Bai tap giai tich - Tap 2

8/14/2019 Bai tap giai tich - Tap 2

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Mc lc

i


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ii


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Li ni u

Bn ang c trong tay tp I ca mt trong nhng sch bi tp gii tch (theochng ti) hay nht th gii .

Tr-c y, hu ht nhng ng-i lm ton ca Vit Nam th-ng s dng hai cunsch ni ting sau (bng ting Nga v -c dch ra ting Vit):

1. Bi tp gii tch ton hc ca Demidovich (B. P. Demidoviq; 1969,Sbornik Zadaq i Upranenii po Matematiqeskomu Analizu, Izdatel~stvo

"Nauka", Moskva)

v

2. Gii tch ton hc, cc v d v bi tp ca Ljaszko, Bojachuk, Gai,Golovach (I. I. Lxko, A. K. Boquk, . G. Ga, G. P. Golobaq; 1975, Matem-atiqeski Analiz v Primerah i Zadaqah, Tom 1, 2, Izdatel~stvo Vixa

Xkola).

ging dy hoc hc gii tch.Cn ch rng, cun th nht ch c bi tp v p s. Cun th hai cho li

gii chi tit i vi phn ln bi tp ca cun th nht v mt s bi ton khc.Ln ny chng ti chn cun sch (bng ting Ba Lan v -c dch ra ting

Anh):

3. Bi tp gii tch. Tp I: S thc, Dy s v Chui s (W. J. Kaczkor, M.T. Nowak, Zadania z Analizy Matematycznej, Czesc Pierwsza, Liczby Rzeczy-wiste, Ciagi i Szeregi Liczbowe, Wydawnictwo Universytetu Marii Curie -Sklodowskiej, Lublin, 1996),

iii


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iv Li ni u

4. Bi tp gii tch. Tp II: Lin tc v Vi phn (W. J. Kaczkor, M.

T. Nowak, Zadania z Analizy Matematycznej, Czesc Druga, Funkcje JednejZmiennejRachunek Rozniczowy, Wydawnictwo Universytetu Marii Curie -Sklodowskiej, Lublin, 1998).

bin dch nhm cung cp thm mt ti liu tt gip bn c hc v dy gii tch.Khi bin dch, chng ti tham kho bn ting Anh:

3*. W. J. Kaczkor, M. T. Nowak, Problems in Mathematical Analysis I,Real Numbers, Sequences and Series, AMS, 2000.

4*. W. J. Kaczkor, M. T. Nowak, Problems in Mathematical Analysis II,Continuity and Differentiation, AMS, 2001.

Sch ny c cc -u im sau:

Cc bi tp -c xp xp t d cho ti kh v c nhiu bi tp hay.

Li gii kh y v chi tit.

Kt hp -

c nhng t-

ng hay gia ton hc s cp v ton hc hin i.Nhiu bi tp c ly t cc tp ch ni ting nh-, American Mathemati-cal Monthly (ting Anh), Mathematics Today (ting Nga), Delta

(ting Balan). V th, sch ny c th dng lm ti liu cho cc hc sinhph thng cc lp chuyn cng nh- cho cc sinh vin i hc ngnh ton.

Cc kin thc c bn gii cc bi tp trong sch ny c th tm trong

5. Nguyn Duy Tin, Bi Ging Gii Tch, Tp I, NXB i Hc Quc Gia HNi, 2000.

6. W. Rudin, Principles of Mathematical Analysis, McGraw -Hil BookCompany, New York, 1964.

Tuy vy, tr-c mi ch-ng chng ti trnh by tm tt l thuyt gip bn cnh li cc kin thc c bn cn thit khi gii bi tp trong ch-ng t-ng ng.


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Li ni u v

Tp I v II ca sch ch bn n hm s mt bin s (tr phn khng gian

metric trong tp II). Kaczkor, Nowak chc s cn vit Bi Tp Gii Tch cho hmnhiu bin v php tnh tch phn.

Chng ti ang bin dch tp II, sp ti s xut bn.Chng ti rt bit n :- Gio s- Phm Xun Ym (Php) gi cho chng ti bn gc ting Anh tp I

ca sch ny,- Gio s-Nguyn Hu Vit H-ng (Vit Nam) gi cho chng ti bn gc ting

Anh tp II ca sch ny,- Gio s- Spencer Shaw (M) gi cho chng ti bn gc ting Anh cun sch

ni ting ca W. Rudin (ni trn), xut bn ln th ba, 1976,- TS D-ng Tt Thng c v v to iu kin chng ti bin dch cun

sch ny.Chng ti chn thnh cm n tp th sinh vin Ton - L K5 H o To C

Nhn Khoa Hc Ti Nng, Tr-ng HKHTN, HQGHN, c k bn tho v sanhiu li ch bn ca bn nh my u tin.

Chng ti hy vng rng cun sch ny s -c ng o bn c n nhn vgp nhiu kin qu bu v phn bin dch v trnh by. Rt mong nhn -c s chgio ca qu v bn c, nhng kin gp xin gi v: Chi on cn b, Khoa

Ton C Tin hc, tr-ng i hc Khoa hc T nhin, i hc Quc giaH Ni, 334 Nguyn Tri, Thanh Xun, H Ni.

Xin chn thnh cm n.H Ni, Xun 2002.

Nhm bin dchon Chi


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Cc k hiu v khi nim

R - tp cc s thc

R+ - tp cc s thc d-ng Z - tp cc s nguyn N - tp cc s nguyn d-ng hay cc s t nhin Q - tp cc s hu t (a; b) - khong m c hai u mt l a v b [a; b] - on (khong ng) c hai u mt l a v b [x] - phn nguyn ca s thc x Vi x 2 R, hm du ca x l

sgn x =

8>:

1 vi x > 0;

1 vi x < 0;0 vi x = 0:

Vi x 2 N,

n! = 1 2 3 ::: n;(2n)!! = 2 4 6 ::: (2n 2) (2n);

(2n 1)!! = 1 3 5 ::: (2n 3) (2n 1):

K hiu nk

= n!

k!(nk)! ; n; k 2 N; n k, l h s ca khai trin nh thcNewton.

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viii Cc k hiu v khi nim

Nu A

R khc rng v b chn trn th ta k hiu supA l cn trn ng

ca n, nu n khng b chn trn th ta quy -c rng supA = +1. Nu A R khc rng v b chn d-i th ta k hiu infA l cn d-i ng

ca n, nu n khng b chn d-i th ta quy -c rng infA = 1. Dy fang cc s thc -c gi l n iu tng (t-ng ng n iu gim)

nu an+1 an (t-ng ng nu an+1 an) vi mi n 2 N. Lp cc dy niu cha cc dy tng v gim.

