1
第六节 极限的存在准则与两个重要极限
习题 1-6
1,计算下列极限,
(1)
0
sin
lim ( 0)
tan
x
x
x
α
β
β
→
≠ ; (2)
0
lim cot
x
x x
→
＋;
(3)
π
lim 3 sin
3
n
n
n→∞; (4)
0
1cos2
lim
sin
x
x
x x
→;
(5)
0
lim
1cos
x
x
x
+
→; (6)
sin
lim
2cos
x
x x
x x
→∞
+
,
解 (1) 若0α ≠,
00 0
sin sin sin
lim lim cos lim
tan sin sin
xx x
xx xx
x
x
α ααβα
β
β βαβ
→→ →
=?==;
若0α =,易知
0
sin
lim 0
tan
x
x
x
α α
β β
→
==,
(2)
00 00
lim cot lim cos lim lim cos 1
sin sin
xx xx
xx
xx x x
→→ →→
=?=?=
＋＋＋＋
,
(3)
π
sin
π
3
lim 3 sin lim ππ
π
3
3
n
n
n
nn
n
→∞ →∞
=?=,
(4)
2
00
1cos2 2sin
lim lim 2
sin sin
xx
xx
xx xx
→→
==,
(5)
000
2
lim lim lim 2 2
1cos
2 sin sin
22
xxx
x
xx
xx
x
+++
→→→
==?=
,
(6)
sin
1
sin 1
lim lim
cos
2cos 2
2
xx
x
xx
x
x
xx
x
→∞ →∞
==?
+
+
,
2,计算下列极限,
(1)
0
lim(1 ) (,0)
b
x
x
ax a b
→
+ > ; (2)
1
lim ( )
1
x
x
x
x
→∞
+;
2
(3)
0
lim 1 2
x
x
x
→; (4)
2sec
π
2
lim (1 cos )
x
x
x
→
+ ;
(5)
1
lim (1 )
kn
n
n
→∞
( k为正整数); (6)
1
lim ( )
1
n
n
n
n
→∞
+
,
解 (1)
1
00
lim(1 ) lim(1 ) e
b
ab
ab
xax
xx
ax ax
→→
+=+ =,
(2)
12
()()
2
21
12
lim ( ) lim (1 ) e
11
xx
x
x
xx
x
+

+
→∞ →∞
=? =
++
,
(3)
1
()(2)
2
2
00
lim 1 2 lim(1 2 ) e
x
x
xx
xx

→→
=? =,
(4)
2
2sec 2
cos
ππ
22
lim(1 cos ) lim (1 cos ) e
x
x
xx
xx
→→
+ =+ =,
(5)
11
lim (1 ) lim[(1 ) ] e
kn n k k
nn
nn

