8/9/2019 Chuong 3-X

1/22

1

1

Gii tch ton hc. Tp 1. NXB i hc quc gia H Ni 2007.

Tkho: Gii tch ton hc, gii tch, Hm lin tc, im gin on, lin tc, lin tc

u, hm scp.

Ti liu trong Thvin in tH Khoa hc Tnhin c th c sdng cho mc

ch hc tp v nghin cu c nhn. Nghim cm mi hnh thc sao chp, in n phc

v cc mc ch khc nu khngc schp thun ca nh xut bn v tc gi.

Mc lc

Chng 3 Hm lin tc mt bin s................................................................................. 3

3.1 nh ngha s lin tc ca hm s ti mt im......................................................3

3.1.1 Cc nh ngha .................................................................................................. 3

3.1.2 Hm lin tc mt pha, lin tc trn mt khong, mt on kn ......................4

3.1.3 Cc nh l v nhng php tnh trn cc hm lin tc......................................5

3.1.4 im gin on ca hm s.............................................................................. 7

3.2 Cc tnh cht ca hm lin tc............................................................................... 10

3.2.1 Tnh cht bo ton du ln cn mt im....................................................10

3.2.2 Tnh cht ca mt hm s lin tc trn mt on...........................................10

3.3 iu kin lin tc ca hm n iu v ca hm s ngc...................................14

3.3.1 iu kin lin tc ca hm n iu..............................................................14

3.3.2 Tnh lin tc ca hm ngc .......................................................................... 15

3.4 Khi nim lin tc u...........................................................................................16

Chng 3. Hm lin tc mt bin s

L Vn Trc

8/9/2019 Chuong 3-X

2/22

2

3.4.1 Mu............................................................................................................16

3.4.2 nh ngha ...................................................................................................... 16

3.4.3 Lin tc ca cc hm s scp ...................................................................... 18

3.5 Bi tp chng 3.................................................................................................... 19

8/9/2019 Chuong 3-X

3/22

3

3

Chng 3

Hm lin tc mt bin s

Khi nim lin tc ca hm s l khi nim rt cbn, ng mt vai tr rt quan trngtrong vic nghin cu hm s c v l thuyt v ng dng. Trc ht, ta hy tm hiu v tnhlin tc ca hm s.

3.1nh ngha slin tc ca hm s ti mt im

3.1.1 Cc nh ngha

nh ngha 1Cho :f A v 0x A . Ta ni rng hm flin tc tiim 0x nu vibt k s 0 > cho trc c thtm c s 0 > sao cho x A m 0| |x x < ta c

0| ( ) ( )|f x f x < .

Hmflin tc ti mi im x A th ta niflin tc trn A.

Nu fkhng lin tc tix0, ta ni rng fgin on tix0.

Quay trvnh ngha gii hn ca hm s ta c th pht biu s lin tc ca hmftix0nh sau:

Tnh cht

Gi s A f : v 0x A . Khi hmflin tc tix0 khi v ch khi

=

00l im ( ) ( )

x xf x f x .

nh ngha 2Hmflin tc ti im 0x A nu nh mi dy {xn} nm trong A, m 0nx x ta u c 0( ) ( )nf x f x khi n .

V d 1: Xt hm s ( ) | | .f x x= Ly bt k 0x

Do 0 0 0| ( ) ( )| | | | | | | | |f x f x x x x x = .

Cho nn 0 > , chn = , th

8/9/2019 Chuong 3-X

4/22

4

Trc ht, nu chn 1 < th | ( ) 10 | 3 .5 15f x < =

Ty, nu ta chn15

< th | ( ) 10|f x < .

Cui cng ta hy chn = min{1, }15 th khi | 2|x < ta c | ( ) 10|f x < , iu phi

chng minh.

V d 3: Dng ngn ng ( ) hy chng minh hm sy = x2 lin tc ti mi im. Gi sx0 l tu , cho 0 > v 0| |x x < . Ta thy

20 0 0

0 0 0 0 0 0

| ( ) ( )| | | | | | |

| | | 2 | | | (| | 2| | )

y x y x x x x x x x

x x x x x x x x x x

= = + =

= + +

nn 0 0| | (2| | )y y x < + . Nu chn 1, < theo trn ta c 0 0| | (2| | 1)y y x < + . Mt khc,

nu 02| | 1x

< + th 0| |y y < . Vy hy chn 0min{1, },2| | 1x

= + th khi 0| |x x < ta c

0| |y y < . Vy hm s lin tc tix0

V d 4: Xt hm sf(x)=sinx. Lyx0 bt k thuc . Ta thy

0 00

0 00

| sin sin | | 2cos .sin |2 2

2| sin | 2| | | |2 2

x x x x x x

x x x x x x

+ =

=

(ta bit| sin | | |x x< vi 0x )

Cho trc >0 chn = , khi 0| |x x < ta c 0| sin sin |x x < . Vy hmflin tc timi x .

