1
第一章总习题
1,填空题,
(1) 设
e,0,
()
cos,0,
x
x
fx
xx
≤?
=
>?
则(1) ef?=,
2
1
2
2
e,1,
(1 )
cos(1 ),1;
x
x
fx
xx
≥
=
<
(2) 设函数( ) lg arcsin
23
x x
fx
x
=+
,则它的定义域是[3,0) (2,3]? ∪ ;
(3) 若() ()f xgx<,且
0
lim ( )
xx
f xA
→
=,
0
lim ( )
xx
g xB
→
=,则A和B的关系是
A B≤ ;
(4) 设函数()f x在点
0
x的某邻域内有定义,则()f x在
0
x x=处连续的充分必要条件是
000
() () ()fx fx fx
+?
==,
解 (1) 略,
(2) 由0
2
x
x
>
,且11
3
x
≤ ≤得,函数定义域为[3,0) (2,3]? ∪,
(3) 略,
(4) 略,
2,下列四个命题中正确的是( B ),
(A) 有界数列必定收敛; (B) 无界数列必定发散;
(C) 发散数列必定无界; (D) 单调数列必有极限,
解 略,
3,设lim
n
n
x
→∞
=+∞,lim ( 0)
n
n
yyy
→∞
= ≠,求lim sin
n
n
n
n
y
x
x
→∞
,
解
sin
lim sin lim
n
nn
nn
n
n
n
y
yx
x yy
y
x
x
→∞ →∞
=?=,
4,求下列极限,
(1)
sin sin
lim ;
xa
x a
xa
→
(2)
22
lim ( 1 2 3)
x
xx x x
→+∞
+ + ;
2
(3)
1
1
2
1
1
lim( )
1
x
x
x
x
x
→
; (4)
2
0
1sin1
lim
e1
x
x
xx
→
+?;
(5)
2
2cot
0
lim(1 3tan )
x
x
x
→
+ ; (6) lim ( 0)
!
n
n
c
c
n
→∞
> ;
(7)
22
lim (1 )(1 ) (1 ),1
n
n
aa aa
→∞
+ ++<null ;
(8)
12
lim ( )
2! 3! ( 1)!
n
n
n
→∞
+++
+
null,
解 (1)
2cos sin sin
sin sin
22 2
lim lim lim cos lim
2
2
xa xa xa xa
x axa xa
xa xa
x a
xa xa
→→ →→
+
+
==?
cos 1 cos
2
aa
a
+
=?=,
(2)
22
22
34
lim ( 1 2 3) lim
123
xx
x
xx x x
xx x x
→+∞ →+∞
+ + =
+?+? +
22
4
3
3
lim
2
11 23
x
x
xx
→+∞
==
+? +?+
,
(3)
1
1 1
11 2
2
11
lim( ) lim( ) ( )
121
2
x
xx
xx
x
xx
+?
→→
===
+?
,
(4)
2
2
00
1
sin
1sin1 1
2
lim lim
2
e1
x
xx
xx
xx
x
→→
+?
==
,
(5)
2
2
1
2cot 2 3 3
3tan
00
lim(1 3tan ) [lim(1 3tan ) ] e
x
x
xx
xx
→→
+ =+ =,
(6) 设k为任一个大于2c的自然数,则当nk>时,
1(2)
0( )( )()
!12 1 2 2 2
kn
knk
n
ccccccc c
c
nkkkn
<= <? =
++
nullnull,
由于
(2 )
lim 0
2
k
n
n
c
→∞
=,由夹逼准则,故lim 0
!
n
n
c
n
→∞
=,
(7)
22 22
1
lim (1 )(1 ) (1 ) lim (1 )(1 )(1 ) (1 )
1
n n
nn
aa a aaa a
a
→∞ →∞
++ +=?++ +
nullnull
1
2
11
lim (1 )
n
n
a
aa
+
→∞
=?=
,
3
(8)
12 11 11 1 1
lim ( ) lim[( ) ( ) ( )]
2! 3! ( 1)! 1! 2! 2! 3! ! ( 1)!
nn
n
nnn
→∞ →∞
+++ =? +? +?nullnull
11
lim ( ) 1
1! ( 1)!
n
n
→∞
=? =
+
,
5,已知当0x→时,
1
2
3
(1 ) 1ax+?与cos 1x?为等价无穷小,求数a,
解 由已知,
1
2
2
3
00
2
1
(1 ) 1 2
3
lim lim 1
1
cos 1 3
2
xx
ax
ax
a
x
x
→→
+?
= =? =
,故
3
2
a =?,
6,确定常数a及b的值,使下列极限等式成立,
(1)
2
lim ( ) 8
x
x
xa
xa
→∞
+
=;
(2)
2
lim ( 1 ) 0
x
xx axb
→∞
+ =,
解 (1)
33
lim
3
3
23
lim( )lim(1 )lim(1 ) e e 8
x
xa ax ax
xx a
axa xa
xx x
xa a a
xa xa xa
→∞
→∞ →∞ →∞
+
=+ =+ = ==
,
所以ln 2a =,
(2)
22
2
2
(1)()
lim ( 1 ) lim
1
xx
x xaxb
xx axb
x xaxb
→∞ →∞
+? +
+ =
++ +
22 2
2
(1 ) (1 2 ) (1 )
lim
1
x
ax abx b
xx axb
→∞
++?
