一、内容提要与基本要求
本章介绍了函数的概念、性质与表示法;数列的极限、函数的极限;函数的增量;函数的连续性.函数是高等
数学中研究的主要对象,极限方法是高等数学的主要方法.极限是从量变认识质变,从近似认识**,从有限认
识无限的一种数学方法.本章必须掌握下面几方面的内容:
1.正确理解函数的概念、性质,会求函数的定义域,能将复合函数分解为若干简单函数.
2.正确理解函数的极限,能用ε-δ定义刻画函数的极限,理解xl→imx0
f( x)是否存在与f( x) 在x0 是否有定义
无关,了解极限的一些性质.
3.熟练掌握极限运算法则,正确理解并熟练应用两个重要极限lxi→m0 sinx
x = 1 ,xli→m∞ 1 + 1
x
x
= e .
4.了解无穷小量、无穷大量,掌握函数的极限与无穷小量的关系.
5.正确理解函数y = f( x) 在x0 点处连续的概念,会判断函数的连续性与间断点,了解初等函数的连续
性,掌握闭区间上连续函数的性质.
二、习题一解答
1.判断下列各对函数是否相同,并说明理由:
(1) y = x + 1 与y = x2 - 1
x - 1 ; (2) y = lnx2 与y = 2lnx ;
(3) y = f( x) 与x = f( y) ; (4) y = 1 + 1
x2 与y = 1 + x2
x ;
(5) y = 3 x4 - x3 与y = x ? 3 x - 1 ; (6) y = ax 与y = ex?lna .
解 (1) 不同. y = x + 1 的定义域为( - ∞ ,+ ∞ ) , y = x2 - 1
x - 1 的定义域为(- ∞ ,1) ∪ (1 , + ∞ ) .
(2) 不同. y = lnx2 的定义域为( - ∞ ,0) ∪ (0 , + ∞ ) , y = 2lnx 的定义域为(0 , + ∞ ) .
(3) 相同.定义域和对应关系都相同.
(4) 不同.对应关系不同.
(5) 相同.定义域和对应关系都相同.
(6) 相同.定义域和对应关系都相同.
2.设f( x) = x
x + 1 ,求f 1
2 , f 3
2 , f 1
x ,[ f( x)]2 , f[ f( x)] , f{ f[ … f( x)]}
n个f
.
解 f 1
2 = 1
3 , f 32
= 3
5 , f 1
x = 1
x + 1 , [ f( x)]2 = x2
( x + 1)2 .
由于
f[ f( x)] =
x
x + 1
x
x + 1 + 1
= x
2 x + 1 ,
又
f{ f[ f( x)]} =
x
2 x + 1
x
2 x + 1 + 1
= x
3 x + 1 ,
假定对n = k 均有
f{ f[ … f( x)]}
k个f
= x
kx + 1 ,
对于n = k + 1 ,
f{ f[ … f( x)]}
k+ 1个f
=
x
kx + 1
x
kx + 1 + 1
= x
( k + 1) x + 1 ,
故对于所有的n ,均有
f{ f[ … f( x)]}
n个f
= x
nx + 1 .
3.设f( x) =
1 - x2 , - ∞ < x ≤ 0 ,
- 2x , 0 < x < + ∞ , 求f( - 1) , f(0) , f(1) , f[ f( - 1)] , f[ f(0)] , f[ f(1)] .
解 f( - 1) = 0 , f(0) = 1 , f(1) = - 2 , f[ f( - 1)] = 1 , f[ f(0)] = - 2 , f[ f(1)] = - 3 .
4.求下列函数的反函数及其定义域:
(1) y = 1 - x2 (0 ≤ x ≤ 1) ; (2) y = 2sin3 x x ∈ - π
6 ,π
6 ;
(3) y = 2x
2x + 1 ; (4) y = aln( bx - c) ;
(5) y = ax + b
cx + d ( ad - bc ≠ 0) .
解 (1) y = 1 - x2 (0 ≤ x ≤ 1) . (2) y = 1
3 arcsin x
2 ( - 2 ≤ x ≤ 2) .
(3) y = log2 x
1 - x (0 < x < 1) . (4) y = 1
b (c + ex
a ) ( - ∞ < x < + ∞ ) .
(5) y = b - dx
cx - a x ≠ a
c .
5.求下列各题中所给函数构成的复合函数,再指出其定义域:
(1) y = eu , u = sinx ; (2) y = u - 1 , u = lgx ;
(3) y = u2 , u = cosv , v = x - 1
x2 - 5 x + 6 ; (4) y = au , u = arctanv , v = 3 w , w = t2 - 1 ;
(5) y = arcsinu , u = 1 + ex
解 (1) y = esinx ( - ∞ < x < + ∞ ) .