3.1.2 Hm lin tc mt pha, lin tc trn mt khong, mt on kn

Cho hm :f A R v 0x A

nh ngha 3Hmflin tc bn phi ti im 0x A nu > 0 cho trc, 0 > sao cho x A m 0 0x x x < + ta c

|f(x)f(x0)|< , (3.1.4)

k hiu0

0l im ( ) ( )x x

f x f x +

= (3.1.4)

nh ngha 4 Hmflin tc bn tri ti im 0x A nu > 0 cho trc, 0 > sao cho

x A m 0 0x x x <

ta c |f(x) f(x0)|< , (3.1.5)

k hiu0

0l im ( ) ( )x x

f x f x

= (3.1.5)

Cc hm s lin tc bn phi hoc bn tri c gi l lin tc mt pha.

8/9/2019 Chuong 3-X

5/22

5

5

nh l 3.1.1iu kin cn v hmflin tc ti im 0x A l n lin tc theo c haipha ti 0x .

Chng minh:

iu kin cn l hin nhin.

Ngc li, nu f lin tc theo c hai pha ti 0x , th > 0 , 1 0 > sao chox A , 0 10 x x < ta c

|f(x) f(x0)|<

v 2 0 > sao cho x A , < 2 0 0x x ta c

|f(x) f(x0)|< .

Khi gi 1 2min( , ) = , th 0,| |x A x x < ta c

|f(x) f(x0)|< .Vyflin tc tix0

nh ngha 5 Cho hmfxc nh trn khong (a,b). Ta ni rng hmflin tc trn khong(a,b) nu n lin tc ti mi im ca khong .

By gicho hm fxc nh trn on [a,b]. Nu hm f lin tc trn (a,b), lin tc bnphi ti im a v lin tc bn tri ti im b, th ta ni rng hmflin tc trn [a,b].

V d 5: Xt hm s

1 v i 0 2

( ) 1 v i 2 3

0 v i c c gi tr cn l i .

x

f x x

x

=

<

Ta thy hm s khng lin tc ti cc imx = 0,x = 2,x = 3 v chng hn tix = 2, tac:

2 2l im ( ) 1, l im ( ) 1x x

f x f x +

= = .

V d 6: Hm s1

( )f xx a

=

khng lin tc ti imx =a v ti a hm s khng xc nh.

3.1.3 Cc nh l v nhng php tnh trn cc hm lin tc

a) Tnh lin tc ca tng hiu tch v thng ca hm lin tc

T cc nh l v gii hn ca tng, hiu, tch v thng ca hai hm s m mi hm uc gii hn ta c th chng minh nh l sau.

nh l 3.1.2Nu hai hm sfvgxc nh trn cng mt tp RA v c hai u lin tc

ti im 0x A th ti im cc hm c.f trong c l hng s; ; .f g f g vf

g(vi

0( ) 0g x ) cng lin tc.

V d 7: Hm a thc 10 1 1...n n

n ny a x a x a x a

= + + + + lin tc trn ton tp v phn thc

hu t

8/9/2019 Chuong 3-X

6/22

6

10 1 1

10 1 1

...

...

n nn n

m mm m

a x a x a x a y

b x b x b x b

+ + + +=

+ + + +lin tc ti cc gi trx tr cc gi tr lm cho mu s

trit tiu.

ii) Hm m = > ( 0, 0)x

y a a a lin tc trn ton tp R

iii) Hm lgarity = logax (a>0, a 0) lin tc trong khong (0,+ )

iv) Hm lu tha = ( )ny x n lin tc trong khong ( , + )

v) Cc hmy =sinx, y = cosx lin tc trn tp , cc hmsin 1

tg , seccos cos

xx x

x x= = lin

tc tr ra cc gi tr

+(2 1)2

k v cc hmcos 1

cot g , cosecsin sin

xx x

x x= = lin tc tr ra cc gi

tr k .

b) Tnh lin tc ca hm s hpnh l 3.1.3 Gi s hm : ( , )f A B A B lin tc ti im 0x A cn hm g:

B lin tc ti im 0 0( )y f x B =

Khi hm hp 0 :g f A lin tc tix0.

