2.f(xC[ab]在(ab)內(nèi)可導(dǎo),f(a分析：f(x2f(x)f(x)2f[lnf(x)](lne2x)[lne2xf(x)]證：令(xe2xff(a)f(b)而(x2e2xf(xe2xfe2x[f(x)2fe2

分析:f(x2f1

f(x)f

x[lnf(x)][ln(x1)2][ln(x1)2f(x)]證：令(xx1)2ff(0)f(1C(0,1)使f(c(c)記?fkf(x)ekxf(x)fk(lnf)(lnekx)?fkf(x)xkf(x)fk(lnf)(lnxk)fx方法二：分組?f()ff(x)f(x)f(x)f(x)f(x)f(x)[f(x)f(x)][f(x)f(x)]gg0g10(lng)(lnex)g(x)ex[f(x)ff(x)f(x)f(x)f(x)[f(x)f(x)][f(x)f(x)]gg(x)ex[f(x)f?f()f()1f(x)f(x)1f f

1)f[f(x)][f(x)1]gg(x)ex[f(x)例1.f(xC[0,1]，在(0,1)內(nèi)可導(dǎo).f(0)?C(0,1),f(c)證：?(xf(x(0 0

0,f

1,f(1) (2C1,1)(0,1)使(c)2f(c)?分析：f(x2[f(xx[f(x)x]2[f(x)x]h(x)e2x[f(x)x]h(0)h(c)0ζ、case1.結(jié)論中只有f()、f(找三 兩次?C(0,1)使f(c)1證：?令(x)

f(x)1x,(0)1,(1)C(0,1使(c)0

f(c)1f()f(c)f(0)1 1

1?C(0,1),f(c)2

ff(x)

f證：?(x) ?(0,c),f()f(c)f(0)

f(1)f(c)1

2(1 f

f

2C2(1C)Case2.含ζ、η兩者復(fù)雜度不 [e2xff f (1 )————?

————證令F(x)x2F(x)2xab)f(bf(ab2f(b)f(a)(ab)fb

(a,b)f(b)f(a)b

f例2f(xC[ab]在(ab)內(nèi)可導(dǎo)(a0).證：,(ab)使abf()2f證：令F(x1F(x1 (a,b),使f(b)f(a) f1 abf(b)f(a)2fb例3f(xC[ab]，在(ab)內(nèi)可導(dǎo)f(a)證：令(xexf

f(b)1.證,(a ebf(b)eaf(a) (a,b)，f(a)f(b)

b

e[f( f(左

b

(a,型四Lf(x)?f(b)f(a)f()(b?見三點(diǎn)兩次例1.limf(xelimf(xf(x1)limxC)xC xx解：f(xf(x1)

f (x1右2Clim

x

)2C

x2C xxCee2C C1例2.f(x0,f(x0,dyf(x0)xyf(x0xf(x0x0dyy,0大小解:yf (x0x0f

fx0 f(x0)f又f(x0)xdy

f3fx在ab]上可導(dǎo)f(a f(b 證. (a,b f(c).

f(x)(b0

M,fxab)內(nèi)至少有一個(gè)零a f(c)

f(a) f'(

1)(c a),1 (a,cf(b) f(c) f'(

2)(bc),2 (c,b f(a f (b f

(ca(bc例4.fx)在[a,b]上二階證ab)使f(01(a,c),2(c,

L:yf(f

) f(c)c

f(aaf(2)2

f(b)b

f(cc

(ac A、C、B f'(1) f'(2(1,2) (a,b f() f(n)(x f(n1) f(x)f(x0)f(x0)(xx0) 0(xx0)n (n

(xx0Rn(x)

f(n1)(n

(xx0

(c x x0 中 端 (c 無(wú) (c 端x 中 任一1f(x)C[1,1 f(1)0,f(0)0,f(1)證：111

1,1）， f()f(0f(1) f(0)

(10)2 f(1

(10)3, (1,0

f(1)

f(0) (10)2!

(10!

,

(0,10

(0)

f(0)

f(11 f(0) f(0) f(2) 2f(1)3

f(2)f(x)C[1,2f(x)在[1,2]上 m,2m f(1) f(2)2m3例

1,1),f()fx)在[0,1],f(0)C(0,1),f(c)1,f(c)

f(1)0,0x

f(x)f(0)f(c)f(1)(0c)2,(0, ff(1)f(c)

2f(1)c22f(2) 2①C

0

1c f() 8,②C

2,1

1f()

18, 3.f(xab]上fx在（a,b）PartⅡ單調(diào)性與極一、yf(x)(xD),x0If0,當(dāng)0 xx0f(x)f(x0

f(x)稱x0為極小點(diǎn)

fx0)為Iff(x)