S thc c -c gi l im gii hn ca dy fang nu tn ti mt dy confankg ca fang hi t v c.

Cho S l tp cc im t ca dy fang. Cn d-i ng v cn trn ng cady , k hiu ln l-t l lim

n!1an v lim

n!1an -c xc nh nh- sau

limn!1

an =

8>:

+1 nu fang khng b chn trn;1 nu fang b chn trn v S = ;;supS nu fang b chn trn v S 6= ;;

limn!1a

n = 8>:1 nu fang khng b chn d-i;+1 nu fang b chn d-i v S = ;;infS nu fang b chn d-i v S 6= ;;

Tch v hn1Qn=1

an hi t nu tn ti n0 2 N sao cho an 6= 0 vi n n0 vdy fan0an0+1 ::: an0+ng hi t khi n ! 1 ti mt gii hn P0 6= 0. SP = an0an0+1 ::: an0+n P0 -c gi l gi tr ca tch v hn.

Trong phn ln cc sch ton n-c ta t tr-c n nay, cc hm tang vctang cng nh- cc hm ng-c ca chng -c k hiu l tg x, cotg x,

arctg x, arccotg x theo cch k hiu ca cc sch c ngun gc t Php vNga, tuy nhin trong cc sch ton ca M v phn ln cc n-c chu u,chng -c k hiu t-ng t l tan x, cot x, arctan x, arccot x. Trong cunsch ny chng ti s s dng nhng k hiu ny bn c lm quen vinhng k hiu -c chun ho trn th gii.


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Bi tp

1


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Ch-ng 1

Gii hn v tnh lin tc

1.1 Gii hn ca hm s

Chng ta dng cc nh ngha sau.

nh ngha 1. Hm f gi l tng (t-ng ng, tng thc s, gim, gim thc

s) trn tp khc rng A 2 R nu x1 < x2; x1; x2 2 A ko theo f(x1) f(x2)(t-ng ng f(x1) < f(x2), f(x1) f(x2), f(x1) > f(x2) ). Hm tng hay gim(t-ng ng, tng thc s hay gim thc s) gi l hm n iu (t-ng ng,

n iu thc s)

nh ngha 2. Tp (a "; a + ") n fag, y " > 0 gi l ln cn khuyt caim a 2 R1.1.1. Tm cc gii hn hoc chng minh chng khng tn ti.

(a) limx!0

x cos1

x; (b) lim

x!0x

1

x

;

(c) limx!0

x

a b

x ; a; b > 0; (d) limx!0[x]

x;

(e) limx!1

x(p

x2 + 1 3p

x3 + 1); (f) limx!0

cos(2

cos x)

sin(sin x):

1.1.2. Gi s f : (a; a) n f0g ! R. Chng minh rng(a) lim

x!0f(x) = l nu v ch nu lim

x!0f(sin x) = l,

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4 Ch-ng1. Gii hn v tnh lin tc

(b) limx!0

f(x) = l th limx!0

f(

jx

j) = l. iu ng-c li c ng khng ?

1.1.3. Gi s hm f : (a; a) n f0g ! (0; +1) tho mn limx!0

(f(x) + 1f(x)) = 2.

Chng minh rng limx!0

f(x) = 1.

1.1.4. Gi s f -c xc nh trn ln cn khuyt ca a v limx!a

(f(x)+ 1jf(x)j) =

0. Tm limx!0

f(x).

1.1.5. Chng minh rng nu f l hm b chn trn [0; 1] tho mn f(ax) =

bf(x) vi 0 x 1a

v a; b > 1 th limx!0+

f(x) = f(0).

1.1.6. Tnh

(a) limx!0

(x2(1 + 2 + 3 + + [ 1jxj ]));

(b) limx!0+

(x([ 1x

] + [ 2x

] + + [ kx

])); k 2 N.

1.1.7. Tnh limx!1

[P(x)]P(jxj) , y P(x) l a thc vi h s d-ng.

1.1.8. Ch ra bng v d rng iu kin

limx!0

(f(x) + f(2x)) = 0(

)

khng suy ra f c gii hn ti 0. Chng minh rng nu tn ti hm ' sao

cho bt ng thc f(x) '(x) -c tho mn trong mt ln cn khuyt ca0 v lim

x!0'(x) = 0 , th () suy ra lim

x!0f(x) = 0.

1.1.9.

(a) Cho v d hm f tho mn iu kin

limx!

0(f(x)f(2x)) = 0

v limx!0

f(x) khng tn ti.

(b) Chng minh rng nu trong mt ln cn khuyt ca 0, cc bt ng

thc f(x) jxj; 12

< < 1; v f(x)f(2x) jxj -c tho mn, thlimx!0

f(x) = 0.


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1.1.10. Cho tr-c s thc , gi s limx!1

f(ax)x

= g(a) vi mi s d-ng a.

Chng minh rng tn ti c sao cho g(a) = ca.

1.1.11. Gi s f : R ! R l hm n iu sao cho limx!1

f(2x)f(x)

= 1. Chng

minh rng limx!1

f(cx)f(x)

= 1 vi mi c > 0.

1.1.12. Chng minh rng nu a > 1 v 2 R th

(a) limx!1

ax

x= +1; (b) lim

x!1ax

x= +1:

1.1.13. Chng minh rng nu > 0, th limx!1lnx

x = 0.-

1.1.14. Cho a > 0, chng minh limx!0

ax = 1. Dng ng thc ny chng

minh tnh lin tc ca hm m.

1.1.15. Chng minh rng

(a) limx!1

1 +

1

x

x= e; (b) lim

x!1

1 +

1

x

x= e;

(c) limx!1

(1 + x)1x = e:

1.1.16. Chng minh rng limx!0

ln(1+x) = 0. Dng ng thc ny, suy ra hm

logarit lin tc trn (0; 1).

