2018基礎視頻講義需要版

2018考研數(shù)學講歡迎使高等數(shù)TOC\o"1-1"\h\z\u引 引任務（3.4.5.6月份第一講：極limfx=A00,當0x-

時，有fxAxx0x0x0x x xfxAfxfxfxlimfxM0,X0,xX時,有fx n為自然數(shù),n專指n，而略去“+”不limxA0N0，當nNxAx 對{x}，xa（常熟，x0，n1, ，若limxn0，則limx（x x

x （C）非0常 【分析】寫出定義：0，N0，當nN時， 0xnxn1xn1

11xn 1

1 取=12

（n1）2）xnfxn1若limfxAA【證】假設limfxBABAB0,0,

0,00,0

fxAfxB取Af A，BfxB取AB，則有ABfxAB e 1 aarctanx存在，求a、Iex x0x0兩個方向進行分析： limexaarctan10x0 2 x x0 x ex aaa1,I 若limfxA，則M0,0，當0

xx0時，恒有fxMfx f AfxA

fxA取2018（0的數(shù)M2018A，證畢A（

B.(0,1

在（）xsin(xxsin(x【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局部保號若limfxA0xx0fx若limfxA0xx0fx【分析】0,xx0,fxA Af A取A2

Afx3A 【例】設limfx=f0且 fx2，則x0是（ x01cossin(sin(3)(), xsin(x sin(2)(2)x0x(x1)(x (1)f1cos

2fx0

f0limfxlimf x01cos

(1cosx)0

0 0①七種未定式（ ,0,,,0,10若limfx0,limg(x 且

f

，則

f

fx x x sin cos 【注】如 sinxlim

x sixarcsixntanxarctaxnexxln )(1x) 12112

0 ,001tanx1sin1tanx1sinx1sin2x ex2e22cos 【例2】 （ e22cos(ex22cosx21)

x22cosx 2x2sin

1cos lim lim lim lim

【分析】碰到01 1 如limxlnx=lim =lim limxln2x行不 ln

x01

ln2x

換一種=

lnx limx0x0 x0 簡單x、ex3：原式=limln(1x1)ln(1x)lim(x1)ln(1x)（換元=limtlntt

cos2【例】lim( x0sin2 xsinxcos

x21sin24

2xsin2xcostantanxsin1x((1sin2x)211tanx1sinlimtanx(1cosx)11 sin212 =lim lim x2sin2 1=lim22cos4xlim 【例】limx2ex1x（令x x et1t et =lim2e1 lim =lim lim tt0 t t0 t t0 第三組（000,1U(x)V(x)eV(x)lnU(lim(x

1x2)x（01ln(x1x2=lim

=

ln(x1x2ln(x1x21lim(tanx)cosxsinx（1x4公式limuvelimvlnuelimv(u1)（u1limtan xcosxsin cosx(1tan2原式=e ex 2fxax0sinxx1x36arcsinxx1x36tanxx1x3o(x3)arctanxx1x33cosx11x21x4o(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fxx0的某領域內(nèi)有定義，且

fxtanxsin4x

0

fx fx

fx

=

fxtanx4tanx1tanIlimsin4x4tanxlimsin4x4x4x41tan limsin4x4xlim4x4tan 1 41x3 =

xn

fxA，則limfntanntann1n2tanxtanxlimx2(xtan1 limtant = （作倒代換）=et0 若xn不易于連續(xù)化，用 【例1】求lim xn nn nn n(n n(n n2n i1n2n n2nnnarctan【例2nnarctannn4nn4nn2nnnnarctan給出xn，若xn單增且有上界或者單減且有下界limxnxn 【例設常數(shù)a0,1,x , n，n1, n

aa,

=a

a，猜測上界為a1 1 不妨設xa，則 n a，則有上界 a x n n1) (x )(x (1)n1(x 2 (xx)(xx

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

AAa

A 11舍去A ，故A11若limfxfx0fxxx00xx0

fx0xx0

fxfx00xx0

fx（2）0xx0

fx（3）fx0（1（2）在若不存在=無窮間斷點若不存在=震蕩震蕩間斷點fx1

x,x 在x0處連續(xù)，則ab limfxlimfxf 應用limfx0xfx0fx limfx0xfx0fx fx0xfx0fx左導數(shù) fx0存在fx0fx0fx