Chng minh:

Cho mt s tu 0 > . V hm lin tc ti y0, nn vi 0 > c th tm c 1 0 > saocho 0 1, | |y B y y < ta c 0| ( ) ( )|g y g y < .

Mt khc v f lin tc ti x = x0 nn vi 1 ni trn ta c th tm c 0 > sao cho

0,| |x A x x < suy ra

0 0 1| | | ( ) ( )|y y f x f x = < .

Tm li, theo cch chn s suy ra 0,| |x A x x < ta c

0 0 0 0 0| ( )( ) ( )( )| | ( ( )) ( ( ))| | ( ) ( )|g f x g f x g f x g f x g y g y = = < .

Vy hm 0g f lin tc ti imx0.

V d 8:

i) Do hm lu tha x (x >0) biu din c di dng ln xx e = l hp ca hm logarit

v hm m nn lin tc.ii) Hm xx lin tc ti im bt kx >0. Tht vy ln ln( )x x x x x x e e= = .

V d 9:

Xt s lin tc ca hm s+

+=

+

2

l im1

nx

nxn

x x ey

e

Vix >0 ta c+ +

++

=+ +

22

l im lim11 1

nx nx

nxn n

nx

xx

x x e e

e

e

8/9/2019 Chuong 3-X

7/22

7

7

Vix >0, khi ch l nxe khi n + ta c

+

+=

+

22l im

1

nx

nxn

x x ex

e

Vix

Hin nhin l khi 0x hm s lin tc. Mt khc ta thy0 0

l im ( ) lim ( ) 0 (0)x x

f x f x f +

= = = ,

vy hm s cng lin tc tix=0.

Do f(x) lin tc x .

V d 10:

Cho1

2

khi 2

1( ) khi 2.

1 x

a x

f x x

e

=

=

+

Tm a hm lin tc x .

D thy khi 2x hm s lin tc. hm s lin tc x th n phi lin tc tix=2.

Ta thy1

2 22

1lim ( ) l im 0

1x x

x

f x

e+ +

= =

+

(bi v1

2xe + khi 2x + )

12 2

2

1lim ( ) lim 1

1x x

x

f x

e

= =

+

(bi v1

2 0xe khi 2x ).

Vy a hm s khng th lin tc tix = 2.

3.1.4 im gin on ca hm s

Hm sf(x) c gi l gin on tix0 nu tix =x0 hmf(x) khng lin tc. Vyx0 lim gin on ca hm sf(x) nu: hocx0 khng thuc tp xc nh caf(x), hocx0 thuctp xc nh caf(x) nhng

0

0l im ( ) ( )x x

f x f x

, hoc khng c0

l im ( )x x

f x

.

im gin on x0 ca hm sf(x) c gi l im gin on loi 1 nu cc gii hnmt pha

0

l im ( )x x

f x+

,0

l im ( )x x

f x

tn ti hu hn v t nht mt trong hai gii hn ny khcf(x0).

im gin on khng phi loi 1 sc gi l im gin on loi 2. Nh vy imx0l im gin on loi 2 nu hm s khng c gii hn mt pha hay mt trong hai gii hn

l v hn.

8/9/2019 Chuong 3-X

8/22

8

Gi sx0 l im gin on ca hm sy =f(x). Nu tha mn ng thc

0 0

lim ( ) lim ( )x x x x

f x f x+

=

th im gin onx0

gi l khc. Nu t nht mt trong cc gii hn mt pha nitrn bng thx0 gi l im gin on v cng.

V d 11: Cho hm s:

khi 0( )

1 khi 0.

x xf x

x x

=

+

8/9/2019 Chuong 3-X

9/22

9

9

1 1

1 1

l im ( ) lim sin sin( 1) sin1,

l im ( ) l im arcsin arcsin( 1)2

x x

x x

f x x

f x x

+ +

= = =

= = =

Suy ra hm s gin on loi 1 tix =1Hn na

1 1

1 1

l im ( ) l im( cos ) cos1,

l im ( ) l im arcsin arcsin1 .2

x x

x x

f x a x a

f x x

+ +

= + = +

= = =

Vy tix = 1 hm s lin tc nu cos12

a

= v gin on loi 1 nu cos12

a

.

