第三讲 函数的连续性与导数.微分的概念
第三讲 函数的连续性与导数、微分的概念
一、单项选择题(每小题4分,共24 分)
1.若f (x )为是连续函数,
且f (0)=1, f (1)=0, 则lim f x sin x →∞⎛⎝1⎫⎪=( ) x ⎭
A . -1 B .0
C .1 D . 不存在
解: 原式
1⎤⎡sin f 连续⎡⎢1⎤⎥=f (1)=0,选B f ⎢lim x sin ⎥=f ⎢lim x →∞⎥x ⎦⎣x →∞⎢⎥x ⎦⎣
2. 要使f (x )=ln (1+kx )在点x =0处连续,应给f (0)补充定义的数值是( )
A . km B . m x k m
km C . ln km D . e
m ⎡⎤解: lim f (x )=ln ⎢lim(1+kx ) x ⎥ x →0⎣x →0⎦
lim kx ⋅m
x =ln e x →0=ln e km =km
∴f (0)=km 选A
3.若lim f (x ) =A ,则下列正确的是 ( ) x →a
A . lim f (x )=A x →a
B .
x →a
=
C . lim f (x )=-A x →a
D . lim f (x ) =A x →a
解:
x →
=选B
⎧f (x ), x ≠0⎪F x =4.设()⎨x
⎪f (0), x =0⎩
且f (x )在x =0处可导,f '(0)≠0,
f (0)=0, 则x =0是F (x )的 ( )
A . 可去间断点 B . 跳跃间断点
C . 无穷间断点 D . 连续点
解: lim F (x )=lim x →0x →0f (x )-f (0)x -0
x →0=f '(0), f '(0)≠f (0)∴F (0)=f (0)≠lim F (0),故x =0是F (x )的第一类可去间断点。选A
1⎧x sin ⎪5.f (x )=⎨x , x ≠0在x =0处 ( )
⎪⎩0, x =0
A . 极限不存在 B .极限存在但不连续
C .连续但不可导 D .可导但不连续
解: lim f (x )=lim x ⋅sin x →0x →01=0,且f (0)=0 x
∴f (x )在x =0连续,又 f '(0)
1x sin -0=lim =不存在,∴f (x )在x =0不可导 选C x →0x -0
⎧x 2+1, x ≤16.设f (x )=⎨在x =1可导,则a , b 为 ( ) ⎩ax +b , x >1
A . a =-2, b =2 B . a =0, b =2
C . a =2, b =0 D . a =1, b =1
解:(1) f (x )在x =1连续,
2∴lim x +1)=2, lim (ax +b )=a +b (-+x →1x →1
故a +b =2⋯(1)
x 2-1=2, f +'(1) (2)f -'(1)=lim -x →1x -1
a (x -1)ax +b -2(1)=lim lim =a x →1+x →1+x -1x -1
∴a =2,代入(1)得b =0,选C
二、 填空题(每小题4分,共24分)
7.设f (x ) 为连续奇函数,则f (0)
解:(1) f (x )为奇函数,∴f (-x )=-f (x )
-f (x )⎤(2) lim f (-x )=lim ⎡⎦ x →0x →0⎣
又 f (x )在x =0连续
∴f (0)=-f (0) 故f (0)=0
8.若f (x )为可导的偶函数,则f '(0)=解:(1) f (x )为偶函数,∴f (-x )=f (x )
(2) f (x )可导,∴-f '(-x )=f '(x ) 故-f '(0)=f '(0)
2f '(0)=0 即f '(0)=0
9.设y =6x +k 是曲线y =3x 2-6x +13的
一条切线,则k =
解: (1) y '=6, y '=6x -6, ∴6x -6=6, x =2
(2)6⨯2+k =3⨯4-6⨯2+13, ∴12+k =12-12+13, 故k =1
10. 若y =f (x ) 满足:f (x ) =f (0)+x
+α(x ),且lim α(x )=0 x →0x
则f '(0)=
解:f '(0)=lim x →0f (x )-f (0)x -0
=lim x →0x -α(x )x =1+0=1