00f(x0

xx

fx0稱0不 00f(a) 0

不存在二、求極值步驟f(x)

0駐點(diǎn)）①yx2,y2x0xyx3,y3x20xyxx0處0,xyx,xf(0)0,f(0)f(0)

xx0,

(x)

xx為極小xx,f(x) ② xx為極xx0② xx為極xx,f(x) 方Th2f(x0) fx00，x0為 0不討論 f(x

) xx

f(x) x 0,當(dāng) f(x x

x

f(x) 0,x (x x0f0 (x 0,x ( , f00 xx0為極小點(diǎn)1.fx連續(xù)

f(x)

x x解x

f(x)1x

2 f(0)xx

f(x)1xf(x)1x

2 2

(0)0,當(dāng)0 x時(shí)f(x)10xx0為2.f(1)0,

f(x)1 f(0f(x)x

3解

f(x

2x

sin300f(x)

x

sin3f(x)f (x)f

0,x0,x

(1,1(1,1x1為極大例3.f(x)連續(xù) f(0)0.x

f(x)f(x)x證fa)1

f(b

M(baC(ab),fc最大，fc)2f(c)f(a)f(1)(c a),1 (a,c2f(b)f(c)f

)(bc),2 (c,b f(a)M(caf(b)M(bc

f(x)f(x)2x

f(0)f(0)0f(0)2lim[f(x)f(0)f(x)f(0) f(0)ff(0)2x0為ylnxyln(1yln(1x)x0f(x)c( , 型例1.f(a)g(a f(a)g(af(x)g(x)(xa證：f(x)g(x)(xa證：（x） f(x)g(x(a)0,(a) 0,(x)0(xa f' (a) (x)0(xa(x)0(xa (a)(x)0(xa)(x) 0(xaCase1關(guān)于ab不等2eab證：abb證：abbablnaalnb令（x xlnaalnx,(a)(x)lnax(a)

(xa(x)0(x a)(x) 0(xaba,(b)：例3.0ab, lnblna：b 2. x的不等式例4.證： ex1x

2a2b證 x）ex1x,(0)(x)(x)

exex

1,(0)0(0)(x)

(x) (x)

0,x0,x x0為x)為最小(0)0,(x)1x例5.證1xln(1x1x證：(x)1xln(1x(0)(x)ln(x

1x2) (1 ) ln(x

1x2),(0) (1 ) (0)0(x)0,x (x) (x)0,x x0為（x）的最小中值定例6 x0,證：1

ln(1x)方法一： f(x)xln(1x),f(0)f(x)1 1

0(x0 f(0) f(x)0(x0f (x)0(x0f令g(x)

1x) 1

,g(0)g(x) 1

(1x)

0(x0 g(0) g(x)0(x0g(x)0(x0令(t) ln(1t)(t0ln(1x)(x (0 () 1

(0 x 1 1 1

ln(1x)例7.0ab,證：lnbln 2 1(a

bx

a2b11

a2bf(x)Mf(x)M型三方程根或函數(shù)零點(diǎn)個(gè)數(shù)22x證：(xlnxe

(x(x)110x (e)1xe為最大點(diǎn)M(e) lim(x)，lim(x) 方程只有2例2.討論xexa幾根(a (x2.(x)exxex(1x)ex0xx M(1)1e?M0,即a1e?M0,即a1時(shí)，方程只有一個(gè)解xe?Mlim(x)lim(x)lim(xex limxxa方程2PartI理論?ax0x1xn[a,b][x0,x1][xn1,xnxixi?i[xi1,xinif(in ?max{x n

0(t)t[T1T2求?T1t0t1tn[T1,T2][t0,t1][tn1,tntiti?i[ti1,tin?max{t n0二、定積分定nyf(x)在[ab]上有界，若limf(i)xi存在，解f(x)在[ab]上可積，極限值稱為f(x)0b[a,b]上的定積分，記afb f(x)dxlimf()ix 0注

f

[a,b]劃與

?0n bax1xnnba

2?例f(x

xxR/nf(x)在[ab]上有界limf()ix0nn0Case2.iR/nn0nn

f(x)在[ab]上不可f(x)在[ab]上連?f(x)在[ab]除有限個(gè)第一類間斷點(diǎn)非連續(xù)，則f(x)在[ab]上一定?設(shè)f(x)在[0,1]

1][1,2][[n1,n n

1n

f()x

f(0

nn 1f( 記 f()f 若干次和或積??定?定積分定義分母比分子高

n

n

1nnlim1 n0

nni11n

1例2.lim1p2pnpp

np1p

x ( 例

lim1nni1in

1(1in 0x定積分 λ

i

f(ξi)ΔΔax0x1xni[xi1,xinni

f(ξi)Δxλ1i

{xi

f注設(shè)f(x)在[0,1]上可[0,1][0,1][1,2][n1,n n

1(0nn nf()x

f(in n01

nn 0fimnf(imnf()f0 例.lim

n2

n2lim1 nni11(i 010arctanx04二、理 f(xc[ab](xxf(t)dt.則(xa

fx xxxx

f fQ2.