fx0狗fx0 狗 limfx0xfx0xfx就是典型錯誤 0xxxlimfxfx0fx00 x01f00fxx0可導的是（f1co f1ehli

lim

fh

limf2hf 【分析】見到fx0：先用定義法寫出來，熟練運用定義法。這是道關于函數(shù)導數(shù)2fx是可導的偶函數(shù)，證明fx是奇函數(shù)【分析】即證明fx f【練習】若fx是可導的奇函數(shù)，證明fx是偶函數(shù)(x)(ax)axln(ex)(lnx)x(sinx)cosx(cosx)sinx(tanx)sec2x(cotx)csc2x(secx)secxtan(cscx)cscxcot1111(arctanx) 1(arccotx) 1x2(ln(x x21))x2x2(ln(x x21))x2【例yx2xxxexxxx0，求y7ye2xlnxexlnxyfxF(xyF(x,y)0兩邊同時對xyy(x（復合求導【例】帶你學y1 d21 1n 所確定的隱函數(shù)的二階導數(shù)1 1nn【注】n

u)u

uuu un ux3(2

(x 【例】dx1證明 yd2x y''d

(y (y xln(1t2 d3求參數(shù)方程ytarctant所確定的函數(shù)的三階導數(shù)dx3d2xtd2設ysint， tfx的定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muM時[a,b].使f()u④（零點定理）ff'(x的定

f(b0時，(a,b使f(0f(xxx0處

f'(x00.（自己證明設f(x)滿足以下三條2）(ab)內(nèi)可導則(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f(x

滿足2）(a,b)內(nèi)可導

f'()

f(b)f.3)g'(x) g'( g(b)【注】若取g(xx,f'()1任何可導函f(xanxn

f(b)fbf(xnf''(x f(n)(xf(x)f(x)f'(x)(xx) 0(xx)2 0(xx)n

f

()(xx xx

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

若f(x)n階可導：f''(x f(n)(xf(x)f(x)f'(x)(xx) 0(xx)2 0(xx)no((xx)n

f(x3f(x)f(x)f'(x)(xx)f''(x0)(xx)2f(x0)(xx)3o((xx f(x)f(0)f'(0)xf''(0)x2f(0)x3 （1）涉及fx的應用（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)羅爾定理的應用（⑥）f(af(bf'(F(xf(f'()1f(x)在[0,1]上連續(xù)0,1)內(nèi)可導且f(1)kkxe1xf(x)dxk0

(I)方程fx=0在(0,1)2 由(uv)'u'vuc'可得1）uf(xvx記F(xf(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

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，g"(x)0,f(a)f(b)g(a)g(b) 1）g(x)0,x2）(a,bf()g(

f"(g"(bf復雜化證明：(ab使bf(baf(af(f'()(b給出相對高階的條件證明低階不等 【例】f''(x0f(00,證明:xx0,有f(xxf(x 給出相對低階的條件 階不等3f(x)二階可導且f(2f(1)f(22f(x)dx.證明(1,3使f''(

b【例】設0ab1，證明arctanbarctana f'()

f(b)fg(b)【例】設f(x)在[ab]上連續(xù),(ab)內(nèi)可導0ab2證明：,(ab使ftanabfsin2 泰勒公式的f(n)”nox0的某個去心領域，xU(x0fxfx0xfx的真正極大值of'(x0,xI，則f(x在I上單調(diào)遞增f'(x0,xI，則f(x在I上單調(diào)遞減of(xxx0處連續(xù)，在U(x0,o當x0x0x0時f'x0,當x0x0x0時f'x0極當xxx時f'x0,當xxx時f'x0極

若fx在xx與xx內(nèi)不變號不是極 f(xxx0處二階可導f'(x0f(x0極小值f(xxx0處二階可導f'(x0f(x0極大值1fx11)x在(0,x2fx連續(xù)fx的圖像如下fx有幾個極小值點，幾個極大值點x1x2I fx1+fx2fx1x2fx是 2 fx1+fx2fx1x2fx是凸 2 判別法，設fxI上二階可導