1.1.17. Tnh cc gii hn sau :

(a) limx!0

ln(1 + x)

x; (b) lim

x!0ax 1

x; a > 0;

(c) limx!0

(1 + x) 1x

; 2 R:

1.1.18. Tm

(a) limx!1

(ln x)1x ; (b) lim

x!0+xsinx;

(c) limx!0

(cos x)1

sin2 x ; (d) limx!1

(ex 1) 1x ;(e) lim

x!0(sin x)

1ln x :


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1.1.19. Tm cc gii hn sau:

(a) limx!0

sin2x + 2 arctg 3x + 3x2

ln(1 + 3x + sin2 x) + xex; (b) lim

x!0lncos x

tg x2;

(c) limx!0+

p1 ex p1 cos xp

sin x; (d) lim

x!0(1 + x2)cotgx:

1.1.20. Tnh

(a) limx!1

(tgx

2x + 1)1x ; (b) lim

x!1x(ln(1 +

x

2) ln x

2):

1.1.21. Gi s rng limx!0+

g(x) = 0 v tn ti 2R , cc s d-ng m; M sao

cho m f(x)x M vi nhng gi tr d-ng ca x trong ln cn ca 0. Chng

minh rng nu limx!0+

g(x) ln x = ; th limx!0+

f(x)g(x) = e. Tr-ng hp = 1hoc = 1, ta gi s e1 = 1 v e1 = 0.1.1.22. Bit rng lim

x!0f(x) = 1 v lim

x!0g(x) = 1. Chng minh rng nu

limx!0

g(x)(f(x) 1) = , th limx!0

f(x)g(x) = e.

1.1.23. Tnh

(a) limx!0+2sin px + px sin 1xx,

(b) limx!0

1 + xe

1x2 sin 1

x4

e 1x2,

(c) limx!0

1 + e

1x2 arctg 1

x2+ xe

1x2 sin 1

x4

e 1x2.

1.1.24. Cho f : [0; +1) ! R l hm sao cho mi dyf(a + n); a 0; hi tti khng. Hi gii hn lim

x!1f(x) c tn ti khng ?

1.1.25. Cho f : [0; +1) ! R l hm sao cho vi mi s d-ng a, dyff(an)g,hi t ti khng. Hi gii hn lim

x!1f(x) c tn ti khng ?

1.1.26. Cho f : [0; +1) ! R l hm sao cho vi mi a 0 v mi b > 0,dyff(a + bn)g; a 0; hi t ti khng. Hi gii hn lim

x!1f(x) c tn ti

khng ?


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1.1.27. Chng minh rng nu limx!0

f(x) = 0 v limx!0

f(2x)f(x)x

= 0 th limx!0

f(x)x

=

0.

1.1.28. Gi s f xc nh trn (a; +1), b chn trn mi khong hu hn(a; b) ; a < b. Chng minh rng nu lim

x!+1(f(x + 1) f(x)) = l, th lim

x!0f(x)x

= l.

1.1.29. Cho f xc nh trn (a; +1), b chn d-i trn mi khong huhn (a; b) ; a < b. Chng minh rng nu lim

x!+1(f(x + 1) f(x)) = +1, th

limx!0

f(x)x

= +1.

1.1.30. Cho f xc nh trn (a; +1), b chn trn mi khong hu hn(a; b) ; a < b. Nu vi s nguyn khng m k , lim

x!+1f(x+1)f(x)

xktn ti, th

limx!+1

f(x)

xk+1=

1

k + 1lim

x!+1f(x + 1) f(x)

xk:

1.1.31. Cho f xc nh trn (a; +1), b chn trn mi khong hu hn(a; b) ; a < b v gi s f(x) c > 0 vi x 2 (a; +1). Chng minh rng nu

limx!+1

f(x+1)f(x)

tn ti, th limx!+1

f(x)1x cng tn ti v

limx!+1

(f(x))1x = lim

x!+1f(x + 1)

f(x):

1.1.32. Gi thit rng limx!0

f

1x

1= 0. T c suy ra lim

x!0f(x) tn ti

khng ?

1.1.33. Cho f : R ! R sao cho vi mi a 2 R, dy f( an) hi t ti khng.Hi f c gii hn ti 0 khng ?

1.1.34. Chng minh rng nu limx!0 fx 1x 1x = 0, th limx!0 f(x) = 0.

1.1.35. Chng minh rng nu f n iu tng ( gim ) trn (a; b), th vi

mi x0 2 (a; b),

(a) f(x+0 ) = limx!x+0

f(x) = infx>x0

f(x) (f(x+0 ) = supx>x0

f(x));


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(b) f(x0 ) = limx!x0

f(x) = supx 0 v p l s nguyn d-ng c nh. K hiu fn l php lp

th n ca f. Chng minh rng nu mp l s nguyn d-ng nh nht sao cho

fmp(0) > 0, th

p

mp lim

n!1

fn(0)

n lim

n!1fn(0)

n

p

mp+

1 + f(0)

mp:

1.1.42. Gi s f : R ! R l hm tng v x 7! f(x) x c chu k 1. Chngminh rng lim

n!1fn(x)n

tn ti v nhn cng gi tr vi mi x 2 R, y fn khiu php lp th n ca f.


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9

1.2 Cc tnh cht ca hm lin tc

1.2.1. Tm tt c cc im lin tc ca hm f xc nh bi

f(x) =

(0 nu x v t,

sin jxj nu x hu t.

1.2.2. Xcnh tp cc im lin tc ca hm f -c cho bi

f(x) =

(x2 1 nu x v t,0 nu x hu t.

1.2.3. Nghin cu tnh lin tc ca cc hm sau:

(a) f(x) =

8>:

0 nu x v t hoc x = 0,1q

nu x = p=q; p 2 Z; q2 N, vp;qnguyn t cng nhau,

(b) f(x) =

8>:

jxj nu x v t hoc x = 0,qx=(qx + 1) nu x = p=q; p 2 Z; q2 N, v

p;qnguyn t cng nhau,

(Hm nh ngha (a) -c gi l hm Riemann.)

1.2.4. Chng minh rng nu f 2 C([a; b]), th jfj 2 C([a; b]). Ch ra bng vd rng iu ng-c li khng ng.

1.2.5. Xc nh tt c cc an v bn sao cho hm xc nh bi

f(x) =

(an + sin x nu x 2 [2n; 2n + 1]; n 2 Z ,bn + cos x nu x 2 (2n 1; 2n); n 2 Z ,

lin tc trn R.

1.2.6. Cho f(x) = [x2]sin x vi x 2 R. Nghin cu tnh lin tc ca f.1.2.7. Bit

f(x) = [x] + (x [x])[x] vi x 12

:

Chng minh rng f lin tc v tng thc s trn [1; 1).