(2) y = lgx - 1 (10 ≤ x < + ∞ ) .
(3) y = cos2 x - 1
x2 - 5 x + 6 ( - ∞ < x < 2) ∪ (2 < x < 3) ∪ (3 < x < + ∞ ) .
(4) y = aarctan 3 x2 - 1 ( - ∞ < x < + ∞ ) .
(5) 由于无论x 取什么值, u = 1 + ex > 1 ,此时u 值对y = arcsinu 没有意义.因此, y = arcsinu 与u =
1 + ex 不能复合成复合函数.
6.设f( x) =
0 , x ≤ 0 ,
x , x > 0 ,
g( x) =
0 , x ≤ 0 ,
- x2 , x > 0 ,求f[ g( x)] ,g[ f( x)] ,f[ f( x)] ,g[ g( x)] .
解 f[ g( x)] = 0 , g[ f( x)] = g( x) , f[ f( x)] = f( x) , g[ g( x)] = 0 .
7.下列函数中,哪些是复合函数? 如是,它们是怎样合成的?
(1) y = arccos(5 + x3 ) ; (2) y = x3 ? 3x ;
(3) y = cos3 x2 + 1
2 ; (4) y = lg x - 1
x + 1 ;
(5) y = x
3 1 - x2 + 12
arcsinx ; (6) y = lnsin 3 x2 + π
4 .
解 (1) 是. y = arccosu , u = 5 + x3 .
(3) 是. y = u3 , u = cosv , v = x2 + 1
2 .
(4) 是. y = lgu , u = v , v = x - 1
x + 1 .
(6) 是. y = lnu , u = sinv , v = w , w = 3 x2 + π
4 .
(2) 、(5)题的函数不是复合函数。
倡8.根据极限定义证明(打“ 倡”的是可选题,以下各章同)
(1) nli→m∞
3 n + 1
2 n + 1 = 3
2 (用“ε-N”语言证明) ; (2) xli→m∞
1 + x3
2 x3 = 1
2 (用“ε-X”语言证明) ;
(3) lxi→m2
(5 x + 2) = 12 (用“ε-δ”语言证明) .
证 (1) 对任意给定的ε> 0 ,总存在正整数N = 1
ε .当n > N 时,
3 n + 1
2 n + 1 - 32
= 1
2(2 n + 1) < 1
n < ε,
nli→m∞
3 n + 1
2 n + 1 = 32
.
(2) 对任意给定的ε> 0 ,总存在正数X =
3 1
2ε .当x > X 时,
1 + x3
2 x3 - 1
2 = 1
2 x3 = 1
2 x 3 < 1
2 X3 = ε,
xli→m∞
1 + x3
2 x3 = 1
2 .
(3) 对任意给定的ε> 0 ,总存在正数δ = ε
5 .当0 < x - 2 < δ时,
5 x + 2 - 12 = 5 x - 2 < 5δ = ε,
lxi→m2
(5 x + 2) = 12 .
9.求下列极限:
(1) xl→im- 1
(3 x3 - 5 x + 2) ; (2) lim x → 2
x2 + 1
x4 - 3 x2 + 1 ;
(3) lxi→m2
x2 - 3
x - 2 ; (4) lxi→m3
x2 - 2 x - 3
x - 3 ;
(5) lxi→m9
4 x - 3
x - 3
; (6) lhi→m0
( x + h)3 - x3
h ;
(7) lxi→m0
x + 1 - ( x + 1)
x + 1 - 1
; (8) lxi→m4
2 x + 1 - 3
x - 2
;
(9) lxi→m1
xm - 1
xn - 1 ( m ,n 为自然数) ; (10) nli→m∞
2 n + 1
n2 + n
;
(11) xli→m∞
(2 x2 + 1)2
x2 + 3 ; (12) xli→m∞
(2 x + 1)3 ( x - 3)2
x5 + 4 ;
(13) xli→m∞
2 x2 - 6 x + 5
x3 - 8 x2 + 1 ; (14) xl→im+ ∞ eax - 1
eax + 1 ( a > 0) ;
(15) nli→m∞
1
n2 + 2
n2 + … + n
n2 ; (16) nli→m∞
( n + 1 - n) ;
(17) xl→im- 1
1
x + 1 - 3
x3 + 1 ; (18) x l→im+ ∞ x x2 + 1 - x .
解 (1) xl→im- 1
(3 x3 - 5 x + 2) = 3 xl→im- 1 x3 - 5 xl→im- 1 x + 2 = 4 .
