2015考研數(shù)學(xué)基礎(chǔ)習(xí)題講義高數(shù)

函 極限連續(xù)fx

xx1g(xexfg(x)gfx設(shè)a0

1，求lim(aararn13n1 n

nn3

n n

2n

k1

n

22

2n(n1)設(shè)0x13xn11設(shè)a11,an1

x(3x，證明limx n0，證明數(shù)列{a收斂，并求lim nlim(2x1)4(x1)65x(x8

x2(3x83x

x

(x ln(13x cosx1arcsin23x2 1tanx x1(5)lim

ex21lim x0xln(12x)sin2x2enxcos

1x2cos x(exf(x x

，求limfxx2x2ax

2

cxd0abcdx0,x設(shè)fx4x 0

x ，求出f(x)的間斷點，

x1ex 設(shè)f(x) 1x，求f(x)的間斷點并判定類型 x2第二講已知fxsinx x0，求f(x) xyx4x

315x

yy3x34x5ey3cotxcscx

yln(cosxtan223ln3ysinxcos (15)ye2x(34xyx3ln2exsinlny xexey yx3lnxsiny yln(x a2x2

ey eex2yarctan x3ycos3xsin(x5yy 設(shè)f(x)在x2處可導(dǎo)，且limf(x)2，則f(2) ，f(2) x2x2f

f(x)設(shè)f(x)二階連續(xù)可導(dǎo)，且lim 0,f(0)4，則lim[1 ]x

f(xx1x2f(x1x2f(x1f(x2f(0)1f(xf(xf(2x)f(2設(shè)f(x)連續(xù)可導(dǎo)，f(2)1，且lim 1，則曲線yf(x)在點(2,f π已知曲線的極坐標(biāo)方程r1cosθ，求曲線上對應(yīng)于θ處的切線與法線的直角坐標(biāo)方程6f(x5x0f(1sinx3f(1sinx8xα(x其中l(wèi)im 0，f(x)在x1處可導(dǎo)，求曲線yf(x)在點(6,f(6))處的切線方程 設(shè)函數(shù)yy(x)由方程xyexey0所確定，則y'0 f(x

txarctan 設(shè)yeyln(et2

若f(x)是可導(dǎo)函數(shù)，且fxsin2sinx1，f(0)4，則f(x)的反函數(shù)xφ(y)當(dāng)自變量取4時的導(dǎo)數(shù)值為 yy(xxyyxdyf(xaf(xbf1c，其中a、b、cab x f(xf(xf(xf(x第三講f(x在[0，3]上連續(xù)，在(0，3)f(0)f(1)f(2)3f(3)f(x在[0,1]上連續(xù)，在(0，1)f(0)1

f(1)0，f

2

2f(x)在[0,1]上連續(xù)，(0,1)內(nèi)可導(dǎo)，f(0)0k為正整數(shù),求證：存在ξ01)ξf(ξ)kf(ξ)f(ξ)f(x),g(x在ab]g(x0f(a)求證:(1)在ab)g(x)0

f(b)g(a)g(b)0(2)存在ξab，使f(ξ)

.f(x在[0,1]上連續(xù)，(0,1)f(0)0f(1)1存在ηζ(0,1)，ηζf(ηf(ζξ2ξ設(shè)ξ為f(x)arctanx在[0,a]上使用微分中值定理的中值，則lim a0f(x在(,limf(xe2，又limxa)xlim[f(xf(x xx ae

(2)x0 2

cos2 2

sin3n x(axbx)(a0,b0)為常數(shù)

sin

x0

xsin2

(a0,b0)為常數(shù)試確定方程exax2a0)fn(xxx2xn(n(1)證明方程f(x)1有唯一的正根x (2)求limx

nf(xxlimf(xf(x0)2f(xx0 (x0

f(x二階導(dǎo)數(shù)連續(xù),且(x1)f(x2(x1)f(x)1e1xxa(a1)是極值點時,是極小值點還時極大值點x1是極值點時,是極大值點還是極小值點曲線yxlnx的斜漸近線 x第四講 ln1x 1x 1 x arctan1x

x1 1cosxsinx11sin

(9) x x

1cos

2x(14xsinxcosx(x

xcos2sec3(lnx

a2a2

max{x2,x2}dx 1nn(n1)(n2)(n1nn(n1)(n2)(n

n n nn

設(shè)f(x)在[0,1]上連續(xù)，且f(x) 1xf(x)dx，則f(x) 0x0

sin2 ln1t2)dt是(1cosx3的幾階無窮小f(x在[0,f(0)0g(x若xfx)g(tx)dtx2ln(1xf(xx設(shè)f(x)二階連續(xù)可導(dǎo)，且f(0)1,f(2)3,f(2)5，則1xf(2x)dx 05 ,x5設(shè)f(x)