Cui cng, nux0 l im gin on loi 1 ca hm s, ta gi hiu

0 0

0 0| ( ) ( )| | l im ( ) l im ( )|x x x x

f x f x f x f x +

+

=

(3.1.6)

l bc nhy ca hm s tix0.

nh l 3.1.4Mi im gin on ca hm sn iu xc nh trn [a,b] u l im ginon loi mt.

Chng minh:

Gi s : [ , ]f a b l mt hm tng v 0 [ , ]x a b l im gin on caf.

t

0sup{ ( )| [ , )}f x x a x = (3.1.7)

0inf{ ( )| ( , ]}f x x x b = . (3.1.8).

Trong trng hp c bit nu x0=a ta ch xt , nu x0= b ta ch xt . Theo (3.1.17)v (3.1.8) , l hu hn.

nh l c chng minh nu ta chng minh c rng

0

0

l im ( )x xx x

f x

= (3.1.10)

Tht vy, theo nh ngha supremum 00, x x > < sao cho ( )f x < . Chn

0x x = .Ta thy x m 0 0x x x < < th 0x x x < < ,ta c

( ) ( )f x f x < (3.1.11)

hay 0 0| ( ) | ( , )f x x x x < (3.1.12)

Vy0

0

l im ( )x xx x

f x

= .

3.2 Cc tnh cht ca hm lin tc

3.2.1Tnh cht bo ton du ln cn mt im

nh l 3.2.1Gi s hm sf(x) xc nh trn tpA, lin tc ti im 0x A . Khi

i) Nuf(x0)> th 0 > sao chof(x) > 00 ( )x x A ,

trong

0 00 ( ) { :| | }x x x x = < . (3.2.1)

ii) Nu 0 sao cho

0( ) 0 ( ) .f x x x A< (3.2.2)

Chng minh:

i) Do hm sf lin tc ti 0x A nn 0, 0 > > sao cho x A 0| |x x < th

0( ) ( )f x f x < < , c ngha l 0( ) ( )f x f x > 00 ( )x x A .

By gihy chn = >0( ) 0f x khi

0 0( ) ( ) ( )f x f x f x > + = 00 ( )x x A .

ii) Tng t hy chn 0( )f x = .

ngha:

Hm lin tcf(x) bo ton du ca n trong mt ln cn ca imx0.

3.2.2 Tnh cht ca mt hm s lin tc trn mt on

nh l 3.2.2 (nh l Weierstrass th nht)

Nu hmfxc nh v lin tc trn on [a,b] th n b chn, tc l tn ti cc hng smv Msao cho

( ) [ , ]m f x M x a b . (3.2.3)

Chng minh:

Ta hy chng minh nh l bng phn chng. Tht vy, ta gi s rng hm s khng bchn. Khi vi mi s t nhin n ta tm c trn [a,b] gi trx =xn sao cho:

( )nf x n> (3.2.4)

Theo b Bolzan - Weierstrass t dy {xn} c th trch ra mt dy con { }kn

x hi tn

mt gii hn hu hn: 0knx x khi k + , trong hin nhin 0 .a x b

8/9/2019 Chuong 3-X

11/22

11

11

V hm lin tc ti x0 nn 0( ) ( )knf x f x . Nhng khi t (3.2.4) ta suy ra

( )kn

f x + , khc f(x0) l mt hm s hu hn. Mu thun ny suy ra nh l c chng

minh.

Ta ch rng nh l khng cn ng i vi nhng khong khng ng. V d nh hm1

xlin tc trn khong (0,1) nhng trong khong ny hm s khng b chn.

nh l 3.2.3 (nh l Weierstrass th hai)

Nu hm : [ , ]f a b lin tc th n t cn trn ng v cn di ng trong [a,b], tc ltn ti 1 2, [ , ]c c a b sao cho

1[ , ]

sup ( ) ( )x a b

f x f c

= v 2[ , ]

inf ( ) ( )x a b

f x f c

= . (3.2.5)

Chng minh:

Theo nh l trn, do hm lin tc nn n b chn. Ta c

M=[ , ]

sup ( )x a b

f x

< + .

Theo nh l supremum ta c mt dy {xn} [a,b] sao cho l im ( ).nn

M f x

= Dy {xn} b

chn nn n cha mt dy con {kn

x } hi t, c th 1knx c khi k .

Mt khc t cc bt ng thckn

a x b suy ra l imknk

a x b

tc l 1 [ , ]c a b . theo gi

thit hmflin tc ti c1, nn 1l im ( ) ( )knkM f x f c

= = .

