分析选论作业解答：作业三

T<3s?
T< 
? 01λ< <
O lim 0
n
n
b
→∞
= b¨?l￡
ü


2
12 0
lim ( ) 0
n
nn n
n
bb b bλλ λ

→∞
++ ++=" b
￡
 0ε?>
 y
#lim 0
n
n
b
→∞
= N
+
∈]
?
H
nN> (1 )
2
n
b
ε
λ<?b,
01
max{,,,} 1
N
Mbbb=+" lim 0
n
n
λ
→∞
by =
#

ε

?
H
μ
1
N
+
∈]
1
nN N?>
1
1
2
1
nN
N
M
ε λ
λ
λ
+
<?
b?
^?
H
μ
1
nNN>+
21
12 0 1 1
1
0
1
(1 )(1 ) (1 )
2
11
(1 )
21 122
nn
nn n n n N
nN n nN nN N
N
nN N
nN
bb b bb b b
bb M
M
λλ λ λ λ
ε
λ λλλλλλ
ελ λε
λλ ε
λ
λ λ

+
+
+ + ++ ≤ + ++
++ ≤? +++ + +++

= + <+=
""
""
"
#
2
12 0
lim ( ) 0
n
nn n
n
bb b bλλ λ

→∞
++ ++=" b
T< 

! ()f x ??
 O? 1x>?
H

2
0
()[ () 1]
2(1 )
x
x
xe
fx ftdt
x
+=
+
∫


p ()f x b
3

7
0
() () 1
x
gx ftdt= +
∫

5 () ()f xgx′= b?
^
2
()()
2(1 )
x
xe
gxgx
x
′ =
+


Hs¤
2
()()
2(1 )
x
xe
g x g x dx dx
x
′ =
+
∫∫

'
2
() ()
2(1 )
x
xe
gxdgx dx
x
=
+
∫∫
V7
2
2
()
2
2(1 )
x
gx xe
dx
x
=
+
∫

'
2
2
()
(1 )
x
xe
gx dx
x
=
+
∫
b
1
2
2
11 1
(1 )
xx
x x
xe x x e x
dx xe d dxe
xx x
x
==?
++ +
+
∫∫ ∫
=
2
1
x
x
xe
xedx
x
+
∫
=
2
(1)
1
x
x
xe
x eC
x
+
+
=
1
x
e
C
x
+
+

? (0) 1g =
# 0C = b
?
^ ()
1
x
e
gx
x
=
+
b
[
2
()
2(1 )
1
x
x
xe
fx
e
x
x
=
+
+
b
4ù3s

9
3
3
2
1
lim
1
n
n
k
k
k
→∞
=
+
∏
b
3

3
3
2
1
lim
1
n
n
k
k
k
→∞
=
+
∏
=
22
22
(1)( 1) 1 1
lim lim ( )
1
(1)( 1) 1
nn
kkk
kkk k kk
k
kkk kk
→∞ →∞
==2
n
=
++? ++
=?
+
+?+?+
∏∏∏
=
2
2
123 1 713 1
lim ( )( )
345 1 3 7
1
n
nnn+
n
nn
→∞
+

+
+
""
2
12 1
lim
(1) 3n
nn
nn→∞
++
+
=
2
3
b =
2