，則1f(x1)dx 2,x4 xe 0(1ex)2 (2)0(x21)(x211arcsin11arcsin

yx3xx1y軸分成兩部分,求這兩部分面積之比3y2

1在點

x2y2a2xb(ba0)旋轉(zhuǎn)所成旋轉(zhuǎn)體體積第五講微分方程yysinx的一個特解具有形式 y*asinC．y*xasinxbcos

y*acosD．y*acosxbsin二階常系數(shù)非齊線性微分方程y2y3y(2x1)ex的特解形式為 (axb)ex2(axb)e

x(axb)e設(shè)φ1(x),φ2(x),φ3(x)為二階非齊線性方程ya1(x)ya2(x)yf(x)的三個線性無關(guān)解，則該 C1[φ1(x)φ2(x)]C2φ3C1[φ1(x)φ2(x)]C2φ3C1[φ1(x)φ2(x)]C2[φ1(x)φ3C1φ1(xC2φ2xC3φ3x，其中CC2C315(1)x yy 0 (2)xdyylnyln15x

2ysin

dy

2x

dy

xy

(x2xyy2)dyy2dx*○(1x2)y''(y')21

*○yy(y)21y7y6y (2)y6y9y(3)y6y13y(5)yy2x2(7)y2y3y

(4)yy2y(6)y2yy(8)yyxcos f(x二階連續(xù)可導(dǎo)，f(0)1fx3xf(t)dt2x1f(tx)dtex0f(x y1xexe2xy2xexexy3xexe2xex是某二階線性非齊次常系數(shù)微分方第六講

xy 1tan 設(shè)zxsiny y

x，則 設(shè)zesin2xy，則dz 2設(shè)zf(x2y2,)，且f(u,v)具有二階連續(xù)的偏導(dǎo)數(shù)，則 zxf(xyg(xyx2y2fg分別二階連續(xù)可導(dǎo)和二階連續(xù)可偏導(dǎo)，則2.設(shè)yy(x,z)是由方程exyzx2y2z2確定的隱函數(shù)，則y x2yx2y f(xy在點(0,0)的某鄰域內(nèi)連續(xù)且滿足

f(xyf(0,0)3f(xyx21xsin(0,0)處 取極大 B.取極小C.不取極 D.無法確定是否有極sin設(shè)函數(shù)f(x)在[11]上連續(xù)，則xcosyf(t)dt f(sinx)f(cos B.f(sinx)cosxf(cosy)sinC.f(sinx)cos f(cosy)sin設(shè)f(x,

sin(x2y2 (x,y)

(x,y) f(xy，xy)2(x2y2ex2y2f(xyf(xy x2z2yφz,其中φ為可微函數(shù),zy zf(exsinyx2y2,f具有二階連續(xù)偏導(dǎo)數(shù),

2xy設(shè)uf(xyzzz(xyxexyeyzez所確定，求duexyxy2和exxzsintdtdu zx2y2xyxyDx0,y0xy3上的最大值與最小值第七講111.011

dy f(xy)f(xyxyf(xy)dσDy0,yx2x1Df(x,y) D計算ydxdyDx2y22axa0x軸圍成上半圓區(qū)域D4.計算

|yx2|dxdy計算|x||y|dxdyx2y2計 |xy|dxdy設(shè)f(x)連續(xù)，且f(0)1，令F(t) f(x2y2)dxdy(t0)，求F(0)x2yV*f(xF(tz2fx2y2dv，其中Vxyz|x2y2t20zhVt0)，求limF(tt0t*計算xy2x2y2x2y2

其中zxyyxx1,z0所圍的區(qū)域dxdydz其中x2y2z2z所圍的區(qū)域*計算(x2y2dxdydz其中x2y2

z2所圍的區(qū)域

計算

)dV其中由曲線y22zozz2xz8第八講曲線積分與曲面積分L是上半圓周x2y22x，求xdsLIxdsLx2y2a2A(0,a經(jīng)點C(a,0)B(x2x2

,aLL

y x2y2dsLx2y1)2IL3x2ydxx3dyL是從點(0,0)經(jīng)過點(1,0到點(0,0)LLx2y22x，計算ydxxdyLIy2xdyx2ydxLx2y2a2LI(xy)dxxy)dyLyx22A(2,2)B(2,2) x2yI[eycosxay]dxeysinxb(xy)]dyL為4x29y236L的部分，方向為從點(3,0到(0,2)Ixyzy2z2z2x2x2y2dS，其中S是球面x2y2z2a2SIxdydzSz1

zdxdySzx2y2z0計算曲面積分I

2dydzy2dzdxz2dxdy{(xyz)0z 4x2y2,x2y21}ILy2z2dx2z2x2dy3x2y2dz,其中L是平面xyz2與柱面第九講無窮級數(shù) 設(shè)(1)n1a2,a2n15，則an

n

冪級數(shù)2n(3)n 冪級數(shù)

(xn 的收斂域 n(n1)n1(an1)(an)(an(n1)n1

3n

(5)