—1 sinn

lim1n

sinn1n1x求xx0x xx0

xx0x

xxx1求lime

xsin

limex

limexx0xsin1sin原式=lim 1x01sinx二af(x

xa x解

fx

ex

y2sinxln

y2cosxlnx2sinx解 已知yf(exef(x)，求 xex2

yexfexefxfexfxefxefxexfexfexfxex2dx1ex2dx21ex2解 四、若2xtan(xy)xysec2tdt ysin2xyxy2xyx2 —1 cosncosn

lim1 2n n222

2

limx2

2222

2

x22

2

21 求lim2 lim lim

2x0

2x2x0x

lim2x 求limx2sinx0x3sin12sin原式=lim xx013sin xsin

x f(x)

x

x=0解：

fxsinxx

fxsinxx fxx0處不連續(xù)，0 y

1解

lnxxyyx求y解：ylnxxlnylnxy1lnyx兩邊分別求導 yyxln

xxyln1 exex1ex 解：原式e2x1dxex21arctane

2xx

10解：原式

2xcosx42x10sin9x

y0xy2 求y22x5和0xy2x

y22x5

yx

yx1（7，3）

2xx S 六(x21dy2xy解：原式

dy

yx2

4x2x22xdx

4x2 yx1y

x211n

求

xxxx

xx

xa

ax

x

xax

x

xax

xxx

1求lime2limsinx0sin

lime2x

lime2x原式

mx

x0二

xf(x)

f（x）x0x解

fxx0 已知ytan3lnx)求y3tan2lnxsec2lnxx已知yf(x2，求y2xfx2 求x2cosx解：原式cos1d1sin1xa

1四

0xf

)dx20xf(x)dx(a0)f(xx2t,

2xdx

x0a,則t0a2左邊

a2

1

xf 0 2五

dx100

0

1

1

01

u1

六求(1y2dxarctanyx)dydxarctany 1ydx arctan 1y 1y1代入：x1

1

dy

arctany1y

dyC

arctany

arctan1y2

earctan

dy令arctanyt,

xettdtdtCettetetCt1arctany1Cearctan以下為模擬卷部分內(nèi)容，供參求

xxx

xx

1x

xx

x1

x

x1x

x

x1x1limex11求lim 解：

x1x x1x1limex1求lim1cos x2解 lim1cosx

sinx x2x

x0 ) 求lim x0[ln(1x2

sin

1

2

11

2cos

1x 求 )x01

1x2x2x

lim2

1x

ex01x1已知yln(x x2a2)求 y

y

x x2x x2a1(x1)((x1)(x3x

2x2x2ax2a解：

x 3

x 1ln2 1ln2xdxln2xdlnx1ln3x 求x3ex2x3ex2dx1x2d解 求sin3 求xf8解：原式xdfxxfxfxdx

lnlnxdxxlnxxdlnxxlnx11十、計算

1x21 111x2

11

darcsinxarcsin五、y'ytanxsin2x解：ye解：ye sin二

n

n2 n2解

1

n2

n2

n2

n2n2n2n2n2n21n1n2n

n2