1
第五节 极限的运算法则
习题 1-5
1,下列运算是否正确,为什么?
(1)
11 1
lim ( )
1
n
nn nn
→∞
+++"
11 1
lim lim lim
1
nn n
nn nn
→∞ →∞ →∞
=+ ++
++
"
00 00=++ +=";
(2) lim ( 1 1) lim 1 lim 1 0
xxx
xx x x
→+∞ →+∞ →+∞
+ = +=∞?∞=;
(3)
000
11
lim sin lim lim sin 0
xxx
xx
→→→
=? =,
解 (1) 不正确,因为只有有限个数列和的极限(且这有限个数列的极限都存在)才等于它们极限的和,
(2) 不正确,因为只有当两函数极限都存在时,才有两函数差的极限等于它们极限的差,
(3) 不正确,因为
0
1
lim sin
x
x
→
不存在,
2,计算下列各极限,
(1)
2
123 ( 1)
lim
n
n
n
→∞
+ ++ +?"; (2)
11 1
lim (1 )
24 2
n
n→∞
++++";
(3)
2
3
523
lim
3
n
nn
nn
→∞
++
+; (4)
11 1
lim ( )
12 23 ( 1)
n
nn
→∞
+++
+
";
(5)
22
lim ( 1 2 )
n
nnn
→∞
+ ; (6)
3
( 1)(2 1)(3 1)
lim
3
n
nnn
n
→∞
+ ++
,
解 (1)
2
123 ( 1)
lim
n
n
n
→∞
+++ +?"
2
(1)
1
2
lim
2
n
nn
n
→∞
= =,
(2)
1
1
1
1(1 ( ) )
11 1 1
2
lim (1 ) lim lim 2(1 ( ) ) 2
1
24 22
1
2
n
n
n
nn
+
+
→∞ →∞ →∞

+ +++ = =? =
",
2
(3)
2
23 23
3
23 23
52 3 52 3
lim ( )
523
lim lim 0
13 13
3
1lim(1)
n
nn
n
nn
nnnn nn
nn
nn nn
→∞
→∞ →∞
→∞
++ ++
++
== =
+
+?+
,
(4)
11 1 111 11
lim ( ) lim (1 )
12 23 ( 1) 2 2 3 1nn n n
→∞ →∞
+++ =?+?++?
+ +
""
1
lim (1 )
1
n
n
→∞
=?
+
1=,
(5)
22
22
2
1
2
12
lim ( 1 2 ) lim lim
12
12
11
nnn
n
n
nnn
nnn
nn
→∞ →∞ →∞
+
+
+ = =
++?
+ +?
1=,
(6)
3
111
(1 )(2 )(3 )
( 1)(2 1)(3 1)
lim lim 2
33
nn
nnn
nnn
n
→∞ →∞
+++
+++
= =,
3,计算下列各极限,
(1)
0
2
lim(1 )
3
x
x
→
; (2)
2
2
371
lim
523
x
xx
xx
→∞
+
+?;
(3)
2
20
lim
11
x
x
x
→
+; (4)
3
1
1
lim
1
x
x
x
→
;
(5)
2
11
lim (1 )(2 )
x
x x
→∞
+?; (6)
2
2
1
1
lim
21
x
x
xx
→

,
解 (1)
00
2225
lim(1 ) 1 lim 1
333
xx
xx
→→
=? =?=

,
(2)
2
22
2
22
71 71
3lim(3)
371 3
lim lim
23 23
5523
5lim(5)
x
xx
x
xx
xx
xx
xx
→∞
→∞ →∞
→∞
+?+
+
== =
+?
+? +?
,
(3)
2
2
200
lim lim(1 1 ) 2
11
xx
x
x
x
→→
=? + + =?
+
,
(4)
3
2
11
1
lim lim( 1) 3
1
xx
x
xx
x
→→
=++=
,
(5)
22
11 1 1
lim (1 )(2 ) (1 lim )(2 lim ) 2
x
xx
→∞ →∞ →∞
+?=+? =,
(6)
2
2
11
112
lim lim
21321
xx
xxx
→→
+
==
+
,
3
4,计算下列各极限,
(1)
22
0
()
lim
h
x hx
h
→
+?; (2)
22 2
11 1
lim (1 )(1 ) (1 )
23
n
n
→∞
",
解 (1)
22
00
()
lim lim(2 ) 2
hh
xh x
x hx
h
→→
+?
=+=,
(2)
22 2
11 1 1111 11
lim (1 )(1 ) (1 ) lim (1 )(1 )(1 )(1 ) (1 )(1 )
223323
nn
nnn
→∞ →∞
=?+?+?+""
111
lim (1 )
22
n
n
→∞
= +=,
5,表述并证明x→∞时函数极限的四则运算法则,
解 若lim ( )
x
f xA
→∞
=,lim ( )
x
g xB
→∞
=,则
(1) lim[ ( ) ( )] lim ( ) lim ( )
xxx
f xgx AB fx gx
→∞ →∞ →∞
±=±= ± ;
(2) lim[ () ()] lim ()lim ()f xgx AB fx gx
→∞ →∞ →∞
==? ;
(3) 若0B≠,则有
lim ( )
()
lim
() lim ()
x
x
x
f x
fx A
g xB gx
→∞
→∞
→∞
==,
证明如下,
(1) 仅证明和的形式,
由lim ( )
x
f xA
→∞
=,lim ( )
x
g xB
→∞
=知,0ε? >,
12
0,0XX? >>,当
1
x X>
时,有()
2
fx A
ε
<; 当
2
x X>时,有()
2
gx B
ε
<,
取
12
max{,}XXX=,则当x X>时,[() ()] ( )f xgx AB+?+ ()f xA≤?
()
22
gx B
ε ε
ε+?<+=,因此lim[ () ()] lim () lim ()
xxx
f xgx AB fx gx
→∞ →∞ →∞
+=+= +,
(2) 由于() () () () () ()f xgx AB fxgx Bfx Bfx AB= +?
() ()f xgxB≤ ()Bfx A+,
由lim ( )
x
f xA
→∞
=及函数极限的局部有界性得,0ε? >,
1
0X? >及0M >,当
1
x X>时,有()
2
fx A
C
ε
<,且()f xM≤,其中max{,}CMB=,
又lim ( )
x
g xB
→∞
=,故
2
0X?>,当
2
x X>时,有()
2
gx B
C
ε
<,
4
取
12
max{,}XXX=,则当x X>时,有
() ()f xgx AB () ()f xgxB≤ ()Bfx A+
2222
MBCC
CCCC
ε εεε
ε<? +? ≤? +? =,
因此lim[ () ()] lim ()lim ()
xxx
f xgx AB fx gx
→∞ →∞ →∞
==?,
(3) 因为
() 1
lim lim[ ( ) ]
() ()
xx
fx
fx
g xgx
→∞ →∞
=?,故由(2),只需证当0B≠时,有
11 1
lim
() lim ()
x
x
g xB gx
→∞
→∞
==,
11 ()11
()
() () ()
Bgx
g xB
gx B gx B B gx
= =
,
由lim ( )
x
g xB
→∞
=及函数极限的局部有界性知,0ε? >,0X? >及0M >,当
x X>时,有()
B
gx B
M
ε?<,且
1
()
M
gx
≤,所以,
1111 1
()
() ()
B
gx B M
gx B B gx B M
ε ε?=< =,
即
11 1
lim
() lim ()
x
x
g xB gx
→∞
→∞
==,