=
++ +
0=,故必有
2
1a =,且
2
2
2
(1 2 ) (1 ) (1 2 )
lim ( 1 ) lim 0
1
1
xx
ab x b ab
xx axb
a
xx axb
→+∞ →+∞
+ ++
+ = = =
+
++ +
,
故1a =,
1
2
b=?,
7,已知
0
()
ln(1 )
sin
lim 3
21
x
x
fx
x
→
+
=
,求
2
0
()
lim
x
f x
x
→
,
解 因为
0
lim(2 1) 0
x
x→
=,
0
()
ln(1 )
sin
lim 3
21
x
x
fx
x
→
+
=
,故必有
0
()
lim ln(1 ) 0
sin
x
fx
x
→
+ =,
因为21ln2
x
x～,
() ()
ln(1 )
sin sin
f xfx
x x
+ ～,所以
4
2
0000
() ()
ln(1 )
1()1()
sin sin
lim lim lim lim 3
ln2 ln2 sin ln221
x
xxxx
fx fx
fx fx
xx
xxx x
→→→→
+
== = =
,
故
2
0
()
lim 3ln 2
x
fx
x
→
=,
8,写出下列函数的连续区间与间断点,并指出间断点的类型,
(1)
1
2
1
1
() e
1
x
x
fx
x
=;
(2)
ln(e )
() lim ( 0)
nn
n
x
fx x
n
→∞
+
= >,
解 (1) 易知连续区间为(,1)(1,)?∞+∞∪,因为
11
2
11
1
lim e lim ( 1)e
1
xx
xx
x
x
x
++
→→
=+=+∞
,
11
2
11
1
lim e lim ( 1)e 0
1
xx
xx
x
x
x
→→
= +=
,
故1x=是第二类间断点,
(2) 当0ex<<时,
ln[e (1 ( ) ]
ln(e )
e
() lim lim
nn
nn
nn
x
x
fx
→∞ →∞
+
+
==
ln(1 ( ) )
e
lim[1 ]
n
n
x
n
→∞
+
=+
()
e
lim[1 ] 1
n
n
x
n
→∞
=+=;
当ex=时,
ln(e ) ln(2e )
() lim lim 1
nn n
nn
x
fx
→∞ →∞
+
===;
当ex >时,
e
ln[ (1 ( ) ]
ln(e )
() lim lim
nn
nn
nn
x
x
x
fx
→∞ →∞
+
+
==
e
ln(1 ( ) )
lim[ln ]
n
n
x
x
n
→∞
+
=+
e
()
lim[ln ] ln
n
n
x
x x
n
→∞
=+=,
即
1,0 e,
()
ln,e,
x
fx
xx
<≤?
=
>
在ex=处,lim ( ) ln e 1
xe
fx
+
→
= =,lim ( ) 1
xe
fx
→
=,
故()f x在ex=处连续,故函数连续区间为(0,)+∞,
9,设
cos
,0,
2
()
,0,
x
x
x
fx
aax
x
x
≥
+?
=
<
要使()f x在(,)?∞+∞内连续,应如何选择数a?
5
解 易知函数在(,0)(0,)?∞ +∞∪上连续,要是函数在0x=处也连续,则
00 0 0
1
lim ( ) lim lim lim
()
xx x x
aax x
fx
x
x aax aax
→→ → →
== =
+?+?
11
(0)
2
2
f
a
= ==,
故1a =,
10,设常数0a >,0b>,证明方程sinx axb= +至少有一个正根,并且它不超过ab+,
证 令函数() ( ) sinf xxbax=,(0) (0 ) sin 0 0fbab= =?<,
()( )sin() sin()f ab abb a ab aa ab+=+ +=? +,
当sin( ) 1ab+<时,( ) 0fa b+>,由零点定理,至少存在一点(0,)abξ∈ +,使得() 0f ξ =,即ξ为原方程的根,它是正根且不超过ab+ ;
当sin( ) 1ab+=时,( ) 0fa b+=,则ab+是原方程的正根,且不超过ab+,
11,设函数()f x在[0,2 ]a上连续,且(0) (2 )f fa=,证明,在[0,2 ]a上至少存在一点ξ,使() ( )f faξ ξ=+,
证 构造函数() () ( )Fx fx fx a=?+,则()Fx在[0,]a上连续,且
(0) (0) ( )Fffa=?,() () (2) () (0)Fa fa f a fa f=?=?,
若(0) ( )f fa=,则0ξ =即是满足() ( )f faξ ξ= +的点；
若(0) ( )f fa≠,则必有(0)F与()Fa异号,故由零点定理,至少存在一点
(0,)aξ∈,使得() 0F ξ =,即() ( )f faξ ξ= +,
综上,至少存在一点[0,] [0,2 ]aaξ∈?,使() ( )f faξ ξ= +,
12,设
12
,,,
n
λ λλnull是n个正数,并且它们的和等于1,证明,如果函数()f x在闭区间[,]ab上连续,那么对于区间[,]ab上的任意n个点
12
,,,
n
x xxnull,至少有一点
[,]abξ∈,使得
1
() ( )
n
kk
k
f fxξλ
=
=
∑
,
证 因为()f x在[,]ab上连续,故存在最大最小值,不妨设
max{ ( ) [,]}M fxx ab=∈,min{ () [,]}mfxab=∈,
则
11 1
()
nn n
kkkk
kk k
mmfx MMλλ λ
== =
=≤ ≤ =
∑∑ ∑
,
由介值定理,至少存在一点[,]abξ∈,使得
1
() ( )
n
kk
k
f fxξλ
=
=
∑
,