11. 设f (x ) 在x =2连续,且f (2)=4, 则lim f (x ) x →24⎫⎛1-2⎪= ⎝x -2x -4⎭
x →2解: 原式=f (2)lim x +2-4 2x -4
=4lim x →211=4⋅=1 x +24
12.f (x ) =sin x ⋅(x -1)的间断点个数为 x 5-x
52解: 令x -x =0, x (x -1)(x +1)x +1=0 ()
x =0, x =-1, x =1为间断点,
故f (x )有三个间断点
三 、计算题(每小题8分,共64分)
⎧sin 2x +e 2ax -1, x ≠0⎪13. 已知f (x ) =⎨ x ⎪a , x =0⎩
在(-∞, +∞)上连续,求a 的值
解: f (x )在x =0连续
sin 2x +e 2ax -1sin 2x e 2ax -1∴lim f (x )=lim =lim +lim =2+2a x →0x →0x →0x →0x x x
且f (0)=a , ∴2+2a =a
故a =-2
⎧1
⎪e x , x
⎪ln x ⎪, x >1⎩x -1
e =0, lim 0=0 解:(1)在x =0处, lim +-x →0x →01x
且f (0)=0
∴f (x )在x =0处连续
0=0, (2)在x =1处, lim -x →1
x →1lim +ln (1+t )ln x x -1=t =lim =1 +x →0x -1t
∴f (x )在x =1不连续
⎧f (x )+a sin x , x ≠0⎪15. 设f (x ) 有连续的导函数,且f (0)=0, f '(0)=b 若F (x )=⎨在x =0连续,求常数A 。 x ⎪A , x =0⎩
解: lim F (x )=lim x →0x →0f (x )-f (0)+a sin x x
=lim x →0f (x )-f (0)x -0+lim a sin x =f '(0)+a x →0x
且F (0)=A ,∴a +b =A 答A =a +b
⎧e x -1, x
⎪kx +b , x ≥0⎩
e x -1解:(1) f (x )在x =0连续,∴lim =1 x →0-x
x →0+lim (kx +b ) =b 故有b =1
(2) f (x )在x =0可导
e x -1-1x f -'(0)=lim - x →0x -0
⎛0⎫ ⎪e x -1-x ⎝0⎭e x -11=lim lim = 2x →0-x →02x 2x
kx +1-1=k , x →0x
11∴k =,答k =, b =1 22f +'(0)=lim
⎧ln(1+ax ) , x ≠0⎪17.设f (x ) =⎨在x =0可导,求a 与f '(0) x ⎪⎩-1, x =0
解:(1) f (x )在x =0连续,
∴lim f (x )=lim x →0ln (1+ax )ax =lim =a x →0x x →0x
且f (0)=-1,故有a =-1
(2) f (x )在x =0可导
ln(1-x ) +1x f '(0)=lim x →0x
⎛0⎫1 ⎪+1ln (1-x )+x ⎝0⎭ =lim lim 2x →0x →02x x
=lim 1+x -11=- x →02x x -12
1 2答:a =-1, f '(0)=-
18. 讨论f (x ) =x -a ϕ(x )在x =a 是否可导,其中ϕ(x )在x =a 连续。 解:(1)f -'(a )=lim -x →a (x -a )ϕ(x )-0x -a
=lim -x →a -(x -a )ϕ(x )x -a ϕ(x ) =-lim -x →a ϕ连续-ϕ(a )
(2)f +'(a )=lim +x →a (x -a )ϕ(x )-0x -a
=lim ϕ(x )-x →a =lim +x →a (x -a )ϕ(x )x -a ϕ连续ϕ(a )答: 当ϕ(a )=0时,f (x )在x =a 连续,
当ϕ(a )≠0时,f (x )在x =a 不连续
19. 求f (x ) =1的间断点,并指出间断点类型 ln x
解:(1) 间断点:x =0, x =-1, x =1
(2) 在x =0处: lim 1=0 x →0ln x
∴x =0是f (x )的第一类间断点。