Hon ton tng t 2c sao cho: 2[ , ]

inf ( ) ( )x a b

f x f c

= .

c) Ch :

i) Cui cng ta ch rng gi trc1, c2 ni trn khng phi l duy nht. V d trn hnh3.2.1

Hnh 3.2.1

V d trn hnh 3.2.1 hmf(x) trong on [a,b] nhn gi tr ln nht ti hai im c1, c2 vnhn gi tr b nht ti nhiu v hn im ca on [d1, d2].

8/9/2019 Chuong 3-X

12/22

12

nh l khng cn ng i vi nhng khong khng ng, v d hmf(x) = 2x2 nh xkhong [0,1) ln khong [0,2), do sup ( ) 2f x = nhng hmf(x) khng nhn gi tr 2 trongkhong [0,1).

ii) Nu hmf(x) khi bin thin trn mt khongXno d l b chn th ta gi dao ngca n trong khong l hiu M m = gia cn trn ng v cn di ng ca n. Nicch khc

,

sup {| ( ) ( )| }.x x X

f x f x

= Nu xt hmf(x) lin tc trn on hu hnX=[a,b], th

theo nh l va chng minh trn, dao ng ca hmf(x) s chn gin l hiu gia gi trln nht v b nht ca hm trn on .

nh l 3.2.4 (nh l Bolzano - Cauchy th nht)

Gi s hmf(x) xc nh v lin tc trn [a,b] v f(a).f(b) < 0. Khi tn ti t nht mtim c(a,b) sao chof(c)=0.

Chng minh:

xc nh ta gi sf(a)< 0 cnf(b)>0. Ta chia i on [a,b] bi im2

a b+ . C th

xy ra lf(x) trit tiu ti im , khi nh l c chng minh, ta c tht c =2

a b+.

By gi gi sf(2

a b+) 0, khi ti cc u mt ca mt trong cc on [a,

2

a b+],

[2

a b+,b] hm s ly cc gi tr khc du nhau (c th gi tr m ti mt tri v gi tr dng

ti mt phi). Gi on l [a1,b1], ta c (xem hnh 3.2.2)f(a1)< 0, f(b1)> 0

Hnh 3.2.2

By gita li chia i on [a1,b1]. C th xy ra hai kh nng:

Hoc lf(x) trit tiu ti trung im 1 12

a b+ca on , khi ta c th chn im c =

1 1

2

a b+, nh l c chng minh.

Hoc l ta thu c on [a2,b2] l mt trong hai na ca on [a1,b1] sao chof(a2) < 0,f(b2) > 0. (Xem hnh 3.2.2)

Ta tip tc qu trnh lp cc on . Khi hoc sau mt s hu hn bc ta s gptrng hp im chia l im ti hm trit tiu v khi nh l c chng minh.

8/9/2019 Chuong 3-X

13/22

13

13

Hoc c mt dy v hn cc on cha nhau. Khi i vi on th n, [an,bn]

(n=1,2,3) ta s cf(an) 0 v di ca on r rng bng bnan=2n

b a.

Dy cc on ta lp c tho mn cc iu kin ca b v dy cc on lng nhau, bi vtheo trn l im( ) 0n n

nb a

= . V vy, c hai dy {an}, {bn} dn ti gii hn chung

= =l im limn n

n na b c, m r rng c[a,b]. Ta hy chng minh im c ny tho mn yu cu

ca nh l.

Tht vy, do tnh lin tc ca hm s tix = c, ta c

( ) l im ( ) 0nn

f c f a

= v

= ( ) l im ( ) 0nn

f c f b .

Vyf(c)=0, nh l c chng minh.

nh l 3.2.5 (nh l Bolzano - Cauchy th hai)

Gi s hmf(x) xc nh v lin tc trn on [a,b] v ti cc u mt ca on hmf(x) nhn cc gi tr khng bng nhauf(a) = A, f(b) = B.

Khi vi sCbt k nm trung gian giaA vB, ta c th tm c im ( , )c a b saochof(c)=C.

Chng minh: Khng mt tnh tng qut ta c th gi thit rng A

8/9/2019 Chuong 3-X

14/22

14

Hnh 3.2.3

s2

5nm trung gian gia s

1(0)

3f = v

2(1)

3f = nhng khng c gi tr c no trn (0,1)

sao cho 2( ) .5

f c =

3.3iu kin lin tc ca hm n iu v ca hm s ngc

3.3.1 iu kin lin tc ca hm n iu

nh l 3.3.1Chof(x) l hm n iu. iu kin cn v hmf(x)lin tc trn on[a,b] l tp gi tr ca n chnh l on vi hai u mtf(a) vf(b).