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1.2.8. Nghin cu tnh lin tc ca cc hm sau y v v th ca chng

(a) f(x) = limn!1

nxnxnx+nx ; x 2 R;

(b) f(x) = limn!1

x2enx+xenx+1

; x 2 R;

(c) f(x) = limn!1

ln(en+xn)n

; x 0;

(d) f(x) = limn!1

n

q4n + x2n + 1

x2n; x 6= 0;

(e) f(x) = limn!12n

pcos2n

x + sin2n

x; x 2 R:1.2.9. Chng minh rng nu f : R ! R lin tc v tun hon th n c gi

tr ln nht v gi tr nh nht.

1.2.10. Cho P(x) = x2n + a2n1x2n1 + + a1x + a0, chng minh rng tn tix 2 R sao cho P(x) = inffP(x) : x 2 Rg. Cng chng minh rng gi trtuyt i ca mi a thc P c gi tr nh nht; tc l, tn ti x 2 R saocho jP(x)j = inffjP(x)j : x 2 Rg.

1.2.11.

(a) Cho v d v hm b chn trn [0; 1] nh-ng khng c gi tr nh nht,

cng khng c gi tr ln nht.

(b) Cho v d v hm b chn trn [0; 1] nh-ng khng c gi tr nh nht

trn mi on [a; b] [0; 1]; a < b.

1.2.12. Cho f : R! R; x0 2 R v > 0, t

!f(x0; ) = supfjf(x) f(x0)j : x 2 R; jx x0j < g

v !f(x0) = lim!0+

!f(x0; ). Chng minh rng f lin tc ti x0 nu v ch nu

!f(x0) = 0.

1.2.13.


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11

(a) Cho f; g

2C([a; b]) v vi x

2[a; b], t h(x) = min

ff(x); g(x)

gv

H(x) = maxff(x); g(x)g. Chng minh rng h; H 2 C([a; b]).

(b) Cho f1; f2; f3 2 C([a; b]) v vi x 2 [a; b], t f(x) l mt trong ba gi trf1(x); f2(x) v f3(x) m nm gia hai gi tr cn li. Chng minh rng

f 2 C([a; b]).

1.2.14. Chng minh rng nu f 2 C([a; b]), th cc hm -c xc nh bi

m(x) = infff() : 2 [a; x]g v M(x) = supff() : 2 [a; x]g

cng lin tc trn [a; b].

1.2.15. Gi f l hm b chn trn [a; b]. Chng minh rng cc hm -c xc

nh bi

m(x) = infff() : 2 [a; x)g v M(x) = supff() : 2 [a; x)g

cng lin tc trn (a; b).

1.2.16. Vi cc gi thit ca bi ton tr-c, kim tra cc hm

m(x) = infff() : 2 [a; x]g v M(x) = supff() : 2 [a; x]g

c lin tc tri trn (a; b) hay khng ?

1.2.17. Gi s f lin tc trn [a; 1) v limx!1

f(x) hu hn. Chng minh rng

f b chn trn [a; 1).

1.2.18. Cho f l hm lin tc trn R v t fxng l dy b chn. Cc btng thc sau

limn!1

f(xn) = f(limn!1

xn) v limn!1

f(xn) = f(limn!1

xn)

c ng khng ?

1.2.19. Cho f : R ! R l hm lin tc, tng v gi fxng l dy b chn.Chng minh rng


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(a) limn!1

f(xn) = f(limn!1

xn);

(b) limn!1

f(xn) = f(limn!1

xn):

1.2.20. Cho f : R ! R l hm lin tc, gim v gi fxng l dy b chn.Chng minh rng

(a) limn!1

f(xn) = f(limn!1

xn);

(b) limn!1

f(xn) = f(limn!1

xn):

1.2.21. Gi s f lin tc trn R; limx!1

f(x) = 1 v limx!1

f(x) = +1. Xcnh g bng cch t

g(x) = supft : f(t) < xg vi x 2 R:

(a) Chng minh rng g lin tc tri.

(b) g c lin tc khng ?

1.2.22. Cho f : R! R l hm tun hon lin tc vi hai chu k khng thng-c T1 v T2; tc l T1T2 v t. Chng minh rng f l hm hng. Cho v d

hm tun hon khc hm hng c hai chu k khng thng -c.

1.2.23.

(a) Chng minh rng nu f : R! R l hm lin tc, tun hon, khc hmhng, th n c chu k d-ng nh nht, gi l chu k c bn.

(b) Cho v d hm tun hon khc hm hng m khng c chu k c bn.

(c) Chng minh rng nu f : R! R l hm tun hon khng c chu k cbn, th tp tt c cc chu k ca f tr mt trong R.

1.2.24.


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13

(a) Chng minh rng nh l trong mc (a) ca bi ton tr-c vn cn ng

khi tnh lin tc ca f trn R -c thay th bi tnh lin tc ti mt

im.

(b) Chng minh rng nu f : R! R l hm tun hon khng c chu k cbn v nu n lin tc ti t nht mt im, th n l hm hng.

1.2.25. Chng minh rng nu f; g : R ! R l hm lin tc, tun hon vlimx!1

(f(x) g(x)) = 0 th f = g.

1.2.26. Cho v d hai hm tun hon f v g sao cho mi chu k ca f khng

thng -c vi mi chu k ca g v sao cho f + g

(a) khng tun hon,

(b) tun hon.

1.2.27. Cho f; g : R ! R l cc hm lin tc v tun hon ln l-t vi chuk c bn d-ng T1 v T2. Chng minh rng nu T1T2 =2 Q, th h = f+ g khngl hm tun hon.

1.2.28. Cho f; g : R ! R l cc hm tun hon .Gi s f lin tc v khngc chu k no ca g thng -c vi chu k c bn ca f. Chng minh rng

f + g khng l hm tun hon.

1.2.29. Chng minh rng tp cc im gin on ca hm n iu f : R!R khng qu m -c.

1.2.30. Gi s f lin tc trn [0; 1]. Chng minh rng

limn!11

n

nXk=1

(1)k

f(

k

n ) = 0:

1.2.31. Cho f lin tc trn [0; 1]. Chng minh rng

limn!1

1

2n

nXk=0

(1)k

n

k

f(

k

n) = 0:


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1.2.32. Gi s f : (0;

1)

!R l hm lin tc sao cho f(x) f(nx) vi mi

s d-ng x v mi s t nhin n. Chng minh rng limx!1

f(x) tn ti (hu

hn hoc v hn).

