微積分上冊答案第六章習(xí)題6 3

6.3按x4fxx45x3x23xf4f'x4x315x22x3,f'4f''x12x230x2,f''4f''x24x30,f'''4fxx45x3x23x45621x437x4211x43x44f0f''x6x23x12x326x23x12,f''0f''x62x3324x23x12x312x23x12x3,f"'0f4x362x32242x3248x23x1122xf5x1442x3962x3482x3482x242x3,f50fxx23x1319x30x245x330x49x5f4

xf'x ,x2

'4f''x1x23,f''41 f''x3x25,f'''43 x2所以fx 21x41x421x435ξ27x44.其中ξ 解 f'x1x

繼而可歸納 fnx1n1n 于,fxlnxln2

k1x2 x2k1 k x2nοx2nxxfx11x1x1kx1n1n1ξn2x1n1.ξx于1xfxtanxf0f'xsec2x,f'0f''x2sec2xtanx,f''0f''x4sec2xtan2x2sec4x,f''0fxtanxx1x3οx33求函數(shù)fxxex的帶有 型余項的n階 解：歸納可得fnxnexxex,fn0n,所以fxxexxx21x32

k

xk n

xnοxnex1x

xοxn，所以xexx

οxn2x 2

n

1x οx chxexex1x2x4

οx2n1 1 1x

n1 解：ln1xx

οx

，所以ln1xln1xln1x2x2x32x5 x2n1οx2n.1cos2

2n cosx

23

2m!ο

1cos 2 1m22m1x2mοx2m1x 1 x32xx ?x32x1 x xx341 13x3x4x4x4xοx.cos cosx

2m!ο cosx1 求arcsinx的局部

ο .

1x2111x23x42n

2nοx2n2 x1 22 x x分得arcsinxx1x33x5 x 2n!2nexcosxx4

ο

2n1 4 4ecosx1x 1xx3

οx4

arctanxx3arctan00,arctanx' 1

,1x2,arctanx'''

21x228x21x21x2所以arctanx所以arctanxx 3sinsinxx3解 sin xο sinsinxsinx οsinx3x οx3 x3οx3?解：

2x3x

x3 ' 6x2

22xx 2x3x

12x2x3x1226x2122x3x 2xx 2x3x 所 x1xx2ο xx2x3 2x3x1x ?

1x

2x1x 1x1xx212x2x21x

1x1xx2

112x2x2x31x3οx412x2x22x4οx4 1x

2解 x xx xxοx2x2 οx421x （1）lim3x33x24x4

1 lim3x33x24x42x3limx133x124 x x x 1 1 limx ο x1 ο x x cosxe x0xxln1

x x

1 οx1 οx

2cosxe

x0

xln1

x2xx οx2 121x0 211x2 x0cosxex2sin 11x2 11x211x21x4 112

x0

2

1

1 οx1xοx lim x03 2

axax2a ln2

ln2 axax

lim lim limx2ln2aln2

ln1xx2ln1xx2

xsinln1xx2ln1xx2

xsin xx2 xx2 xx2 οx2xx2 οx2 lim limx0

ex316 sin

6 ex31

1x3 ο2

1

1

lim sin6 26 x026 lim1 x0 sinx 3 lim1 limsinxxlimx6οxxlim6

sinx

x0xsin

00limfx0hfx0h2fx0

f"x hlimfx0hfx0h2fx0h

f"

f"x0設(shè)fnx存在，且fxf'xfn fxοxxnxx a和bfxxabcosxsinxx0x的5階無窮小。fxxabcosxsin xabbx2bx4οx4x1x31x 1abxa4bx3a16bx5οx5 1ab a4所以a4b

即 a16b b i設(shè)對xabfx0,求證xa,b,i12,,nixxx1f 2 n nfxi xx 證明：記 nx，則fxfxf'xxx ixx nf"ξ iix ifxlnxx