2014競賽輔導老師的講義

極 1

20149elimn1ne 求極限

exe2xenx

ex，其中n是給定的正整數(shù)解法1

exe2xenx)

lim(1exe2xenxn) )e Alimexe2xenxn exe2xenxe elimex2e2x

e12nn1 exe2xenx

e因

)xeAen解法2 lim

exe2xenx )

elimln(exe2xenx)lnx0 ex2e2x 12 nex0

e2

enx)xe

eee x

1

lnex(1

)x

x2ln(1

)

x

lim

lim (ln(1t)t 1/(1t 1

lime2t

lime2(1t ee求

1

xsin

sinxxsinxxx1coslimx lim1

1sin limsin limcos lim方法

limex1cosx

12

x03

x03

esinx sinx sin 1 lim limcosx limsinx1cos limx

1

1

3

x03 lim x0

ex01cosx

elimexex2x2(11)xx xln(11

()x(111(1 111(1 11()(1)xx

xex2x2 x3e2x3 x2ee2x e[1

1(1)2(1 x2(11)xex2exee

lim[ x k求極限lim enk1nk e e

e

e 1解：n1 n limn0edxekk

k1n kkk

nkk e

nene lim elim

，故

enk1n nn1k1 nk1nk4.f(xx0f(0)0x00x之間的一點f(xf(0)2f(，求limx0解：對f(x)f(0)2f()兩邊取極限得f(0) f(x)f 2f( f()f(0)2f(0)limf(0)

lim2

x0得 x0 f(x)是(0,)上遞減的連續(xù)函數(shù)，且f(x)0，證明：數(shù)列an收斂 anf(k) f(x)dxk n n證明：an1anf(n1) f(x)dx [f(n1)f(x n nyn1kykanf(n)k

[f(k)f(x)]dx故an單調(diào)遞減有下界，故an收斂 23 n-

xn 收斂準則(不可替代性)，兼談放縮技巧證明：若存在常數(shù)cnNx2則xn收斂

x3

xnxn1anx2x1x3x2xnxn1，則an單調(diào)遞增有上界an收斂任給0,存在N，nN時，pNanpanxnpxnxn1xnxnpxnp1anpanxn收斂fR滿足：在f(

f(xfy)kxy(0k1)f(x證明：任x0xnf(xn1(n1,2,)，則得到數(shù)列xn，由xn1xnf(xn)f(xn1)kxnxn1(nxn1xnknx1x0(n1,2,（*）xnpxnxnpxnp1xn1xn1k1(knp1kn)x1x0 x11k1

k1

x1x00Cauchyxn收斂，記limn

f(xn)xnknx1 nf(xnxnf(xnf( an0nN且an發(fā)散記Snai證明：S發(fā)散

n1an0,an發(fā)散，故Sn(n令uan, nun0，N,當nN時，對任意的pN,un1un2un

aN1

aNN N N

SN SNaN1aNp1a1SN對任一正整數(shù)N,因故un發(fā)散

SN,故存在正整數(shù)p1a1SN

12lim(1111lnnc（歐拉常數(shù) 比較與鑒別：上一題與下面題 S設an0(n1),Snak.證明：級數(shù) 2收斂SkanSn2

n12 S2SnSn111 故有

12 n1

n2S Sn2lim12 n 因此正項級數(shù)an收斂n1Sn2012預賽第七題 設an與bn lim( （1） 1)lim( ，則

an（2）

0bn發(fā)散，則an

證明：（1）設lim( 1)20，則存在正整數(shù)N，當nN時，

1，anan1 ， 1(anan1

n n

m m1(anan1)1(aNam1)1n

n

故an的部分和有上界，所以an （2）若lim( 1)0，則存在正整數(shù)N，當nN時，n anbn

bn1

bb

bn

bnbN1

NaNN

bn

bn

n所以，由bn發(fā)散，得到ann 導數(shù)的定義1.f(0)0,f(0)xf(1f(2f(n，求limx n n n n1f(0)2.f(xx1f(1)0,ysin2xcosx,當x0時，y

(1)2，求

f(sin2xcos.x2xtan原式

f(y)f(1)limcosxcos2

1y

y

x2xtan 3.f(x)在(,)有定義，對任x,y(,f(xy)f(x)f(y)1f(x)f(f(0)1f(x的表達式f(xy)f(x)y

f(y)

