1
第八节 函数的连续性
习题 1-8
1,讨论下列函数的连续性,并画出函数的图形,
(1)
3
1,0 1,
()
3,1 2;
xx
fx
xx
+≤<?
=
≤≤?
(2)
2
1,0,
()
1,0.
xx
fx
xx
<?
=
≥?
解 (1) 易知()f x在[0,1)和(1,2]上连续,在1x =点处,
11
lim ( ) lim (3 ) 2 (1)
xx
f xxf
++
→→
=?==;
3
11
lim ( ) lim ( 1) 2 (1)
xx
f xx f

→→
=+==,
故()f x在1x =处也连续,即函数在定义域[1,2]上连续,如图1.5,
(2) 函数定义域为[1,1]?,易知函数在[1,0)?和(0,1]上连续,在0x =点处,
2
00
lim ( ) lim 1 1 (0)
xx
f xxf
++
→→
=?==;
00
lim ( ) lim ( 1) 1 (0)
xx
f xx f

→→
=?=?≠,
故0x =是()f x的跳跃间断点,如图1.6,
2,指出下列函数的间断点及其类型,如果是可去间断点,则补充或改变函数的定义使之连续,
(1)
1
siny
x
= ; (2)
arcsin x
y
x
= ;
(3)
2
2
1
32
x
y
xx
=
+; (4)
2
1,0,
()
2,0.
xx
fx
xx
+ >?
=
≤?
解 (1) 函数在0x =处无定义,且当0x →时,函数值在1?和1之间无限次的变动,称0x =是函数的振荡间断点,
(2) 因为
0
arcsin
lim 1
x
x
x
→
=,所以0x =是可去间断点,补充定义(0) 1y =,则函数连续,
(3) 函数在1x =和2x =处无定义,
O
y
x1
2
2
1
图1.5
x
1?
O
y
1?
1
图1.6
1
2
因为
2
2
11
11
lim lim 2
232
xx
xx
xxx
→→
+
==?
+
,所以1x =是函数的可去间断点,补充定义
(1) 2y =?,则函数连续;
因为
2
2
2
1
lim
32
x
x
xx
→
=∞
+
,所以2x =是无穷间断点,
(4) 因为
2
00
lim ( ) lim ( 1) 1
xx
fx x
++
→→
=+=;
00
lim ( ) lim (2 ) 2
xx
fx x

→→
=?=,所以0x =是函数的跳跃间断点,
3,设函数()f x在点
0
x处连续,证明它的绝对值()f x亦在点
0
x处连续,
证 由()f x在
0
x x=连续,故0ε? >,0δ? >,当
0
xx δ? <时,恒有
0
() ( )fx fx ε?<,故
00
() ( ) () ( )fx fx fx fx ε? ≤? <,
即()f x在
0
x也连续,
4,讨论函数
2
2
1
() lim
1
n
n
n
x
f xx
x
→∞
=
+
的连续性,若有间断点,判断其类型,
解 易知
2
2
1,
1
() lim 0 1,
1
1,
n
n
n
xx
x
fx x x
x
xx
→∞
>
===
+
<
当当当
在1x =?处,
11
lim ( ) lim 1
xx
fx x
++
→? →?
==?,
11
lim ( ) lim ( ) 1
xx
fx x

→? →?
=?=,所以1x =?为跳跃间断点;
在1x =处,
11
lim ( ) lim ( ) 1
xx
fx x
++
→→
=?=?,
11
lim ( ) lim 1
xx
f xx

→→
= =,所以1x =为跳跃间断点,
5,计算下列极限,
(1)
1
limsin(2 1)
x
x
→; (2)
π
4
lim ln(tan )
x
x
→;
(3)
22
lim ( 2 )
x
x xx
→+∞
+ ; (4)
2
22
lim
2
x
x
x
→
+?;
(5)
2
0
1
sin
lim
sin 2
x
x
x
x
→; (6)
1
lim ( 1 1)
x
x
x
→+∞
+?,
解 (1)
1
limsin(2 1) sin1
x
x
→
=,
3
(2)
π
4
π
lim ln(tan ) ln(tan ) 0
4
x
x
→
==,
(3)
22
22
2
2
1
2
lim ( 2 ) lim lim
21
2
11
xxx
x
x
xxx
xxx
xx
→+∞ →+∞ →+∞
+
+
+ = =
++?
+ +?
1
2
=,
(4)
22 2
22 2 1 1
lim lim lim
24
(2)( 22) 22
xx x
x
xx x
→→ →
+
===
++ ++
,
(5)
22
000
11
sin sin
1
lim lim lim sin 0
sin 2 2 2
xxx
xx
x
xx
x
→→→
===,
(6)
11
lim(1 1)lim( 1)lim ( 1 )
xxx
x
x xxxx
x
x
→+∞ →+∞ →+∞
+
+?=?= +?
11
lim lim
2
11
11
xx
x
xx
x
→+∞ →+∞
===
++
++
,
6,计算下列极限,
(1)
0
tan 2
lim
x
x
x
→; (2)
2
21
lim e
x
x
x
+
→∞;
(3)
2
2cot
0
lim(1 2 tan )
x
x
x
→
+ ; (4)
2
2
2
1
lim ( )
1
x
x
x
x
→∞
+;
(5) lim [ln(1 ) ln ]
n
nn
→∞
+? ; (6)
1
cot
1
lim ( )
x
x
x
x
→∞
,
解 (1)
00
tan 2 2
lim lim 2
xx
xx
→→
==,
(2)
22
21 21
lim
0
lim e e e 1
x
xx
x
→∞
++
→∞
===,
(3)
2
2
1
2cot 2 2 2
2tan
00
lim(1 2 tan ) [lim(1 2 tan ) ] e
x
x
xx
xx
→→
+ =+ =,
(4)
22 2
2
22
12 2
2
lim
2
2 11
22
12
lim ( ) lim (1 ) e e
11
x
xx x
x
xx
xx
x
→∞
+

++
→∞ →∞
=? = =
++
,
4
(5) 因为
11
lim [ln(1 ) ln ] lim ln(1 ) ln lim (1 ) 1
xx
xxx
xxx
→∞ →∞ →∞
+? = + = + =,所以
lim [ln(1 ) ln ] 1
n
nn
→∞
+?=,
(6)
1
1
lim ( )cot
11
cot ( )( ) tan
1
lim ( ) lim (1 ) e e
x
x
x
x
xx
xx
x
xx
→∞

→∞ →∞
=? = =,
7,设函数
11
,0,
(),0,
arcsin
,0.
2
x
x
x
fx b x
ax
x
x
+?
>
= =
<
试求a、b,使()f x处处连续,
解 ()f x处处连续,则必在0x =处连续,故
00 0
1
11 1
2
lim ( ) lim lim (0)
2
xx x
x
x
f xfb
xx
++ +
→→ →
+?
====,即
1
2
b = ;
00
arcsin
lim ( ) lim (0)
22
xx
ax a
f b
x

→→
====,故21ab= =,