(3) 在x =±1处: lim 1=∞ x →±1ln x
∴x =±1为f (x )的第二类无穷间断点。
⎧x 1⎪e -1, x >0f (x ) =20. 设指出f (x ) 的间断点,并判断间断点的类型。 ⎨⎪⎩ln (1+x ), -1
解:(1)x =1为间断点,x =0可能是间断点。
(2)在x =1处:
lim e -x →11x -1=e -∞=0, lim e +x →11x -1=∞
∴x =1是f (x )的第二类无穷间断点
(3)在x =0处:
lim e +x →01x -1=e -1, lim ln (1+x )=0 -x →0
∴x =0是f (x )的第一类跳跃间断点
四、 综合题(每小题10分,共20分)
11-
21. 求f (x ) =的间断点,并判别间断点的类型。 -x -1x
解: (1)间断点:x =0,x =-1, x =1
(2)在x =0处:
f (x )=x (x -1)x -11 ⨯=x (x +1) 1x +1
x →0 lim f (x )=lim x →0x -1=-1 x +1
∴x =0是f (x )的第一类可去间断点
(3)在x =1处: lim f (x )=lim x →1x →1x -1=0 x +1
∴x =1是f (x )的第一类可去间断点
(4)在x =-1处: lim
x -1=∞ x →-1x +1
∴x =-1是f (x )的第二类无穷间断点
⎧x 2+x , x ≤0⎪22.已知f (x ) =⎨ax 3+bx 2+cx +d ,0
解:(1) f (x )在x =0连续,
∴x lim →0+(ax 3+bx 2+cx +d )=d
lim x →0-(x 2+x )=0, f (0)=0
故d =0⋯(1)
(2) f (x )在x =0可导
f (0)=lim x 2+x
-'x →0-x =1,
f 0)=lim ax 3+bx 2+cx
+'(x →0x =c
故有c =1⋯(2)
(3) f (x )在x =1连续,
lim x →1-(ax 3+bx 2+x )=f (1)
即a +b +1=f (1)=0
∴a +b +1=0⋯(3)
(4) f (x )在x =0可导:
x 2
∴f -x
+'(1)=lim +x -1=1 x →1
(ax 3
f 1)=lim +bx 2+x
-'x -1 x →1-
⎛ 0⎫
⎝0⎪⎭2
x lim →1-(3ax +2bx +1)
=3a +2b +1
故有3a +2b =0⋯(4)
由(3)(4)解得a =2, b =-3
答:a =2, b =-3, c =1, d =0
五、证明题(每小题9分,共18分)
23. 证明x -2x -4=0在区间(-2,2)内至少有两个实根。 4
证:(1) f (x ) 在[-2,0]连续,
且f (0)=-40
∴由零点定理知,
f (x ) =0在(-2,0)上至少有一个实根。
(2) f (x ) 在[0,2]连续, 且
f (0)=-40
∴由零点定理知,
f (x ) =0在(0,2)上至少有一个实根
(3)综上所述,f (x ) =0在(-2,2)上至少有两个实根
1⎧u ⎪x sin , x ≠024. 设f (x )=⎨,证明(1)当u >0时f (x )在x =0连续,当u >1时,f (x )在x =0x ⎪⎩0, x =0
可导
解:(1) lim x sin x →0u 1u >0时0 x
⎛1u ⎫sin ≤1,lim x 0⎪ x →0x ⎝⎭
∴当u >0时,f (x )在x =0连续 1
=lim x u -1sin 1u >1时0 (2) lim x →0x →0x -1x x u sin
⎛1u -1⎫ sin ≤1,lim x 0⎪ x →0x ⎝⎭
当u >1时,f (x )在x =0可导
总之,当u >0时,f (x )在x =0连续
当u >1时,f (x )在x =0可导
选做题
设对于任意的x ,函数满足f (1+x )=
af (x )且f '(0)=b , 证明f '(1)=a ⋅b
证:(1)令x =0,f (1+0) =af (0),即f (1)=af (0)
(2) f '(1)=lim x →0f (1+x )-f (1)x
=lim x →0af (x )-af (0)x =af '(0)=a ⋅b
证毕