1
第七节 无穷小与无穷大
习题 1-7
1,利用等价无穷小替换定理求下列极限,
(1)
0
tan 5
lim
2
x
x
x
→; (2)
0
sin( )
lim (,)
(sin )
n
m
x
x
mn
x
→
∈
*
N ;
(3)
3
0
(1 cos )
lim
sin
x
x x
x
→; (4)
0
arcsin 3
lim
sin 2
x
x
x
→;
(5)
1
lim arctan
x
x
x
→∞; (6)
1
2
3
0
(1 ) 1
lim
cos 1
x
x
x
→
+?
,
解 (1)
00
tan 5 5 5
lim lim
222
xx
xx
→→
==,
(2)
00
0,
sin( )
lim lim 1,
(sin )
.
nn
mm
xx
nm
xx
nm
xx
nm
→→
>?
== =
∞ <
当当当
(3)
2
33
00
1
(1 cos ) 1
2
lim lim
2sin
xx
xx
xx
xx
→→
==,
(4)
00
arcsin 3 3 3
lim lim
sin 2 2 2
xx
xx
xx
→→
==,
(5)
11
arctan
1
lim arctan lim lim 1
11xxx
xx
x
x
xx
→∞ →∞ →∞
===,
(6)
1
2
2
3
00
2
1
(1 ) 1 2
3
lim lim
1
cos 1 3
2
xx
x
x
x
x
→→
+?
==?
,
2,当0x→时,试确定下列无穷小关于x的阶数,
(1) sinx x+ ; (2)
32
10x x+ ;
(3)
2
1cos2x? ; (4)
2
tan 2x,
2
解 (1) 因为
0
sin
lim 1
x
xx
x
→
+
=,所以阶数为1,
(2) 因为
32
2
0
10
lim 10
x
xx
x
→
+
=,所以阶数为2,
(3) 因为
22
2
44
00
1
(2 )
1cos2
2
lim lim 2
xx
x
x
xx
→→
==,所以阶数为4,
(4) 因为
22
00
tan 2 2
lim lim 2
xx
xx
→→
==,所以阶数为2,
3,当0x→时,
k
x与
23
tan (2 )x是同阶无穷小,则k等于多少?
解 因为
23 6
66
00
tan (2 ) 2
lim lim 2
xx
xx
→→
==,即
23
tan (2 )x与
6
x是同阶无穷小,故
6k =,
4,当,mn∈
*
N,证明,当0x→时,
(1) ( ) ( ) ( ),min{,}
mnl
ox ox ox l mn+= = ;
(2) ( ) ( ) ( )
mn mn
ox ox ox
+
= ;
(3) 若α是0x→时的无穷小,则()
mm
x oxα = ;
(4) ( ) ( )( 0)
nn
okx ox k= ≠,
证 (1)
000
() () () ()
lim lim lim 0
mn m n
lll
xxx
ox ox ox ox
xxx
→→→
+
= +=,故
() () ()
mnl
ox ox ox+=,
(2)
000
()() () ()
lim lim lim 0
mn m n
mn m n
xxx
ox ox ox ox
xxx
+
→→→
=?=,故()() ( )
mn mn
ox ox ox
+
=,
(3)
00
lim lim 0
m
m
xx
x
x
α
α
→→
==,故()
mm
x oxα =,
(4)
00
() ()
lim lim 0
nn
xx
okx okx
k
xk
→→
=?=,故()()( 0)
nn
okx ox k= ≠,
5,函数sinyx x=在(,)?∞ +∞内是否有界? 这个函数是否为x→+∞时的无穷大?
解 sinyx x=在(,)?∞ +∞内无界,因为0M? > (无论它多么大),总能找到
3
π
2 π ()
2
xk k=+∈N,使得当
π
2
2π
M
k
>时,
π
2 π
2
yk M= +>,
但当x→+∞时,sinyx x=不是无穷大,例如,取2 π()x kkN= ∈,当k →+∞时,
x→+∞,但0y =,