→∞ →∞
=? =,
(6)
12
2
21
12
lim ( ) lim (1 ) e
11
nn
n
n
nn
n
→∞ →∞
+
=+ =

,
3,利用夹逼准则证明下列极限,
(1)
22 2
11 1
lim ( ) 1
12
n
nn nn
→∞
+++ =
++ +
";
(2)
22 2
12 1
lim ( )
212
n
n
nn nn
→∞
+++ =
++ +
";
(3)
22 2
ππ π
lim (sin sin sin ) π
12
n
nn n
→∞
+++ =
+
";
(4)
0
lim 1 1
n
x
x
→
+ =,
证 (1) 因为
222 22
11 1
12 1
nn
nn n n nn n
<+++<
+++ ++
",
又
2
1
lim lim 1
1
1
nn
n
nn
n
→∞ →∞
= =
+
+;
2
2
1
lim lim 1
1
1
1
nn
n
n
n
→∞ →∞
= =
+
+;
所以
22 2
11 1
lim ( ) 1
12
n
nn nn
→∞
+++ =
++ +
",
3
(2) 因为
222 222 2
1
(1)
12 1
2
12
nn
nn
nn nnnn nnn n nn
+
= + ++ < + ++
+++ + + +
""
22 2 2
1
(1)
12
2
11 1 1
nn
n
nn n n
+
<+++=
++ + +
",
又
22
11
(1) (1)
1
lim lim
21
nn
nn nn
nn n
→∞ →∞
++
= =,故
22 2
12 1
lim ( )
212
n
n
nn nn
→∞
+ ++ =
++ +
",
(3) 因为
222 2 2
πππ π π
sin sin sin sin sin
12 1nn n n nn n
≤++ ≤
++ + +
",
又
2
22
2
π
sin
ππ
lim sin lim π
πnn
n
nn
n
nn nn
nn
→∞ →∞
+
=?=
++
+
,同理
2
π
lim sin π
1
n
n
n
→∞
=
+
,故
22 2
ππ π
lim (sin sin sin ) π
12
n
nn nn
→∞
+ ++ =
++ +
",
(4) 当0x >时,11 1
n
x x<+<+,故
0
lim 1 1
n
x
x
+
→
+ = ;
当10x?< <时,111
n
x x+< +<,故
0
lim 1 1
n
x
x
→
+ =,
故
0
lim 1 1
n
x
x
→
+ =,
4,利用单调有界准则证明下面数列存在极限,并求其极限值,
(1)
12
2,22,,22 2
n
aa a== ="" (n次复合);
(2)
11
12
11
1,1,,1
n
n
n
xx
xx x
xx
==+ =+
++
",
证 (1) 易知
1
2( 1,2,)
nn
aan
+
==",下证此数列单调有界,
当1n=时,
1
22a =<,假设nk=时,2
k
a <,则当1nk= +时,
1k
a
+
=
2
k
a 2<,即2( 1,2,)
n
an<=",即此数列有界;
4
因为
2
1
2(2)
2
22
nn nn
nn nn
nn nn
aa aa
aa aa
aa aa
+

=?= =
++
,由2
n
a <,故
1
0
nn
aa
+
>,
即
1nn
aa
+
>,
综上,lim
n
n
a
→∞
存在,令lim
n
n
aA
→∞
=,
又
1
2
nn
aa
+
=,故
2
1
2
nn
aa
+
=,因此
2
1
lim 2 lim
nn
nn
aa
+
→∞ →∞
=,即
2
2A A=,
解得
1
2A =,
2
0A = (舍去),故lim 2
n
n
a
→∞
=,
(2) 易知0
n
x >,先证此数列单调有界,
当1n=时,
1
12x =≤,当1n>时,
1
1
12
1
n
n
n
x
x
x
= +≤
+
,即2( 1,2,)
n
xn≤=",即此数列有界;
又
21
1
0
2
xx?=>,故
11
1
(1 ) (1 )
11(1)(1)
nnnn
nn
xxxx
xx

+
=+?+ =
++++
21
121
0
( 1)( 1) ( 1)( 1)
nn
xx
xx xx
== >
++ ++
"
"
,
即
1nn
x x
+
>,
综上,lim
n
n
x
→∞
存在,令lim
n
n
x A
→∞
=,
又
1
1
1
1
n
n
n
x
x
x
=+
+
,因此
1
1
lim
lim 1
lim 1
n
n
n
n
n
n
x
x
x
→∞
→∞
→∞
=+
+
,即1
1
A
A
A
=+
+
,
解得
1
15
2
A
+
=,
1
15
2
A
= (舍去),故
15
lim
2
n
n
x
→∞
+
=,
5,记(2 1)!! 1 3 5 7 (2 1)nn? =",(2)!! 2468 (2)nn=",
设
(2 1)!!
(2 )!!
n
n
x
n
=,试证明
11
421
n
x
nn
<<
+
,并求极限lim
n
n
x
→∞
,
证 易知
1
(2( 1) 1)!! 2 1
(2( 1))!! 2 2
nn
nn
x x
+
+? +
==?
++
,
当1n=时,
1
111
2
42
x<=<
+
,假设nk=时,
11
421
k
x
kk
<<
+
,则当
1nk=+时,
5
1
1 211 21 21 1 1
22 22 22
4( 1) 4 2 1 2( 1) 1
kk
kkk
xx
kk kk
+
+++
<?<=?<? <
+
,
故
11
421
n
x
nn
<<
+
,由
11
lim lim 0
421
nn
nn
→∞ →∞
= =
+
,故lim 0
n
n
x
→∞
=,