2018歡迎使高等數(shù)TOC\o"1-1"\h\z\u引 引第一講：極limfx=A00當(dāng)0x-

時(shí)，有fxAxx x x x x xfxAfxlimfxM0X0,xX時(shí),有fx n為自然數(shù) n專指n，而略去“+”不limxA0N0，當(dāng)nN時(shí)xAx x對(duì){x}，xa（常熟，x0，n1, ，若limxn0，則limx（x xn （C）非0常 0N0nxn0xnxn1xn1

1xn

1 取12

（nlimfxAA【證】假設(shè)limfxBABAB0,0,

0,00,0

x0x時(shí)，x0x

fxAfxB取Af A，BfxB取AB，則有ABfxAB e 1 aarctanx存在，求a、Iex x0 lim

a 0 x0 2 x

a

1

x0 x ex aaa1,I limfxA，則M0,0，當(dāng)0

xx0時(shí)，恒有fxM

，有fxAfx f AfxAA取2018（0的數(shù)M2018A，證畢xsin(xxsin(xA（

在（） D.(2,3【fx在I①若I為a,b，用“連續(xù)函數(shù)fx在a,b上必有②若I為a,b，則用 ,x x- xsin(x x1+x(x1)(x若limfxA0xx0fx若limfxA0xx0fx【分析】0,xx0,fxA AfxAA取A2

Afx3A 【例】設(shè)limfx=f0且 fx2，則x0是（ x01cosf

D.無法判sin(3)(), xsin(x sin(2)(2)x0x(x1)(x (1)1cosx2fx0

f0limfxlimf x01cos

(1cosx)0

0 0①七種未定式（ ,0,,,0,10若limfx0,limg(x f f f 且 ，則 x x x sin cos 【注】如 sinxlim

x sinx arcsixn arctaxnex ln ) 111

0 ,00【例1】 （0 ex2e22cos 【例2】 （ e22cos(ex22cosx21) x22cosx 2x2sin 1 【分析】碰到01 1 如limxlnx=lim =lim limxln2x行不通 ln

x01

ln2x

換一種

lnx limx0x0 簡(jiǎn)單：x、ex等3：原式=limln(1x1)ln(1x)lim(x1)ln(1x)（換元=limtlntt

cos2【例】lim( x0sin2 sin2sin2x4111tanx1sin xx42 =1=lim22cos4xlim 【例】limx2ex1x（令x x et1t et = 2e1 lim =lim tt0 t第三組（0,00,1

t0 t t0 U(x)V(x)eV(x)lnU(lim(x

1x2)x（01ln(x1x2=lim

limln(x1x2=

e01lim(tanx)cosxsinx（1x4：limuvelimvlnuelimv(u1)（u1limtanx1 limtanxcosxsin cosx(1tan原式=e e fxx0sinxx1x36arcsinxx1x36tx1arctanxx1x33cosx11x21x4o(x4 ln(1x)xx2x3x4o(x ex1xx2x3o(x 11

1xx2x3 （x1】limarcsinxarctan Barcsinxx1x36arctanxx1x33arcsinxarctanx1x3o(x3即原式

2與cxkc、cosx11x21x4o(x4 x2 e21 ) o(x4 所以cosxe2 x4 x4o(x4) x4o(x4 cosx

14【例3】設(shè)fxx0的某領(lǐng)域內(nèi)有定義，且

fxtanxsin4x

0

fx

fx

=

fxtanx4tanxtanIlimsin4x4tanxlimsin4x4x4x4tan limsin4x4xlim4x4tan 1 41x3 =

xn易于連續(xù)化，轉(zhuǎn)化為函數(shù)極限計(jì)

fxA，則limfnntan1tan1n2tantanlimx2(xtan1 limtant = （作倒代換）=et0 若xn不易于連續(xù)化，用“準(zhǔn)則”或定積分定 【例1】求lim xn nn nn n(n n(n

n2n i1n2n n2n nnarctan【例2nnarctannnnnnnnnnnnarctanxnxnfxn1給出，用“單調(diào)有界準(zhǔn)給出xn，若xn單增且有上界或者單減且有下界limxnxn 【例設(shè)常數(shù)a0,1,x , n，n1, ，證明xn收斂并求lim n