0xIfx 0xIfx3若f(x)x點的左右f''(x 號x(,f(x))拐

2（帶你學，P13914）yf(xf''(x00,f'''(x0證明(x0,f(x0為拐點3】yx1)(x2)2(x3)3(x4)4，則其某一個拐點為（A（1,0） B x

f(xxx0為f(x)的一條limf(xA，則稱yAf(x的一條x-

f(x)ByB為f(x)的一條limf(x)a0,limf(xaxb,則yaxb為一條 x2x2x(x1)(x ）

f'(x0x0若給出[a,b]

f'(x)不x不可導點，是可疑點 比較f(x0、f(x1)、f(a)、f(b)取其最大（?。┱邽樽畲螅ㄐ。? 1位于第一象限的部分求一點P，使該點處的切線、橢圓及兩坐 設f(x)定義在某區(qū)間I上，若存在可導函數(shù)F(xF'(x)f(x對xI都成立F(xf(x在I上的一個原函數(shù)f(x)dxF(xCb定積af(x)dxbNL公式:bf(x)dxF(x)xbF(b xkdx

1dx1 xxk1C,kxk

1dx xx1dxln|x|C,axdx1axC,a0,axexdxex

lnsinxdxecosxC,cosxdxsinxtanxdxln|cosx|C,cotxdxln|sinx|secxdxln|secxtanx|C,cscxdxln|cscxcotx|sec2xdxtanxC,csc2xdxcotxsecxtanxdxsecxC,csc1cotxdxcscxdxarcsinx dx

1x dx

1ln

a

| a2

a dxln(x a2x2)

dx a2

xdxln(x x2a2)

dx

ln

|x2 x2 ax2xa2a2arctanxa2a2x2dx2 1】(2x)1(2x2)1(2x2)1cos2xsin)3】cosx(1cosxesinx)【分析】對于F(x)dx，F(xiàn)(x)越復雜我們就越高34（xlnx)2(lnxx2三角換元--當被積函數(shù)含有a2x2,a2x2x2a2令xasintta2 a2令xatantta2x2令xax2

x0,0t2 x0,t ax2bxax2bxk222(xk2,2(xk22倒代換--xt

復雜部分代換—令復雜部分axcxnaxb aebxcaxcxax,ext（指數(shù)代換lnxt（對數(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)Q將Qm

ax

(x分解出axbk產(chǎn)生k(ax(ax，k1,(ax分解出px2qxrk產(chǎn)生kA1xBpx2qx

A2xB AkxAkx，k1,24x26x1】計算(x1)(2x1)2x【例2】計算 xxxxbaf(x)dxF(b)b -1 對于af(x)dx=1(a) (t)dt（令 (t)且要求(t)連續(xù)x(t不超過區(qū)間ax【例1】 x1

2sinnxdxn13】1

1x2yy1(xyy2(xxaxb,(abbSa|y2(x)y1(x)b

(a,b

1x(1ln2

， yy(x)與xaxb,(abxxVby2ayy(x)與xaxb,(abxybVya2b【例】設平面圖形yx22xy0,x1,x3圍成，求y軸旋轉(zhuǎn)一周所得的by(x在[abyb 1

在區(qū)間[0,]2o0,0,當P(x,y) U(P0,)時，恒o|f(x,y)A|y

f(x,y)xy1xy1y

f(xyf(x0y0f(xy在(x0y0處連續(xù)【注】若偏導數(shù)（必考）zf(xlimf(x0limf(x0x,y0)f(x0,y0(x0,y0

limf(xlimf(x0,y0x)f(x0,y0(x0,y0

x2【例】設f(x,y) ，求fx'(0,0),fyx2zf(uvw),uuy),vv(xywzz|vz| v w高階偏zf(uv 2zf(exsin

),f

無條件的zf(x

zf(xy在點(x0y0處

，則fx'(x0y00fy'(x0y00 f''(x, 記f''(xyB，則B

''(x0,y0)

0該法失效，別謀他法（概念題 f(xyx2y2xlnx

(x,y,z)提法：求目標函數(shù)uf(xyz在約束條件(xyz)0下的極值 Fx0Fy0F