Chng minh:

iu kin cn: Gi sf(x) l n iu tng v lin tc trn [a,b], ta phi chng minh:=([ , ]) [ ( ), ( )]f a b f a f b . (3.3.1)

Tht vy, ly bt k ([ , ])f a b , khi [ , ]x a b sao cho ( )f x = . Do f n iu tng,nn khi

=

( ) ( ) ( ) ( ) [ ( ), ( )]

([ , ]) [ ( ), ( )].

a x b f a f x f b f x f a f b

f a b f a f b

Ngc li, ly [ ( ), ( )]f a f b do f(x) lin tc trn on [a,b] nn [ , ]c a b sao( ) ([ , ]).f c f a b = Suy ra [ ( ), ( )] ([ , ])f a f b f a b v h thc (3.3.1) c chng minh.

iu kin

Gi sf(x)l n iu tng vf([a,b])= [f(a),f(b)]. Ta hy chng minhf(x)lin tc trn[a,b].

Gi s ngc li, f(x) gin on ti 0 [ , ]x a b . Nu a < 0x < b, ta hy t

0 0

l im ( ) , l im ( )x x x x

f x f x +

= = (Hnh 3.3.1).

Khi hoc < 0( )f x , hoc 0( )f x < .

Nu 0( )f x < , th f([a,b]) khng cha khong 0( , ( ))f x . Nu f(x0)< th f([a,b])

khng cha khong (f(x0), ), iu ny tri vi gi thitf([a,b])=[f(a),f(b)].Trng hpx0 = a hocx0 = b chng minh tng t.

8/9/2019 Chuong 3-X

15/22

15

15

Hnh 3.3.1

3.3.2Tnh lin tc ca hm ngc

nh l 3.3.2Gi sf(x) l hm tng thc s v lin tc trn [a,b]. Khi f(x)c hm ngcf-1 xc nh trn tp [f(a),f(b)], ng thi 1f cng tng thc s v lin tc trn [f(a),f(b)].

Chng minh:

Theo nh l trn f([a,b])=[f(a),f(b)], nn

[ ( ), ( )], [ , ]y f a f b x a b sao cho 1 1 2 2( ) ( )y f x f x y = < = ( )f x y= . Phn tx ni trn lduy nht. Tht vy, ta gi s ,x [ , ],x a b x x < sao cho ( ) ( )f x f x y = = , iu ny v l

do hmfthc s tng.By gicho tng ng phn t [ ( ), ( )]y f a f b vi phn t duy nht [ , ]x a b ni trn ta

thu c hm ngc n tr

1 : [ ( ), ( )] [ , ]f f a f b a b . (3.3.2)

By gita hy chng minhf-1 l hm tng thc s.

Tht vy 1 2 1 2, [ ( ), ( )],y y f a f b y y < khi 1 2, [ , ]x x a b sao cho

1 1 2 2( ), ( ).y f x y f x = = do, v doftng thc s nn

1 11 2 1 2( ) ( )x x f y f y

< < .

Cui cng ta thy f 1 l hm tng v c min gi tr l

[a,b]=[ f 1 (f(a)), f 1 (f(b))]. (3.3.3)

Vy f 1 l hm lin tc.

Nhn xt:

nh l cn ng khifl hm gim thc s hoc thay [a,b] bng (a,b).

8/9/2019 Chuong 3-X

16/22

16

V d 1: Hm sin x: [ , ] [ 1,1]2 2

n iu tng v lin tc, nn theo nh l trn hm

arcsinx: [1,1] [ , ]2 2

cng n iu tng v lin tc.

Hm cosx: [0, ] [1,1] n iu gim v lin tc , nn hm arccosx: [1,1] [0,] cng n iu gim v lin tc.

Hm tg x: ( , ) ( , )2 2

+ n iu tng v lin tc nn hm arctg x:

( , ) ( , )2 2

+ n iu tng v lin tc.

vi) Tng t hm cotgx: (0, ) ( , ) + n iu gim v lin tc nn hm ngcarccotgx: ( , ) + (0, ) n iu gim v lin tc.

