华工-数学分析范例

《数学分析》中的一题多解汇编

（《数学分析》课程辅导材料—部分初稿）

《数学分析》课程组 编

华南理工大学 理学院 数学系

1

√n

n

n

,

Gn=

√n

(am+1Amm +11)

1/mm

=G.Am+1=

Am+AAm+1 Gm+1,

(1)

2

(1)

a1=a2=···=an.

An=

a1+a2+···+an

n 1

=An,

2

m

2

√n

n

n 1

a1+a2+···+an

,

a1+a2+···+an

n 1

a1+a2+···+an

n 1

a1a2

a3+a4+···+an

n

√n

n

=nlim→∞

an.

n 1

.

n 1

(3)

3

n

=lim

xn

yn yn 1

n→∞

=liman.

n→∞

2

δn=

a1+a2+···+an

n

n

A=|

k=1

n0

n0

n n0|(ak a)|+k=1

n0

1|(ak a)|+k=1

n

+ ,

(ak a)|.

n→∞

lim

A

n

n

=

sn0

n n0

(1

n0

n

+3M(1

n0

→0,1

n0

n

n

<

M

n

>

1

>M.sn

n→∞

lim

n

=+∞.

3.

f

[a,b]

c∈[a,b],

x∈[a,b],

f(c) f(x).

1

M={x∈[a,b]:f(x) f(u) u∈[a,x]}.

4

2(an

+bn).

u∈[an,cn]

v∈[cn,bn]

f(v) f(u),

an+1=an,bn+1=cn.

u∈[an,cn],

v∈[cn,bn],f(v)>f(u),

an+1=cn,bn+1=bn.

1

(i)an an+1 bn+1 bn.(ii)bn+1 an+1=

5

6

7

n}

1.

√2

n

1

n≥2√n

1 n·

√

0

√n

n

<<1+

2

,n

√

√n

2

√n

2

]

.

n 1< .

yn=

2

2

yn.

yn,

n>1

√n

]+1

.

2

2

6.

n→∞

lim

n

n!nn!

=e.

1

1.

ln(

Cauchy

nnlnn (ln2+ln3+···+lnn)=n!

n

.

n→∞

lim(bn+1 bn)=l lim

n)

bn

n→∞

=ln(1+

1

8

nn

√n

n!

,

n→∞

lim

an+1

(n+1)!

·

n!

n

)n=e.

7.

f

[a,b]

f

[a,b]

.

1(

)f

[a,b]

x∈[a,b],

O(x)

Mx,

|f(x)|<Mx

x∈O(x)

].

x∈[a,b]

(x)

OM

[a,bx,

[a,b]

.

O1,O2,···,On,

.

M1,M2,···,MnM=max{M1,M2,···,Mnn},

[a,b]

Oi,

i

=1x∈[a,b]

|f(x)|<M,f

[a,b]

.2(

Bolzano

.

f

[a,b]

)

a1=a,b1=b.

(a1+b1)

[a1,b1]

c1=

2n 1

(b a).

.

9

10

2

2)+

π

2

π

2

2

+

π

2

2

],m=[2Nπ+2π],([···]2N+

π

4

)

m>n>N,

4

,2Nπ+π<m<2Nπ+2π

√2

|sinn sinm|≥ε0=

.

|sinn|

3

n→∞

limsinn=A

sin(n+2) sinn=2sin1cos(n+1),

n→∞

lim2sin1cos(n+

n→∞

n→∞

1)=lim(sin(n+2) sinn)=A A=0,

n→∞√lim

n→∞

limcosn=0,A=limsinn=

11

√

k+1 2

√√

√k 2

(n+1)2

√

+1 2

√

√

(n+1)2

2

√

(n+1)2

√

≤lim

√

n+1,

1

n+1

n→∞

k=n2

2n+21≤k

1:(1)

x1=

√

(c+xn)

,(c>1

)

n→∞

limxn.c,

√√

c,f(x)=

c(1+x)

c+x

.c

xn>

c+xn

√

=f(xn)>f(

x1>

√

xn+1 xn=

c.

c(1+xn)

c+xn.

<0

{xn}

,x1<

.

√

c,xn

,xn

,

.

xn+1=

c(1+xn)

c.

2:

xn>0,

x>0

,f′(x)≡(c(1+x)

c2

(c+x)2

>0.

c>1

f′(x)=

c(c 1)

=1

1

c.

4.lim

f(x),g(x)

E

,x0

E

,

lim

n→x0

g(x)≤

n→x0

limf(x)+lim

12

f(xn)+lim(f(xn)+g(x))≤

g(xn).

.

5.

x

xlim

sinx tan→0

sinx

cosx 1

x3

=xlim

→0

cosx

2

.

2

sinx tanx

x3cosx

,

x(x→0),1 cosx～

12x2,

x( 1

1

xlim

sinx tanx

→0

x3cosx

=xlim→0x3

=

2)π,n

∈N+,

2

.

a

b

asinx+bcosx=

sinx～

a=0.

13

π

x′n=2nπ

n→∞,

2√

x

x=0

.

1(

Heine

)

xn=

1

,yn=

1

2

xn

=sin(2nπ+

π

yn

=sin2nπ=0,n∈N+.sin

1

yn

1

Heine

0.

x

x=0

2

.x′=

1

,x′′=

1

2

x3

.

1

3

L’Hospital

xlim

x tanx

→0

3x2=xlim

→0

2sec2xtanx=

1

3

1 sec2x

x3

=xlim

→0

3x2

=

1

x3sinx

.

sin

1

=

1

(1)

Heine

14

x3sinx

=lim

2sinx2

(3 x2)sinx+5xcosx

4cosx2 8x2sinx2

2 2

.

x→0

=lim

x→0

2

x→0