aa,

=a

a，猜測(cè)上界為a1 1 不妨設(shè)xa，則 n a，則有上界 a x n n1) (x )(x

x2x110xn1xn0，xn21再根據(jù)單調(diào)有界準(zhǔn)則，可知xn1記lim

AAa

A 舍去A ，故A若limfxfx0fxxx0

fx

fxfx0

fx（2）

fx（3）fx02（1（2）在若不存在=無窮間斷點(diǎn)若不存在=震蕩震蕩間斷點(diǎn)fx1

x,x 在x0處連續(xù)，則ab limfxlimfxf 第二講：一元函數(shù)微應(yīng)用limfx0xfx0fx fx0xfx0fx右導(dǎo) fx0xfx0fx左導(dǎo) fx0存在fx0fx0fx0

fx0狗fx0狗limfx0xfx0xfx就是典型錯(cuò)誤 xxxlimfxfx0fx x 1f00fxx0可導(dǎo)的是（f1co f1ehli

fh

f2hf【分析】見到fx0：先用定義法寫出來，熟練運(yùn)用定義法。這是道關(guān)于函數(shù)導(dǎo)數(shù)2】若fx是可導(dǎo)的偶函數(shù)，證明fx【分析】即證明fx f【練習(xí)】若fx是可導(dǎo)的奇函數(shù)，證明fx(x)(ax)axln(ex)(lnx)x(sinx)cosx(cosx)sinx(tanx)sec2x(cotx)csc2x(secx)secxtan(cscx)cscxcot(arcsinx) (arccosx) (arctanx) 1(arccotx) 1(ln(x x21)) (ln(x x21)) 【例】yx2xxxexxxx0，求y7ye2xlnxexlnxyfxF(xyF(x,y)0兩邊同時(shí)對(duì)xyy(x（復(fù)合求導(dǎo)【例】帶你學(xué)y1 d2【注】

所確定的隱函數(shù)的二階導(dǎo)數(shù)dx2x3(2

(x 【例】dx1 yd2xy''d

(y3(y'')2y'y (y xln(1t2 d3求參數(shù)方程ytarctant所確定的函數(shù)的三dx3xt 設(shè)ysint

t涉及fx的定設(shè)fx在ab連續(xù)①（有界性定理）M0, |f(x)|M,x②（最值定理）mf(x)M，mf(x)在ab③（介值定理）當(dāng)muM[a,b].f()u④（零點(diǎn)定理）f涉及f'(x

f(b0時(shí)，(a,b使f(01）可f(x)xx0處

f'(x00.（自己證明設(shè)f(x)滿足以下三條2）(ab)內(nèi)可導(dǎo)則(a,b)，使f'()0.（自己證明3)f(a)f【注】若f(af(b),則f'(f(bf(a)=0b

b 設(shè)f(x

f'(

f(b)f滿足2）(a,b)內(nèi)可導(dǎo)， 3)g'(x) g(b)g(xx,f'()1任何可導(dǎo)函數(shù)f(x)anxn

f(b)fb帶拉格朗日余項(xiàng)的泰勒：若f(x)n階可導(dǎo)：f''(x f(n)(x f(n1)(f(x)f(x)f'(x)(xx) 0(xx)2 0(xx)n (xx xx0之

(n f(xf(x)f(x)f'(x)(xx)f''(x0)(xx)2f()(xx xx0

當(dāng)x00時(shí)，泰勒又成為麥克勞林f(x)f(0)f'(0)xf''(0)x2f() 若f(x)n階可導(dǎo)：f(x)f(x)f'(x)(xx)

f

(xx)2

(xx)no((xx)n f(x3f(x)f(x)f'(x)(xx)f''(x0)(xx)2f(x0)(xx)3o((xx f(x) (0)'(0)x''(0)x2f2!3!拉 用于證佩 用于計(jì)fx的應(yīng)用（①-b【例】f(x在ab上連續(xù)，證明a,b使af(x)dxf()(ba)b直接說：mf(x證明：mu直接得出：f()u,(2)羅爾定理的應(yīng)用（⑥）f(af(bf'(F(xf(f'()1f(x)在[0,1]上連續(xù)0,1)內(nèi)可導(dǎo)且f(1kkxe1xf(x)dxk0證明：(0,1使f'((11f(