(2) lim x → 2
x2 + 1
x4 - 3 x2 + 1 = lim x → 2
( x2 + 1)
lim x → 2
( x4 - 3 x2 + 1) = 3
- 1 = - 3 .
(3) lxi→m2
x2 - 3
x - 2 = ∞ 因为lxi→m2
x - 2
x2 - 3 = 0 .
(4) lxi→m3
x2 - 2 x - 3
x - 3 = lxi→m3
( x - 3)( x + 1)
x - 3 = 4 .
(5) lxi→m9
4 x - 3
x - 3
= lxi→m9
4 x - 3
(4 x - 3)(4 x + 3)
= 1
2 3
.
(6) lhi→m0
( x + h)3 - x3
h = lhi→m0
(3 x2 + 3 xh + h2 ) h
h = 3 x2 .
(7) lxi→m0
x + 1 - ( x + 1)
x + 1 - 1
= lxi→m0
x + 1(1 - x + 1)
x + 1 - 1
= - lxi→m0 x + 1 = - 1 .
(8) lxi→m4
2 x + 1 - 3
x - 2
= lxi→m4
( 2 x + 12 - 32 )( x + 2)
( x2 - 22 )( 2 x + 1 + 3)
= lxi→m4
(2 x - 8)( x + 2)
( x - 4)( 2 x + 1 + 3)
= lxi→m4
2( x + 2)
2 x + 1 + 3
= 4
3 .
(9) lxi→m1
xm - 1
xn - 1 = lxi→m1
xm- 1 + xm- 2 + … + 1
xn- 1 + xn- 2 + … + 1 = mn
( m ,n 为自然数) .
(10) nli→m∞
2 n + 1
n2 + n
= nli→m∞
2 + 1
n
1 + 1
n
= 2 .
(11) xli→m∞
(2 x2 + 1)2
x2 + 3 = ∞ .
(12) xli→m∞
(2 x + 1)3 ( x - 3)2
x5 + 4 = 8 .
(13) xli→m∞
2 x2 - 6 x + 5
x3 - 8 x2 + 1 = 0 .
(14) xl→im+ ∞ eax - 1
eax + 1 = xl→im+ ∞
1 - 1
eax
1 + 1
eax
= 1 ( a > 0) .
(15) nli→m∞
1
n2 + 2
n2 + … + n
n2 = nli→m∞
n( n + 1)
2 n2 = 1
2 .
(16) nli→m∞
( n + 1 - n) = nli→m∞
1
n + 1 + n
= 0 .
(17) xl→im- 1
1
x + 1 - 3
x3 + 1 = xl→im- 1
x2 - x + 1 - 3
x3 + 1 = xl→im- 1
( x + 1)( x - 2)
( x + 1)( x2 - x + 1) = - 1 .
(18) x l→im+ ∞ x x2 + 1 - x = x l→im+ ∞ x x2 + 1 - x x2 + 1 + x
x2 + 1 + x
= xl→im+ ∞
x
x2 + 1 + x
= xl→im+ ∞
1
1 + 1
x2 + 1
= 1
2 .
10.下列函数在给定条件下,哪些是无穷小? 哪些是无穷大?
(1) 1 + 2 x2
x ( x → 0) ; (2) sinx
x ( x → ∞ ) ;
(3) lgx ( x → 0+ ) ; (4) 2 x + 5 ( x → - ∞ ) ;
(5) x + 1
x2 - 4 ( x → 2) ; (6) 1 - cos2 t ( t → 0) .
解 (2) ,(6)为无穷小;(1) ,(3) ,(4) ,(5)为无穷大.
11.x2 , x2 - 1
x3 ,e- x 何时是无穷大? 何时是无穷小?
解 x → ∞ 时, x2 → ∞ , x2 - 1
x3 → 0 ;
x → + ∞ 时,e- x → 0 ;
x → - ∞ 时,e- x → ∞ ;
x → ± 1 时, x2 - 1
x3 → 0 ;
x → 0 时, x2 → 0 , x2 - 1
x3 → ∞ .
12.x → 1 时,下列函数中哪个是1 - x 的高阶无穷小? 哪个是1 - x 的等阶无穷小?
(1) (1 - x) 3
2 ; (2) 1 - x
1 + x ; (3) 2(1 - x) .
解 (1) 因为lxi→m1
(1 - x) 3
2
1 - x = lxi→m1
(1 - x) 1
2 = 0 ,所以当x → 1 时,(1 - x) 3
2 较1 - x 为高阶无穷小.
(2) 因为lxi→m1
1 - x
1 + x
1 - x = lxi→m1
1
1 + x = 1
2 ,所以当x → 1 时,1 - x
1 + x 与1 - x 是同阶无穷小.
