高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类(2)

1

xn

{xn}

xn=

13n

xn=( 1)

n

1n

xn=2+

1n3

xn=

n 2n+2

xn=( 1)n

n

11

,,39127,1

LL→081

1111, ,, ,L→02345

1111

2+1,2+,2+,2+,2+,LL→2

8276412544441

xn=1 ,L→1 1 ,1 ,1 ,L1

n+2345100 1,

1,2 3,4,L→∞

2

lim

1=0n→∞nk

k

1+3n3

=

n→∞4n 14lim

lim

n+2

sinn=0

n→∞n2 2

高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类

lim

1=0n→∞nk

ε

1

0<εnk

1

<n ε

lim

1k

1

k1 N=

ε

ε>0

n>N

1

0<εnk

1

=0n→∞nk

1

k1 ε

N

N

N

1+3n3

=

n→∞4n 14lim

ε

3n+137

=<ε4n 144(4n 1)

n>

7+4ε16ε 7+4ε N=

16ε

ε>0n>N

3n+13

<ε4n 14

1+3n3

=

n→∞4n 14

n+2

sinn=0lim2

n→∞n 2lim

n+2n+21

n <sin0=

n 2n2 2n2 2

ε

n+2

sinn 0<ε2

n 2

1

<εn 2

n>

1

ε

+2

n>2

1 N= +2

ε

ε>0

n>N

n+2

sinn 0<εn2 2

lim

n+2

sinn=0

n→∞n2 2

3

{xn}

xn=

1nπcosn2

limxn=?

n→∞

N

n>N

xn

ε

ε=0 001

N

高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类

1nπlimcos=0n→∞n2

1nπ1

cos 0<nn21

ε

1nπ

cos 0<εn2

1<εn

n>

ε

1

N=

ε

N

1 ε

ε>0n>N

1nπ

cos 0<ε

2n

nπ1

limcos=0n→∞n2

ε=0 001N=1000

4

nπ 1

an= 1+ sin

2 n

{an}

A

A

n=2k,k∈N

2kπ1

a2k= 1+ sin

2 2k

n→∞

k→∞

2kπ1

lim 1+=0sin k→∞22k

n=4k+1,k∈N

π 1 (4k+1)π 1 a4k+1= 1+= 1+ sin 2kπ+ sin

2 2 4k+1 4k+1 n→∞

k→∞

(4k+1)π=lim1+1=11

lim 1+sin k→∞k→∞4k+12 4k+1

5

{xn}

limyn=0

n→∞

limxnyn=0

n→∞

xn<M

yn<ε1xnyn<ε

高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类

{xn}

M

n

xn≤M

xnyn≤Myn

limyn=0

n→∞

ε1

N

n>NN

yn<ε1

ε

ε1=ε

M

ε

M=ε

n>N

yn≤ε1=

ε

M

xnyn<M×

limxnyn=0

n→∞

6

{xn}

k→∞

limx2k 1=a

limx2k=a

k→∞

limxn=a

n→∞

ε>0

xn a<ε

n

ε>0

k→∞

limx2k 1=a

N1

2k-1>N1

x2k 1 a<ε

k→∞

limx2k=a

N2

2k>N2

x2k 1 a<ε

N=max{N1,N2}

n>N

n=2k 1

n=2k

xn a<ε

limxn=a

n→∞

1

f(x)

g(x)

f(x)g(x)

f(x)g(x)

f(x)=x

g(x)=sin

1x

limx=0

x→0

limg(x)

x→0

limxsin

x→0

1

=0x

2

2x+32

=

x→+∞3x3lim

x→+∞

lim

sinxx

=0

1lim=1x→2x 1

x2 1lim2=2x→1x x

高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类

ε>0

2x+32

<ε3x3

x

X

δ

X

δ

ε

f(x) A=

2x+321

=<ε,3x3x

x>

1

ε

X=

1

ε

x>X

2x+32

<ε3x32x+32

=

x→+∞3x3lim

ε

f(x) A=

1

sinxx

0≤

1x

1x

<ε

1

x>

sinxx

ε2

0<ε

X=

ε2

x>X

x>0

sinxx

0<ε

x→+∞

lim

sinxx

=0

f(x) A=1

2

1x 2

, 1=

x 1x 1

0<x 2<δ

δ

x

x 2<

x→x0

f(x)

x0

x

f(x)

1

x 2<

2

35<x<22

ε

2

x 2<ε31 2

x 22x 2<=x 2x 133

x 2<ε

2

δ=min ε,

f(x) A=

3 2

0<x 2<δ

x 22

<x 2x 13

2

x 2<ε3

x 22

<x 2<ε,x 13

lim

x→2

1

=1x 1x <

12

x2 1x 1

f(x) A=2 2=

xx x

13<x<22

x 1x 1

<=2x 1x

高等数学(上册)习题答案_吴赣昌_人民大学出版社_高数_理工类

ε

2x 1<ε

x 1<ε/2

1 ε

δ=min ,

2

2

0<x 1<δ

f(x) A=

x 1

<2x 1<εx

x2 1lim2=2x→1x x

3

x→2

y=x2→4

δ

0<x 2<δ

y 4<0 001

x→2

x 2<1

1<x<3

1<x<3

y 4=x2 4=x 2 x+2<5x 2

5x 2<0 001

x 2<

0 0015

δ=

0 001

=0.00025

0<x 2<δ

y 4<0 001

δ

0.0002

y 4<0 001

4

f(x)=lim

nx

n→+∞nx2+2

0,x=0

0,x=0 nx

f(x)=lim2= limx,x≠0= 1

n→+∞nx+2,x≠0 n→+∞x2+2 x n

5

f(x)=

xx

x→0

x→x0

limf(x)=A lim+f(x)=A

x→x0

x→ x0

limf(x)=A

f(x)=

x 1,x>0=

x 1x<0