3.4 Khi nim lin tc u

3.4.1 Mu

Cho hmf(x) xc nh trn tpA (c th l khong ng hay m, hu hn hay v hn) vlin tc ti mi imx0 A . Theo ngn ng ta c th pht biu nh sau: i vi mi >0 ta c th chn c mt s >0 sao cho x A m |xx0| cho trc s tm c mt

s sao cho bt ng thc (3.4.1) c tho mn. Ta thy khi 0x bin thin trn tpA, chod cnh, s ni chung s thay i. Ni cch khc s khng nhng ch ph thucvo m cn ph thuc vo 0x .

Nh vy, i vi hmf(x) lin tc trn tpA, ny ra vn l: vi 0 > cho trc, tnti hay khng mt s >0 ph hp vi mi im 0x A . Ta c nh ngha sau.

3.4.2 nh ngha

Ta ni rng hm s :f A lin tc u trnA, nu 0 > cho trc 0 > ch phthuc vo sao cho ,x x A m | |x x < , th

| ( ) ( )|f x f x < . (3.4.2)

V d 1:

i) Chng minh rng hmf(x) =x lin tc u trn ton trc s. Tht vy 0 > ly = ta thy ,x x m | |x x < th

| ( ) ( )| | |f x f x x x = < .

ii) Hmy =sinx, y =cosx lin tc u trn . Tht vy, chng hn xt hmy = cosx, tathy

8/9/2019 Chuong 3-X

17/22

17

17

| | | cos cos | 2| sin .sin |2 2

| |2 | | .

2

x x x x y y x x

x xx x

+ = =

=

Vi 0 > cho trc bt k, ta ch cn chn = th khi ,x x , | |x x < ta c| |y y < .

V d 2: Chng minh rng hm f(x) = x2 lin tc u trn khong (1,1). Tht vy, ly haiim bt k , ( 1,1)x x , khi

2 2| ( ) ( )| | | | ( )( )|

| | | | 2| |

f x f x x x x x x x

x x x x x x

= = +

= + <

Vi 0 > nh ty , ta ch cn chn2

= , khi x x, ( 1,1) m x x| | < th

f x f x | ( ) ( )| 2 2.2

< = =

Nhn xt: chng minh hm f(x) khng lin tc u trn tp A ta ch cn chng minhmnh sau:

n nx x A0, , > sao cho n nx x| | 0 th

n nf x f x | ( ) ( )| . (3.4.3)

nh l 3.4.1 (nh l Cantor): Nu : [ , ]f a b lin tc th n lin tc u trn [ , ]a b .

Chng minh:Ta hy chng minh nh l bng phn chng. Gi sf(x) khng lin tc u trn [ , ]a b ,

tc tn ti mt s dng sao cho 0, > tn ti , [ , ]x x a b m x x| | < th

0| ( ) ( )|f x f x .

Ln lt ly1

n = (n=1,2,3,) ta s tm c cc dy , [ , ]n nx x a b , m

1| |n nx x

n <

nhng

*0| ( ) ( )|n nf x f x n . (3.4.4).

Theo b Bolzann - Werierstrass dy {xn} b chn, n cha mt dy con hi t:0 [ , ].knx x a b

Khi vi dy nx{ } ta cng c dy con tng ng knx x0{ } . Tht vy,

0 0| | | | | | 0 + k k k k n n n nx x x x x x khi k .

Mt khc theo gi thitf(x) lin tc tix0 ta c

0lim ( ) lim ( ) ( )

= =

k kn nk kf x f x f x ,

suy ra lim | ( ) ( ) | 0

=k k

n nk

f x f x , iu ny mu thun vi (3.4.4). nh l c chng minh.

8/9/2019 Chuong 3-X

18/22

18

Ch rng nh l khng cn ng nu hmf(x) ch lin tc u trn khong (a,b).

V d 3: Hm1

yx

= lin tc trn khong (0,1) nhng khng lin tc u trn khong ny.

Tht vy,0

1 11, ,2

n nx xn n = = = .

Khi 1

| | 0,2

n nx x n = nhng

| ( ) ( ) | | 2 | 1n nf x f x n n n = = = .

3.4.3 Lin tc ca cc hm s scp

Tnh l v cc php tnh ca cc hm lin tc v nh l v tnh lin tc ca hm hp ta

c nh l sau:nh l 3.4.2Cc hm scp lin tc ti mi im thuc tp xc nh ca n.

Ch :

Ta ch rng nuf(x) lin tc ti imx0, th

0 00lim ( ) ( ) (lim ).

= =

x x x xf x f x f x

Nh vy, nu hmf(x) lin tc th c th thay i th t vic ly gii hn v vic tnh gitr ca hm.