1f21f(x)f(y)y0ff(x1ff(0)0f(xtanf(x在(,xyf(xy)f(x)f(y)f(xx0f(0)1f(x的表達式f(xyf(xfy2xyx0y0f(0)f(0)limf(y)f(0)limf(y) y f(x)limf(xy)f(x)limf(y)2xy1 f(xx2xf(0)0C0f(xx2f(x連續(xù)，g(x1f(xt)dt，且limf(x)A，Ag(x0x0處的連續(xù)性f

f解：由 A和函數(shù)f(x)連續(xù)知，f(0)limf(x)limx

g(x)0f(xt)dtg(0)0f(0)dt 因此，當x0時，g(x) f(u)du，x

f(0)0limg(x)

xf0

limf(x)f(0) x，當x0時，g(x) f(u)dufx，x2 1xf xfg(0)limg(x)g(0)limx lim limf(x) x0 limg(x)lim[ f(u)duf(x)]limfx

xf(u)duAA x2 x0x2 g(xx0處連續(xù).介值定理與實數(shù)根個數(shù)f(xxx2x3xn(n2,3,).fn(x1在(0,xn求limxn解：（1）fn(x在[0,1]fn(0)0,fn(1)n1xn(0,1)，fn(xn1x(0,fn(x12xnxn10fn(x在(0,fn(x1在(0,內(nèi)有唯一的實根.fn(xnfn1xn11xnx2xn x )xn1 fn(xxnxn1故limxna，由于0xnx21，故limxn

xn(1xn

nxnxnxn1

1

令n，對上式取極限得： 1，解得a1，即limx1x4axb0

1

n yx4axb，求導得:y4x32命y0得唯一駐點x ，又y12x0，故當x 值

4)3a4a

33

)3a4axyxy當且僅當b

a4（2）當b

a3時，方程無實根4f(x在(,f(x0,

f(x)0，

f(x)0x0f(x00.f(x0在(,恰有兩個實根.證明：存在f(a)0，由f(x)嚴格單增，當xa時f(x)f(a)f()(xa)f(a)f(a)(xa)，(axf(x；同理，xf(x故由介值定理在(,x0),(x0,)各有一個實根，再由反證法證明只有兩個實根設f(x)在[0,1]上連續(xù),f(0)=f(1),求證 對于任意正整數(shù)n，必存在xn[0,1]，f(x)f

1) 證明令x)fxfx1x)在[0,11]上連續(xù)M

mkM((1

k0,1,,n

1n所以mnk

(k)Mn故存在xn[0,1n nn

(n11n 1 [f(0)

f

f

n

)f(1)]

1[f(0)f(1)]( ( ( f(x)f

1) 羅比達法則與泰勒f(x在[0,2]f(1)02f(x)dxM,Mmaxf(x) f(xx01f(x)f(1)f(1)(x1)1f()(x1)22

1xff dx100022f21f(002

f()(x1)2dx

1f()(x1)222022M2(x1)2dx2 f(xf(0)f(0)xf在x軸上截距記為，求lim x0f

f(0)0yf(x在點(x,f(x))(x0)Yf(x

(xXx，令Y0,Xx

f(x)f由f(0)0f(xx0f(x單增，易知limf(0)f(0)0x0f(x)1f()x2(0x)，f()1f()2(0

xf

12

)2lim xflim f

limxf(x)f故lim lim x0f x01f()x x0 xf xf

f f 1x0f(x)xf f(x)fx

f(0)f 設f(x)有連續(xù)的二階導數(shù)，f(0)f(0) f(x)

u( f 求

yf(x在點(x,f(xx軸上的截距.Yf(x)f(xXx)

f0x軸上的截距為：u(x)xf(x)f

于是：u(x)f(x)f(x)[ff(x1f(0)x2o(x22u(

f(xf(0)xo(x)，知u(x)x2x

f(t)dtlimf(u(x))u(x)limf(u(x))f f f [f0f(x)[1f(0)u2(x)o(u2 1[f(0)x f(xx011x0 1f(0)f(0)f(0)及l(fā)im1f(x

e3x01

f(x) f(x)