(I)方程fx=0在(0,1)2 （II）方程fxfxfx=0在(0,1)由(uv)'u'vuc'可得1）uf(xvx記F(x)f(x)xF'(x)f'(x)xf(x)(x)'f'(x)xf則可證f'()f(2）uf(xvex記F(xf(x)exF'(x)f'(x)exf(x)exex[f'(x)f則可證ef'(f(0或者f'(f(3）uf(xve(x),記F(xf(x)e(x)F'(x)f'(x)e(x)f(x)e(x)'(x)e(x)[f'(x)f(x)①將欲證結(jié)論中的改成

f(b)fb

f'()【例2】證明柯西中值定理：g'(

f(b)fg(b)f(x)、g(x)在[a,b]上二階可導(dǎo)，g"(x)0,f(a)f(b)g(a)g(b) 1）g(x)0,x2）(a,bf()g(

f"(g"(f'(f(bf(a)，（a,b）f(bf(a)=f'(ba，bf證明：(ab使bf(baf(af(f'()(b給出相對(duì)高階的條件【例】f''(x0f(00,證明:x1x20,有f(x1x2f(x1f(x2給出相對(duì)低階的條件 階不等3f(x)二階可導(dǎo),且f(2)f(1),f(2)2f(x)dx.證明(1,3使f''(

b【例】設(shè)0ab1，證明arctanbarctana f'()

f(b)fg(b)【例】設(shè)f(x)在[ab]上連續(xù),(ab)內(nèi)可導(dǎo)0ab2證明：,(ab使ftanabfsin2泰 的應(yīng)用-信號(hào)“f(n)”nox0的某個(gè)去心領(lǐng)域，xU(x0fxfx0xfx的真正極大值of'(x)0,xI，則f(x)在I上單調(diào)遞增f'(x)0,xI，則f(x)在I上單調(diào)遞oo當(dāng)x0x0x0時(shí)f'x0,當(dāng)x0x0x0時(shí)f'x0極當(dāng)xxx時(shí)f'x0,當(dāng)xxx時(shí)f'x0極

若fx在xx與xx內(nèi)不變號(hào)不是 f(xxx0處二階可導(dǎo)f'(x0f(x0極小值f(xxx0處二階可導(dǎo)f'(x0f(x0極大值1fx11)x在(0,x2】設(shè)fx連續(xù)，其fx的圖像如下：則fx有幾個(gè)極小值點(diǎn)，幾個(gè)x1x2Ifx1+fx2fx1x2fx是凹曲 fx1+fx2fx1x2fx是凸曲 判別法，設(shè)fxI

3若f(x)點(diǎn)的左右鄰域f''() ,x(f)()

2（帶你學(xué)，P13914）yf(xf''(x00,f'''(x0證明(x0,f(x0為拐點(diǎn)3】y(x1)(x2)2(x3)3(x4)4，則其某一個(gè)拐點(diǎn)為（A（1,0） B x

f(xxx0f(x)的一條鉛垂?jié)u近0或limf(xAyAf(x的一條水平漸近線x-

f(x)ByB為f(x)的一條limf(x)a0,limf(xaxb,則yaxb為一條斜漸近 x2x1x2x1

f'(x0x0若給出[a,b]

f'(x)不x不可導(dǎo)點(diǎn)，是可疑點(diǎn) 比較f(x0、f(x1)、f(a)、f(b)取其最大（?。┱邽樽畲螅ㄐ。┤粼贗上求出唯一極大（?。┲迭c(diǎn)，則由實(shí)際背景此點(diǎn)即為最大（?。? 1位于第一象限的部分求一點(diǎn)P，使該點(diǎn)處的切線、橢圓及兩坐 第三講、一元函數(shù)積設(shè)f(x)定義在某區(qū)間I上，若存在可導(dǎo)函數(shù)F(xF'(x)f(x對(duì)xI都成立，則F(xf(x在If(x)dxF(xCb定積af(x)dxbN :bf(x)dxF(x)xbF(b) xkdx