1.2.33. Hm f xc nh trn khong I R -c gi l li trn I nu

f(x1 + (1 )x2) f(x1) + (1 )f(x2)

vi mi x1; x2 2 I v 2 (0; 1). Chng minh rng nu f li trn khong m,th n lin tc. Hm li trn khong bt k c nht thit lin tc khng ?

1.2.34. Chng minh rng nu dy ffng cc hm lin tc trn A hi t uti f trn A, th f lin tc trn A.

1.3 Tnh cht gi tr trung gian

Ta nhc li nh ngha sau:

nh ngha 3. Hm thc f c tnh cht gi tr trung gian trn khong Icha [a; b] nu f(a) < v < f(b) hoc f(b) < v < f(a); tc l, nu v nm gia

f(a) v f(b), th tn ti c nm gia a v b sao cho f(c) = v.

1.3.1. Cho cc v d cc hm c tnh cht gi tr trung gian trn khong I

nh-ng khng lin tc trn khong ny.

1.3.2. Chng minh rng hm tng thc s f : [a; b] ! R c tnh cht gi trtrung gian th lin tc trn [a; b].

1.3.3. Cho f : [0; 1] ! [0; 1] lin tc. Chng minh rng f c im c nhtrong [0; 1]; tc l, tn ti x0 2 [0; 1] sao cho f(x0) = x0.

1.3.4. Gi s f; g : [a; b] ! R lin tc sao cho f(a) < g(a) v f(b) > g(b).Chng minh rng tn ti x0 2 (a; b) sao cho f(x0) = g(x0).


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15

1.3.5. Cho f : R

!R lin tc v tun hon vi chu k T > 0. Chng minh

rng tn ti x0 sao cho

f

x0 +

T

2

= f(x0):

1.3.6. Hm f : (a; b) ! R lin tc. Chng minh rng, vi x1; x2; : : : ; xn chotr-c trong (a; b), tn ti x0 2 (a; b) sao cho

f(x0) =1

n(f(x1) + f(x2) + + f(xn)):

1.3.7.

(a) Chng minh rng ph-ng trnh (1 x)cos x = sin x c t nht mtnghim trong (0; 1).

(b) Vi a thc khc khng P, chng minh rng ph-ng trnh jP(x)j = exc t nht mt nghim.

1.3.8. Vi a0 < b0 < a1 < b1 < < an < bn, chng minh rng mi nghimca a thc

P(x) =n

Yk=0(x + ak) + 2n

Yk=0(x + bk); x 2 R;u l thc.

1.3.9. Gi s f v g c tnh cht gi tr trung gian trn [a; b]. Hi f + g c

tnh cht gi tr trung gian trn khong khng ?

1.3.10. Gi s f 2 C([0; 2]) v f(0) = f(2). Chng minh rng tn ti x1 vx2 trong [0; 2] sao cho

x2

x1 = 1 v f(x2) = f(x1):

Gii thch ngha hnh hc kt qu trn.

1.3.11. Cho f 2 C([0; 2]). Chng minh rng tn ti x1 v x2 trong [0; 2] saocho

x2 x1 = 1 v f(x2) f(x1) = 12

(f(2) f(0)):


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1.3.12. Vi n

2N, gi f

2C([0; n]) sao cho f(0) = f(n). Chng minh rng

tn ti x1 v x2 trong [0; n] tho mn

x2 x1 = 1 v f(x2) = f(x1):

1.3.13. Hm lin tc f trn [0; n]; n 2 N, tho mn f(0) = f(n). Chng minhrng vi mi k 2 f1; 2; : : : ; n 1g, tn ti xk v x0k sao cho f(xk) = f(x

0k),

y xk x0k = k hoc xk x0k = n k. Hi vi mi k 2 f1; 2; : : : ; n 1g, c tnti xk v x

0k sao cho f(xk) = f(x

0k), y xk x0k = k ?

1

.3.1

4. 6 Vi n 2 N, gi f 2 C([0; n]) sao cho f(0) = f(n). Chng minh rngph-ng trnh f(x) = f(y) c t nht n nghim vi x y 2 N.

1.3.15. Gi s cc hm thc lin tc f v g xc nh trn R giao hon vi

nhau; tc l, f(g(x)) = g(f(x)) vi mi x 2 R. Chng minh rng nu ph-ngtrnh f2(x) = g2(x) c nghim, th ph-ng trnh f(x) = g(x) cng c nghim

( y f2(x) = f(f(x)) v g2(x) = g(g(x)) ).

Ch ra v d rng gi thit v tnh lin tc ca f v g trong bi ton trn

khng th b qua.

1.3.16. Chng minh rng n nh lin tc f : R! R th hoc tng thc s,hoc gim thc s.

1.3.17. Gi s f : R! R l dn nh lin tc. Chng minh rng nu tn tin sao cho php lp th n ca f l nh x ng nht, tc l, fn(x) = x vi

mi x 2 R, th

(a) f(x) = x; x 2 R, nu f tng thc s,

(b) f2(x) = x; x2R, nu f gim thc s.

1.3.18. Gi s f : R ! R tho mn iu kin f(f(x)) = f2(x) = x; x 2R.Chng minh rng f khng th lin tc.

1.3.19. Tm tt c cc hm f : R! R c tnh cht gi tr trung gian v tnti n 2 N sao cho fn(x) = x; x 2 R, y fn k hiu php lp th n ca f.


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1.3.20. Chng minh rng nu f : R

!R c tnh cht gi tr trung gian v

f1(fqg) ng vi mi q hu t, th f lin tc.

1.3.21. Gi s f : (a; 1) ! R lin tc v b chn. Chng minh rng, vi Tcho tr-c, tn ti dy fxng sao cho

limn!1

xn = +1 v limn!1

(f(xn + T) f(xn)):

1.3.22. Cho v d hm lin tc f : R ! R t mi gi tr ca n ng baln. Hi c tn ti hay khng hm lin tc f : R! R t mi gi tr ca nng hai ln ?

1.3.23. Cho f : [0; 1] ! R lin tc v n iu thc s tng mnh. (Hm fgi l n iu thc s tng mnh trn [0; 1], nu tn ti phn hoch ca

[0; 1] thnh hu hn khong con [ti1; ti], y i = 1; 2; : : : ; n v 0 = t0 < t1 0, v tn ti hmg lin tc trn [0; 1] sao cho f + g gim. Chng minh rng ph-ng trnh

f(x) = 0 c nghim trong khong m (0; 1).