Sau y da vo tnh lin tc ca nhng hm scp, chng ta sa ra hng lot giihn quan trng:

a)0

log (1 )lim log

+=a a e . (3.4.5)

Gii hn c dng0

0. Ta c

1log (1 )

log (1 )a a

+= + .

V biu thc nm bn phi di du lgarit tin n e khi 0 , nn theo tnh lin tc,lgarit ca n tin n logae. Cng thc c chng minh, ni ring khi a = e ta c cng thc

0

11

ln( )l im

+= (3.4.6)

b)0

1l im ln

aa

= (3.4.7)

Gii hn ny c dng0

0.

Ta t 1 =a , khi theo tnh lin tc ca hm s m khi 0 th 0 . Ngoi rachng ta c log (1 )a = + , nh vy l

0 0

1 1lim lim ln

log (1 ) loga a

aa

e

= = =

+

.

8/9/2019 Chuong 3-X

19/22

19

19

Ni ring, nu ly1

( 1,2,3...)nn

= = th ta nhn c cng thc

n

nn a al im ( 1) ln

= (3.4.8)

c)a 0

(1 ) 1l im

+ = (3.4.9)

Ta t (1 ) 1 + = . Do tnh lin tc ca hm lu tha khi 0 th 0 . Ly

lgarit hai v ca ng thc (1 ) 1 + = + ta nhn c

ln(1 ) ln(1 ) + = +

Nhh thc ny, ta bin i biu thc cho nh sau

(1 ) 1 ln(1 ). .

ln(1 )

+ += =

+

Theo cc gii hn trn, c hai biu thcln(1 )

+v

ln(1 )

+u c gii hn l 1 khi

0 , 0 , v vy cng thc c chng minh.

3.5Bi tp chng 3

3.1 Cho2 1

( )3

xf x

x

+=

+. Chng minh

1

1lim ( )

2xf x

= bng ngn ng .

3.2 Chng minh hm s 1( )1

f xx

=+

lin tc ti mi im 1x bng ngn ng .

3.3 Kho st lin tc ca cc hm s sau:

1)2

1( ) khi 1

(1 )f x x

x=

+v ( 1)f tu .

2)1

( ) sin khi 0, (0) 0f x x x f x

= =

3)2

1

( ) khi 0, (0) 0xf x e x f

= = .

3.4 Xt xem hm s [ ]: 0,2f c cho bi

2 khi 0 1( )

2 khi 1 2

x xf x

x x

=

<

c lin tc khng?

3.5 Tm a hm s

8/9/2019 Chuong 3-X

20/22

20

khi 0( )

khi 0

xe xf x

a x x

8/9/2019 Chuong 3-X

21/22

21

21

3.12 Chng minh rng nu hmf(x) lin tc trong khong a x < + v tn ti gii hnhu hn l im ( )

xf x

+th hm s ny b chn trong khong cho.

3.13 Chng minh rng hm ( ) sinf x x

= lin tc v b chn trong khong (0,1) nhngkhng lin tc u trong khong .

3.14 Chng minh rng hmf(x) =sin2x lin tc v b chn trong khong v hnx < < + nhng khng lin tc u trong khong .

3.15 Chng minh rng hm khng b chnf(x)=x+sinx lin tc u trn ton trc sx < < + .

3.16 Xt tnh lin tc u ca cc hm sau:

1)2

( ) khi [ 1,1]4

xf x x

x=

2)1

( ) cos khi (0,1)xf x e x x

=

3) ( ) khi [1, )f x x x = +

3.17 Nghin cu tnh lin tc v v phc ho th ca hm s sau

1 1 1arctg

1 2

y

x x x

= + +

3.18 Nghin cu tnh lin tc v v tnh lin tc v v th cc hm s sau:

1)1

lim ( 0)1 nn

y xx+

= +

2)ln(1 )

l imln(1 )

xt

tt

ey

e+

+=

+

3.19 Chng minh rng phng trnh:

1 0xxe =

c t nht mt nghim dng nh hn 1.

3.20 S dng cng thc tng ng tnh gn ng:

1) 105

2) 1632

3) 0,31 .

3.21 Chng minh rng hm| sin |

( )x

f xx

= lin tc u trn cc khong (1,0) v (0,1)

nhng khng lin tc u trn (1,0) (0,1).

8/9/2019 Chuong 3-X

22/22

22

3.22 Xt tnh lin tc ca cc hm s sau:

1)1 khi hu t

( )0 khi v t

xf x

x

=

2)khi hu t

( )0 khi v t.

x xf x

x

=