解因為e3lim1xx0

ex0x

x所以lim

f(x)3. xf(x)lim

3x0 f(x)2o(1)，f(x)2x2o(x2)x2由此可 f(0)0，f(0)0，f(0)421 f(x) 2x1

o(x

)并有l(wèi)im1

lim1 e 中值定理

f(10,f11.2(1)存在1,1)f(；(2)存在0,f(2

f()f(1f(F(xF(x0有根，令G(x)exF(x)即可.f(x在abf(x在(abf(af(b0，bf(x)dx0a在(ab內(nèi)至少有一點，使得f(fb解：存在c(abaf(x)dxf(c)(ba0f(c令G(xexf(x則G(aG(cG(b0，且G(x在ab上連續(xù)，在(ab別在ac和cb上用羅1(a,c),2(cb)，使得G(1)G(2)0而G(xexf(xf(x)f(1f(1),f(2f(2F(x)exf(x)f(x)F(1)F(2)0,F(x在1,2上連續(xù)，在(1,2f2fx在閉區(qū)間[0,1]上連續(xù)，在開區(qū)間(0,1)內(nèi)可導，且有

fx

arctanxdx1 f10，則至少存在一點0,1，使得12arctanf1證明：由積分中值定理知，存在

efarctan

11

又ef1arctan14，故若設xefxarctanx x,10,1，顯然x滿足羅爾理的各個條件，從而至少存在一點,10,1使0. f ef 12從而 12arctanf1定積計算e2xsinx0n2 2 k20

sinxdxk

e(k

sinxdx(k

(1)

kk(k

(1)ke2xsinxdx1e2k(1e25n2xsinxdx1(1

2k1(1

e25故05

k

15nxn1)時ne2xsinxdxxe2xsinxdx(n1)e2xsinx5 2

1e2令n，由兩邊夾定理：0 sinxdx5e2注：若直接用e2xsinxdxlimne2 1e2 n

sinxdx5e21需要判定e2xsinxdx收斂.（類比級數(shù)的絕對值收斂，則原級數(shù)收斂0

2sinx2dx證明：令tx2,原式=2sint dtsint1dt2sint 2 u令ut，則2sint1dtsin(u) dusinuu 2 1故原式=0sin1

)dt22πsin2n dx limsin2n1x2n x0點處有界（x0不是被積函數(shù)的奇點）.sin2n1xsin2n1x2cos2nxsinx

πsin2n1xsin2n In

n1

dx22cos2nxdx0

sin2nx20 0InIn1I0.0I n k 求極限 nk1

nn

1 解：由泰 得sinn2n2on2 lim1ksinklim1

k

3n

o1n nn

n

n

n2k1 1x1xdx5

k1 設f(x)是(,)上的連續(xù)非負函數(shù),且f(x) xf(xt)dtsin40f(x在區(qū)間[0π上的平均值 解令xtu，則0fxt)dt0f記Fx)xf(u)du則FxFx)sin4x π1 π

(x)|π

sin4xdx

3π故F2π)

π，即F(π)

3π 從而f(x)在區(qū)間[0,π]上的平均值為1πf(x)dx 3ππ 變上限；積分的變量代換。不等2

22n1

ln(11) n ln(11 只需證2xln(1x

1x(1x2 11nf(xln(1x2x2

2g(x) 1

0,g由f(x) x 0,g(1x)(2

(1x1)303

f(0)g(0)0即得設常kln21，證x>0x1x1xln2x2klnx10證明fxxln2x2klnx故要證x1xln2x2klnx10

x只需證：當0x1時，fx0；當1x時，fx0fx12lnx2k1x2lnx2k xx2lnx2k，則x12x2 x2時，x0x=2為唯一駐點.又x2，210，所以x2為x 的唯一極小值點，故221ln22k2kln210為x的最小值(x>0)，x>0時fx0，從而fx嚴格單調(diào)遞增.f10，所以當0x1時，fx0；當1x時，fx0f(x在區(qū)間[0,1]上連續(xù)，在(0,1)f(0)00f(x1 10f(t)dt0

xF(x0f(t)dt0f(t)dt,F(x)2f(x)xf(t)dtf3(x)0令G(x)2xf(t)dtf2x0

f(x)[2xf(t)dtf20G(x)2f(x)2f(x)f(x)2f(x)[1fG(x0G(x在[0,1]G(x0，F(xiàn)(x)0,F(0)0F(x)0x0,1]， 1F(1)0即0f(t)dt0f x（2）F(x0f(t)dt,G(x0F F(x)F F( 20f則G(x)G(x)G(0)G()

20f(t)dt20f(t)d