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1.3.28. Chng minh rng mi song nh f : R

![0;

1) c v hn im gin

on.

1.3.29. Nhc li rng mi x 2 (0; 1) c th -c biu din bi s nh phn(binary fraction) :a1a2a3 : : : , y ai 2 f0; 1g; i = 1; 2; : : : . Trong tr-ng hpx c hai khai trin nh phn khc nhau, ta chn khai trin c v hn ch s

1. Tip , gi hm f : (0; 1) ! [0; 1] -c xc nh bi

f(x) = limn!1

1

n

nXi=1

ai:

Chng minh rng f gin on ti mi x 2 (0; 1), tuy nhin, n c tnh chtgi tr trung gian.

1.4 Hm na lin tc

nh ngha 4. H thng s thc m rng R bao gm h thng s thc v

hai k hiu +1,1 vi cc tnh cht sau :(i) Nu x thc, th 1 < x < +1, v x + 1 = +1; x 1 = 1; x+1 =

x1 = 0.

(ii) Nu x > 0, th x (+1) = +1, x (1) = 1.

(iii) Nu x < 0, th x (+1) = 1, x (1) = +1.nh ngha 5. Nu A R l tp khc rng, th sup A (t-ng ng infA) ls thc m rng nh nht (t-ng ng, ln nht) m ln hn (t-ng ng, nh

hn) hoc bng mi phn t ca A.

Cho f l hm thc xc nh trn tp khc rng A R.nh ngha 6. Nu x0 l im gii hn ca A, th gii hn d-i (t-ng nggii hn trn) ca f(x) khi x ! x0 -c nh ngha l inf (t-ng ng sup)ca tp tt c cc y 2 R sao cho tn ti dy fxng cc im trong A khc x0,hi t ti x0 v y = lim

n!1f(xn). Gii hn d-i v gii hn trn ca f(x) khi

x ! x0 -c k hiu t-ng ng bi limx!x0

f(x) v limx!x0

f(x).


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nh ngha 7. Mt hm gi tr thc gi l na lin tc d-i (t-ng ng

trn) ti x0 2 A; x0 l im gii hn ca A, nu limx!x0

f(x) f(x0) (t-ng nglimx!x0

f(x) f(x0)). Nu x0 l im c lp ca A, th ta gi s rng f l na

lin tc trn v d-i ti im ny.

1.4.1. Chng minh rng nu x0 l im gii hn ca A v f : A ! R, th

(a) limx!x0

f(x) = sup>0

infff(x)g : x 2 A; 0 < jx x0j < ;

(b) limx!x0

f(x) = inf>0

supf

f(x)g

: x2

A; 0 0, tn ti x0 2 A sao cho 0 < jx0 x0j < v f(x) < y0 + ":

Thit lp bi ton t-ng t cho gii hn trn ca f ti x0:

1.4.4. Cho f : A ! R v x0 l im ti hn ca A. Chng minh rng

(a) limx!x0

f(x) = 1 nu v ch nu vi mi y thc v vi mi > 0, tn tix0 2 A sao cho 0 < jx0 x0j < v f(x0) < y.

(b) limx!x0

f(x) = +1 nu v ch nu vi mi y thc v vi mi > 0, tn tix0 2 A sao cho 0 < jx0 x0j < v f(x0) > y.


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1.4.5. Gi s f : A

!R v x0 l im gii hn ca A. Chng minh rng nu

l = limx!x0

f(x) (t-ng ng L = limx!x0

f(x)), th tn ti dy fxng; xn 2 A; xn 6= x0,hi t ti x0 sao cho l = lim

n!1f(xn) (t-ng ng L = lim

n!1f(xn)).

1.4.6. Cho f : A ! R v x0 l im gii hn ca A. Chng minh rng

limx!x0

(f(x)) = limx!x0

f(x) v limx!x0

(f(x)) = limx!x0

f(x):

1.4.7. Cho f : A ! (0; 1) v x0 l im gii hn ca A. Chng minh rng

limx!x0

1f(x) = 1limx!x0

f(x) v limx!x0 1f(x) = 1limx!x0

f(x) :

(Ta gi s rng 1+1 = 0 v

10+ = +1.)

1.4.8. Gi s f; g : A ! R v x0 l im gii hn ca A. Chng minh rngcc bt ng thc sau y ng (tr tr-ng hp cc dng bt nh +1 1v 1 + 1):

limx!x0

f(x) + limx!x0

g(x) limx!x0

(f(x) + g(x)) limx!x0

f(x) + limx!x0

g(x)

limx!x0

(f(x) + g(x)) limx!x0

f(x) + limx!x0

g(x):

Cho v d cc hm sao cho 00 00 trong cc bt ng thc trn -c thay bi00


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1.4.10. Chng minh rng nu limx!x0

f(x) tn ti, th (tr tr-ng hp cc dng

bt nh +1 1 v 1 + 1):limx!x0

(f(x) + g(x)) = limx!0

f(x) + limx!x0

g(x);

limx!x0

(f(x) + g(x)) = limx!0

f(x) + limx!x0

g(x):

Ngoi ra, nu f v g l cc hm khng m, th (tr tr-ng hp cc dng bt

nh 0 (+1) v (+1) 0):limx!x0

(f(x) g(x)) = limx

!0

f(x) limx!x0

g(x);

limx!x0

(f(x) g(x)) = limx!0

f(x) limx!x0

g(x):

1.4.11. Chng minh rng nu f lin tc trn (a; b); l = limx!a

f(x) v L =

limx!a

f(x), th vi mi 2 [l; L], tn ti dy fxng gm cc im trong (a; b) hit ti a sao cho lim

n!1f(xn) = .

1.4.12. Tm tt c cc im ti f : R! R xc nh bi

f(x) = (0 nu x v t,sin x nu x hu tl na lin tc.

1.4.13. Xc nh tt c cc im ti f xc nh bi

f(x) =

(x2 1 nu x v t,0 nu x hu t

l na lin tc.


f(x) =

8>:

0 nu x v t hoc x = 0,1q

nu x = pq

; p 2 Z; q2 N,v p; q nguyn t cng nhau,

l na lin tc trn.