f(x,

f

f①dx d 2(dfdx1(例1.f(x)連續(xù)F(xxtf(xt)dt求F0010 x0f(u)du0uf2.F(x)xf(u)duxf(x)xf0F(x)fb (N.L.)af(x)dxF(b)Fb三、性（一）、一般性 a(fg)dxafdxa akfdxka fdxafdxcba1dxbb①f(x)0af(x)dx

(a②fg

fdx

(a a★③|afdx|a

|f|dx

|f (abf(x)c[a,b],則[a,b]使 fdxf()(bb證:fC[abmmf(x)bm(ba)af(x)dxM(bbm bf(x)dxMbaa[abf() bf(x)dxbaaxex2de1fex2d(x2)1x2(t

(xsinf(x)dxf x2sint,t

,2xcost,txtant,t,2udvuvvdbbudvuv|b （二）、特殊性

f(x)C[a,a],af(x)dx0[f(x)f 證：左af(x)dx0f xt f(x)dxf(t)(dt)f(t)dt

f 左0f(xf 4 4( 01sin24sec202tanx4

1sin ?f(xC[0,1]2f(sinx)dx2

f(cos xt2

f2 0f(cost)dt

f(cos例2.2 sin dxI0sinxxt

cos 2sinxcos22II 特例：20

n

20

nxdxIIn1n n I0I1 I13 34 例

22

xdx

2sin4xdx

2X4231 o

22c

xdx

97531 sin220

sin211

sin211

2 )sin2x01 1x2sin2xdxI

1

例5?設(shè)f(x)g(xC[aa],f(xf(x)Ag(x)

f(x)g(x)dxA

g(x)dx,計(jì)數(shù)?2arctan

cos3 2a證：?左0[f(x)g(x)faa0[f(x)faaA0a2arctanexcos322(arctanexarctanex)cos30(arctan

x e arctane)1e2x1e2xarctanexarctanex取x0,A2I 312os3xdx 2 ?f(sinx)dx

f(sinx)dx,

sinxdx 證：左0f(sinx)dxf(sin2 xtf(sin2

f(sin22f(sin0 ncosxdx 2In n2?xf(sinx)dxf(sin2 xt證：左I (t)f(sin0(x)f(sin If(sin2

f(sinx)dx0xf(sinI

xsinx01cos2 sin 2o1cos2 d(cos 2o1cos

cos0 7xt

sin 0

t

2t2 2

20

22

1 設(shè)fx)可

T為周期，則①

a f(x)dx

fx)dx(平如4sin2

2sin2xdxs4in2xdx2 ② f(x)dxn0f(例1.lim( nn n

1 nni11n1 ln(1x)1ln01 例2 lim1nn 0x

1

0sintcos cos dx20sinxcosxt例3limn(n1)(n n1n2

nnlim nn n 1 nn nenn

n111ln(1x)dxxln(1x)11x1 1ln21(11 1ln21ln(1x)0ln41lneIe型二例1.fx)連續(xù)，limfx)

F(x)xtf(x2t2)dt

F(x x x1.F xtf(x2t201xf(x2t2)d(x2t2 x2t2 2x2f(u) x2f(u 2.limF(xx x12x

x2f(u0xx

1f(x2)224x14x

f(x2 x x例2.f(x)連續(xù) f(0)0 0tf(xt)xx.0tf(xt)xxt 0(xu)f(u)(du0x(xt)f(t)

x 0

f(t) x f(t)dt0tf(t) x.I x f(t)dt0tf(t)x

xxx

xf(t)

f(t)xxx0

f(t)dtxf(xx f(t)xx xx0

f(t)dt

f(x.x

0f(t)x

x

f(x)f(0I f(0 f(0)f(0 型三計(jì)算例1.2x 01(x1)1(x1)11(x1)20[(2

1)1]2 d(x1111

(x1) (12xx2 21(1x2 0xsin

2(10

sin2t)cos2 2(10

2t2(2

I42(2

3

22ln2

exex解： txln(1exI2

3t 2 1t3(1 )1t32( )3例3fx)

xet2dt1

0f(x)dx1 1解0

f(x)dxxf(x

xf'(x)1xex201例3.f(x) xet2dt11

0f(x)dx解0

f(x)dxxf(x 1xf'(x0001xex2011ex2d(x2211exd(x21ex11(e1 例4.f(x) xsintdt 0

f(x)dx解: f(x)dxx

f(x0

0xf(x)f() xsin