1dx1 xk1C,kk

1dx 1dxln|x|C,axdx1axC,a0,axexdxex

sinxdxecosxC,cosxdxsinxtanxdxln|cosx|C,cotxdxln|sinx|secxdxln|secxtanx|C,cscxdxln|cscxcotx|sec2xdxtanxC,csc2xdxcotxsecxtanxdxsecxC,csc1cotxdxcscxdxarcsinx dxarctanxdx

1x dx

1ln

a

|a2 a2 aa2 dxln(x a2x2) dx1arctanx a2

xdxln(x x2a2)

dx

ln

|x2 x2 ax2a2a2arctanxa2a2x2dx2 1】(2x)1(2x2)1(2x2)1cos2xsin)3】cosx(1cosxesinx)【分析】對(duì)于F(x)dx，F(xiàn)(x)越復(fù)雜我們34（xlnx)2(lnxx2三角換元--當(dāng)被積函數(shù)含有a2x2,a2x2 x2a2令xasintta2 a2令xatantta2x2令xax2

x0,0t2x0,t 2(xk2,2(xk2 倒代換--xt

復(fù)雜部分代換—令復(fù)雜部分axnaxb aebxcaxax,ext（指數(shù)代換lnxt（對(duì)數(shù)代換arcsinxarctanxt（反三角函數(shù)代換）x2dx,x2 2a2x2udvuvvdu（前面的積分 ex【例1】ex

sin2 cos定義：形如

Pn(xdx,(nmQm將Qm(x)因式分Q將Qm

ax

(x分解出axbk產(chǎn)生k (ax

分解出px2qxrk產(chǎn)生k(px2qxrA1xB1 A(px2qxrpx2qx (px2qxr)4x26x【例1】計(jì)算(x1)(2x1)2x【例2】計(jì)算 xxxxbaf(x)dxF(b)b -1 對(duì)于af(x)dx=1(a) (t)dt（令 (t)且要求(t)連續(xù)，并x(t)不超過區(qū)間a【例1】 1

2sinnxdxn13】1

1x2yy1(xyy2(xxaxb,(abbSa|y2(x)y1(x)b 1所圍圖形的面積。(a,b

x(1ln2

下方（ex，軸上 yy(x)與xaxb,(abxxVby2ayy(x)與xaxb,(abxybVya2b【例】設(shè)平面圖形yx22xy0x1,x3圍成，求y軸旋轉(zhuǎn)一周所得的by(x在[abyb

在區(qū)間[0,]2第四講、多元函數(shù)微o0,0,當(dāng)P(x,y) U(P0,)時(shí)，恒o|f(x,y)A|y

f(x,y)xy1xy1y

f(xyf(x0y0f(xy在(x0y0處連續(xù)偏導(dǎo)數(shù)（必考）zf(x(x0,y0

fx'(x0,y0limf(x0,limf(x0,y0x)f(x0,y0(x0,y0

【例】設(shè)f(x,y) ，求fx'(0,0),fy'(0,0)zf(uvw),uuy),vv(xywzz|vz| v wzf(uvxzf(esin

2y),f有二階連續(xù)偏導(dǎo)數(shù)，求x，xyzf(x

zf(x,y在點(diǎn)(x0y0處f''(x,y)

，則fx'(x0y00fy'(x0y00 A0 記 ''(x,y)B，則

''(x0,y0)

0該法失效，別謀他法（概念題 f(xyx2y2xlnx

(x,y,z)提法：求目標(biāo)函數(shù)uf(xyz在約束條件(x,yz)0 構(gòu)造輔助函F(x,yz)f(x,yz(x,yz(x,y 解方程組Pi(xi,yi,zi)u(Pi，比較取最大、最小者為最大、最小值【例】求uxy2yzx2y2z210第五講、二