(3) 因为lxi→m1
2(1 - x)
1 - x = lxi→m1
2
1 + x
= 1 ,所以当x → 1 时,2(1 - x) 与1 - x 等价,即2(1 - x) ~ (1 - x) .
13.设有函数
f( x) =
( x + a)2 - a2
x , x < 0 ,
x - 2 , 0 < x ≤ 1 ,
x2 - 5 x + 4
x2 + x - 2 , x > 1 .
(1) 求xl→im- ∞ f( x) ,xl→im+ ∞ f( x) ; (2) a 为何值时,lxi→m0 f( x) 存在;
(3) 求lxi→m1 f( x) .
解 (1) x l→im- ∞ f( x) = x l→im- ∞
( x + a)2 - a2
x = x l→im- ∞
x2 + 2 ax
x = x l→im- ∞
( x + 2 a) = - ∞ ,
x l→im+ ∞ f( x) = x l→im+ ∞
x2 - 5 x + 4
x2 + x - 2 = 1 .
(2) 因为
lim x → 0 + f( x) = lim x → 0 +
( x - 2) = - 2 ,
lim x → 0 - f( x) = lim x → 0 -
( x + a)2 - a2
x = lim x → 0 -
( x + 2 a) = 2 a ,
所以
a = - 1 时,lxi→m0 f( x) 存在.
(3) 因为
lim x → 1 + f( x) = lim x → 1 +
x2 - 5 x + 4
x2 + x - 2 = lim x → 1 +
( x - 1)( x - 4)
( x - 1)( x + 2) = - 1 ,
lim x → 1 - f( x) = lim x → 1 -
( x - 2) = - 1 ,
所以
lxi→m1 f( x) = - 1 .
14.已知xli→m∞
x2 + 1
x + 1 - ax - b = 0 ,试确定a ,b的值.
解 xli→m∞
x2 + 1
x + 1 - ax - b = xli→m∞
(1 - a) x2 - (a + b) x - b + 1
x + 1 .
因为极限存在,所以1 - a = 0 ,即a = 1 ,从而
原式= xli→m∞
- (1 + b) x - b + 1
x + 1 = - (1 + b) .
由给定条件知- (1 + b) = 0 ,所以b = - 1 .
15.求下列极限:
(1) lxi→m0 sin3 x
sin4 x ; (2) lxi→m0 tan3 x
sin5 x ;
(3) xli→m∞ xsin 1
x ; (4) lxi→m0 xsin 1
x ;
(5) lxi→mπ sinx
π - x ; (6) lhi→m0
1 - cos2 x
xsinx ;
(7) nli→m∞
2n sin a
2n ( a ≠ 0) ; (8) lxi→m0
x + 2sinx
x + sinx ;
(9) xl→im- ∞ x sin 1
x2 ; (10) xli→m∞ 1 + k
x
x
;
(11) lxi→m0 1 + x
2
x - 1
x ; (12) lxi→m0
(1 + 2tanx)cotx ;
(13) lxi→m0
(cosx) 1
1 - cosx ; (14) xli→m∞
x + 3
x
x + 2
.
解 (1) lxi→m0 sin3 x
sin4 x = lxi→m0
sin3 x
3 x ? 3 x
sin4 x
4 x ? 4 x
= 3
4 lxi→m0
sin3 x
3 x
sin4 x
4 x
= 3
4 .
(2) lxi→m0 tan3 x
sin5 x = lxi→m0
sin3 x
3 x ? 3 x
cos3 x
sin5 x
5 x ? 5 x
= 3
5 .
(3) xli→m∞ xsin 1
x = xli→m∞
sin 1
x
1
x
= 1 .
(4) 因为lxi→m0 x = 0 ,而sin 1
x 为有界函数,即sin 1
x ≤ 1 ,所以
lxi→m0 xsin 1
x = 0
(5) lxi→mπ sinx
π - x = lxi→mπ sin(π - x)
π - x = 1 .
(6) lxi→m0
1 - cos2 x
xsinx = 2 lxi→m0 sin2 x
xsinx = 2 lxi→m0 sinx
x = 2 .
(7) nli→m∞
2n sin a
2n = nli→m∞
sin a
2n
a
2n
? a = a( a ≠ 0) .
(8) lxi→m0
x + 2sinx
x + sinx = lxi→m0
1 + 2 sinx
x
1 + sinx
x
= 3
2 .
(9) xl→im- ∞ x sin 1
x2 = xl→im- ∞
- sin 1
x2
1
x2
= - 1 .
(10) xli→m∞
1 + k
x
x
= xli→m∞
1 + k
x
x
k k
= ek .