微积分(经管类第四版)习题1-9答案

习题1-9

lim1.x0

xxxx

23

2

limx0

x0

当x0时，故xx是xx的高阶无穷小

2

232

1/2(1x)1

lim(1x)1 2.limx1x121x

12

1x与(1x)同阶，为等价无穷小 故，2

lim

axa

x

33

3.x0

lim

(axa)(axa)

x(axa)

3

3

33

x0

lim

1axa

3

3

x0



12a

ax-a与x相比是三阶无穷小122

ln(1x)xo(x) 4.因为cosx12xo(x)，

sinxxo(x)

122

11x-o(x)xo(x)(1cosx)ln(1x)2limlim 1122x0x0

sinxxcosxo(x)xcos

xx

2xlim

x0

xxo(x)2

xo(x)2o(x)o(x)o(x)12

xo(x)xcos

x

2

32

xxo(x)22o(x)2o(x)/xo(x)o(x)/x

lim

1x0

1o(x)/xxcos

x

2

故(1cosx)ln(1x)与sinxx2cos1为同阶无穷小，

x

但不是等价无穷小

arctan3x3xo(x)3o(x)/x3limlimlim 5.(1)x0x0x05x5x55

(2)lim(3)lim

ln(13xsinx)

tanx

3

2

x0

lim

3xsinxx

4

2

x0

3

(sinx)tanx1cosxe

5x

2

x0

x

lim2x014

x2

(4)lim

1

x0

x

5xo(x)5o(x)/xlimlim5x0x0x1

xsinx1(xsinx1)(xsinx1)

(5)limlimx0x0xarctanxxarctanx(xsinx1)lim

xsinx

xarctanx(xsinx1)

1

x0

lim

x

2

2

x0

x(xsinx1)

1limx0xsinx12

(6)lim

5xsinx2xtanx4x

2

2

23

x0

lim

5xxo(x)2x

2

3

x0

2

xo(x)4x

3

2

2

2

lim

5xx2xo(x)o(x)2x

xo(x)4x

2

x0

5x2o(x)o(x)/x2xlimx01o(x)/x4x5

习题1-9

lim1.x0

xxxx

23

2

limx0

x0

当x0时，故xx是xx的高阶无穷小

2

232

1/2(1x)1

lim(1x)1 2.limx1x121x

12

1x与(1x)同阶，为等价无穷小 故，2

lim

axa

x

33

3.x0

lim

(axa)(axa)

x(axa)

3

3

33

x0

lim

1axa

3

3

x0



12a

ax-a与x相比是三阶无穷小122

ln(1x)xo(x) 4.因为cosx12xo(x)，

sinxxo(x)

122

11x-o(x)xo(x)(1cosx)ln(1x)2limlim 1122x0x0

sinxxcosxo(x)xcos

xx

2xlim

x0

xxo(x)2

xo(x)2o(x)o(x)o(x)12

xo(x)xcos

x

2

32

xxo(x)22o(x)2o(x)/xo(x)o(x)/x

lim

1x0

1o(x)/xxcos

x

2

故(1cosx)ln(1x)与sinxx2cos1为同阶无穷小，

x

但不是等价无穷小

arctan3x3xo(x)3o(x)/x3limlimlim 5.(1)x0x0x05x5x55

(2)lim(3)lim

ln(13xsinx)

tanx

3

2

x0

lim

3xsinxx

4

2

x0

3

(sinx)tanx1cosxe

5x

2

x0

x

lim2x014

x2

(4)lim

1

x0

x

5xo(x)5o(x)/xlimlim5x0x0x1

xsinx1(xsinx1)(xsinx1)

(5)limlimx0x0xarctanxxarctanx(xsinx1)lim

xsinx

xarctanx(xsinx1)

1

x0

lim

x

2

2

x0

x(xsinx1)

1limx0xsinx12

(6)lim

5xsinx2xtanx4x

2

2

23

x0

lim

5xxo(x)2x

2

3

x0

2

xo(x)4x

3

2

2

2

lim

5xx2xo(x)o(x)2x

xo(x)4x

2

x0

5x2o(x)o(x)/x2xlimx01o(x)/x4x5