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1.4.15. Tm tt c cc im ti hm xc nh bi

(a) f(x) =

8>:

jxj nu x v t hoc x = 0,qx

qx+1nu x = pq; p 2 Z; q2 N,

v p; q nguyn t cng nhau

(b) f(x) =

8>:

(1)qq+1

nu x 2 Q \ (0; 1] v x = pq

; p; q2 N,v p; q nguyn t cng nhau,

0 nu x 2 (0; 1) v t

khng na lin tc trn, cng khng na lin tc d-i.

1.4.16. Cho f; g : A ! R na lin tc trn (t-ng ng, d-i) ti x0 2 A.Chng minh rng

(a) nu a > 0 th af na lin tc d-i (t-ng ng, trn) ti x0 2 A. Nua > 0 th af na lin tc trn (t-ng ng, d-i) ti x0.

(b) f + g na lin tc d-i (t-ng ng, trn) ti x0.

1.4.17. Gi s rng fn : A!R; n

2N, na lin tc d-i (t-ng ng, trn)

ti x0 2 A. Chng minh rng supn2N

fn (t-ng ng, supn2N

fn) na lin tc d-i

(t-ng ng, trn) ti x0.

1.4.18. Chng minh rng gii hn theo tng im ca mt dy tng (t-ng

ng, gim) cc hm na lin tc d-i (t-ng ng, trn) l na lin tc d-i

(t-ng ng, trn).

1.4.19. Vi f : A ! R v x l im gii hn ca A, nh ngha dao ca fti x bi

of(x) = lim!0+

supfjf(z) f(u)j : z; u 2 A; jz xj < ; ju xj < g

Chng minh rng of(x) = f1(x) f2(x), y

f1(x) = maxff(x); limz!x

f(z)g v f2(x) = minff(x); limz!x

f(z)g:


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1.4.20. Gi f1; f2, v of nhtrong bi ton tr-c. Chng minh rng f1 v of

l na lin tc trn, v f2 l na lin tc d-i.

1.4.21. Chng minh rng f : A ! R l na lin tc d-i (t-ng ng,trn) ti x0 2 A, iu kin cn v l vi mi a < f(x0) (t-ng ng,a > f(x0)), tn ti > 0 sao cho f(x) > a (t-ng ng, f(x) < a) bt c khi

no jx x0j < ; x 2 A.

1.4.22. Chng minh rng f : A ! R l na lin tc d-i (t-ng ng,

trn) ti x0 2 A, iu kin cn v l vi mi a 2 R, tp fx 2 A : f(x) > ag(t-ng ng, fx 2 A : f(x) < ag) l m trong A.

1.4.23. Chng minh rng f : R! R l na lin tc d-i nu v ch nu tpf(x; y) 2 R2 : y f(x)g l ng trong R2.

Lp cng thc v chng minh iu kin cn v cho tnh na lin tc

trn ca f trn R.

1.4.24. Chng minh nh l Baire sau y. Mi hm na lin tc d-i (t-ng

ng, trn) f : A ! R l gii hn ca dy tng (t-ng ng, gim) cc hmlin tc trn A.

1.4.25. Chng minh rng nu f : A ! R na lin tc trn, g : A ! R nalin tc d-i v f(x) g(x) khp ni trn A, th tn ti hm lin tc h trn

A sao cho

f(x) h(x) g(x); x 2 A:

1.5 Tnh lin tc u

nh ngha 8. Hm thc f xc nh trn tp A 2 R -c gi l lin tc utrn A nu, vi " cho tr-c, tn ti > 0 sao cho vi mi x v y trong A m

jx yj < , ta c jf(x) f(y)j < ".


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1.5.1. Kim tra cc hm sau y c lin tc u trn (0; 1) hay khng :

(a) f(x) = ex; (b) f(x) = sin1

x;

(c) f(x) = x sin1

x; (d) f(x) = e

1x ;

(e) f(x) = e1x ; (f) f(x) = ex cos

1

x;

(g) f(x) = ln x; (h) f(x) = cos x cos x

;

(i) f(x) = cotg x:

1.5.2. Hm no trong s cc hm sau y lin tc u trn [0; 1) ?

(a) f(x) =p

x; (b) f(x) = x sin x;

(c) f(x) = sin2 x; (d) f(x) = sin x2;

(e) f(x) = ex; (f) f(x) = esin(x2);

(g) f(x) = sin sin x; (h) f(x) = sin(x sin x);

(i) f(x) = sinp

x:

1.5.3. Chng minh rng nu f lin tc u trn (a; b); a; b 2 R, th limx!a+ f(x)v lim

x!bf(x) tn ti nh- cc gii hn hu hn.

1.5.4. Gi s f v g lin tc du trn (a; b) ([a; 1)). T c suy ra tnh lintc u trn (a; b) ([a; 1)) ca cc hm

(a) f + g;

(b) f g;

(c) x 7! f(x)sin x ?

1.5.5.

(a) Chng minh rng nu f l lin tc u trn (a; b] v trn [b; c) , th n

cng lin tc trn (a; c).


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(b) Gi s A v B l cc tp ng trong R v gi f : A

[B

!R l lin tc

u trn A v B. Hi f c nht thit lin tc u trn A [ B ?

1.5.6. Chng minh rng mi hm lin tc v tun hon trn R th lin tc

u trn R.

1.5.7.

(a) Chng minh rng nu f : R! R lin tc sao cho limx!1

f(x) v limx!1

f(x)

l hu hn, th f cng lin tc u trn R.

(b) Chng minh rng nu f : [a; +1) ! R lin tc v limx!1

f(x) l hu hn,

th f cng lin tc u trn [a; 1).

1.5.8. Kim tra tnh lin tc u ca

(a) f(x) = arctg x trn (1; +1);

(b) f(x) = x sin 1x trn (0; +1);

(c) f(x) = e1x trn (0; +

1):

1.5.9. Gi s f lin tc u trn (0; 1). Hi cc gii hn limx!+0

f(x) v

limx!1

f(x) c tn ti khng ?

1.5.10. Chng minh rng mi hm b chn, n iu v lin tc trn khong

I R l lin tc u trn I.

1.5.11. Gi s f lin tc u v khng b chn trn [0; 1). Phi chng hoclimx

!1

f(x) = +1 , hoc limx

!1

f(x) = 1 ?

1.5.12. Hm f : [0; 1) ! R lin tc u v vi mi x 0, dy ff(x + n)g hit ti khng. Chng minh rng lim

x!1f(x) = 0.

1.5.13. Gi s f : [1; 1) ! R lin tc u. Chng minh rng tn ti sd-ng M sao cho jf(x)j

x M vi x 1.


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1.5.14. Gi f : [0;

1)

!R lin tc u. Chng minh rng tn ti s d-ng

M vi tnh cht sau y :

supu>0

fjf(x + u) f(u)jg M(x + 1) vi mi x 0:

1.5.15. Cho f : A ! R; A R; lin tc u. Chng minh rng nu fxng ldy Cauchy cc phn t trong A, th ff(xn)g cng l dy Cauchy.

1.5.16. Gi s A R b chn. Chng minh rng nu f : A ! R bin dyCauchy cc phn t ca A thnh dy Cauchy, th f lin tc u trn A. Tnh

b chn c A c phi l gi thit ct yu khng ?

1.5.17. Chng minh rng f lin tc u trn A 2 R nu v ch nu vi midy fxng v fyng cc phn t ca A,

limn!1

(xn yn) = 0 suy ra limn!1

(f(xn) f(yn)) = 0:

1.5.18. Gi s f : (0; 1) ! (0; 1) lin tc u. T c suy ra

limx!1

f(x + 1x)

f(x)

= 1 ?

1.5.19. Hm f : R! R lin tc ti 0 v tho mn cc iu kin sau y

f(0) = 0 v f(x1 + x2) f(x1) + f(x2) vi mi x1; x2 2 R:

Chng minh rng f lin tc u trn R.

1.5.20. Vi f : A ! R; A R, ta nh ngha

!f() = sup

fjf(x1

f(x2))

j: x1; x2

2A;

jx1

x2

j<

gv gi !f l m un lin tc ca f. Chng minh rng f lin tc u trn A

nu v ch nu lim!0+

!f() = 0:

1.5.21. Cho f : R ! R lin tc u. Chng minh rng cc pht biu saut-ng -ng.


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27

(a) Vi mi hm lin tc u g : R

!R; f

g lin tc u trn R

(b) Hm x 7! jxjf(x) lin tc u trn R.

1.5.22. Chng minh iu kin cn v sau y f l hm lin tc u

trn khong I. Vi " > 0 cho tr-c, tn ti N > 0 sao cho vi mi x1; x2 2I; x1 6= x2,

f(x1) f(x2)x1 x 2

> N suy ra jf(x1) f(x2)j < ":

1.6 Ph-ng trnh hm

1.6.1. Chng minh rng hm duy nht lin tc trn R v tho mn ph-ng

trnh hm Cauchy

f(x + y) = f(x) + f(y)

l hm tuyn tnh dng f(x) = ax:

1.6.2. Chng minh rng nu f : R! R tho mn ph-ng trnh hm Cauchy

f(x + y) = f(x) + f(y)

v mt trong cc iu kin

(a) f lin tc ti x0 2 R,

(b) f b ch trn khong (a; b) no ,

(c) f n iu trn R,

th f(x) = ax.

1.6.3. Xc nh tt c cc hm lin tc f : R! R sao cho f(1) > 0 v

f(x + y) = f(x)f(y):


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1.6.4. Chng minh rng cc nghim duy nht m khng ng nht bng

khng v lin tc trn (0; 1) ca ph-ng trnh hm

f(xy) = f(x) + f(y)

l cc hm logarit.

1.6.5. Chng minh rng cc nghim duy nht m khng ng nht bng

khng v lin tc trn (0; 1) ca ph-ng trnh hm

f(xy) = f(x)f(y)

l cc hm dng f(x) = xa.

1.6.6. Tm tt c cc hm lin tc f : R! R sao cho f(x) f(y) hu t vix y hu t.

1.6.7. Vi jqj < 1, tm tt c cc hm f : R ! R lin tc ti khng v thomn ph-ng trnh hm

f(x) + f(qx) = 0:

1.6.8. Tm tt c cc hm f : R! R lin tc ti khng v tho m

n ph-

ngtrnh hm

f(x) + f

2

3x

= x:

1.6.9. Xc nh mi nghim f : R ! R lin tc ti khng ca ph-ng trnhhm

2f(2x) = f(x) + x:

1.6.10. Tm tt c cc hm lin tc f : R! R tho mn ph-ng trnh Jensen

fx + y2

= f(x) + f(y)2

:

1.6.11. Tm tt c cc hm lin tc trn (a; b); a;b 2 R, tho mn ph-ngtrnh Jensen

f

x + y

2

=

f(x) + f(y)

2:


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29

1.6.12. Xc nh tt c cc nghim lin tc ti

1 ca ph-ng trnh hm

f(2x + 1) = f(x):

1.6.13. Vi a thc, chng minh rng nu f : R ! R l nghim lin tc caph-ng trnh

f(x + y) = f(x) + f(y) + axy;

th f(x) = a2x2 + bx, y b = f(1) a2 .

1.6.14. Xc nh mi nghim lin tc ti 0 ca ph-ng trnh hm

f(x) = f

x

1 x

; x 6= 1:

1.6.15. Gi f : [0; 1] ! [0; 1] l hm lin tc, n iu gim sao cho f(f(x)) =x vi x 2 [0; 1]. Hi f(x) = 1 x c phi l hm duy nht nh- vy khng ?

1.6.16. Gi s rng f v g tho mn ph-ng trnh

f(x + y) + f(x

y) = 2f(x)f(y); x; y

2R:

Chng minh rng nu f khng ng nht bng khng v jf(x)j 1 vi x 2 R,th ta cng c jg(x)j 1 vi x 2 R.

1.6.17. Tm tt c cc hm lin tc tho mn ph-ng trnh hm

f(x + y) = f(x)ey + f(y)ex:

1.6.18. Xc nh mi nghim lin tc ti khng f : R! R ca

f(x + y) f(x y) = f(x)f(y):

1.6.19. Gii ph-ng trnh hm

f(x) + f

x 1

x

= 1 + x vi x 6= 0; 1:


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1.6.20. Dy

fxn

ghi t theo ngha Cesro nu

C limn!1

xn = limn!1

x1 + x2 + x3 + + xnn

tn ti v hu hn. Tm tt c cc hm lin tc Cesro, tc l

f(C limn!1

xn) = C limn!1

f(xn)

vi mi dy hi t Cesro fxng:

1.6.21. Cho f : [0; 1] ! [0; 1] l n nh sao cho f(2xf(x)) = x vi x 2 [0; 1].Chng minh rng f(x) = x; x 2 [0; 1].

1.6.22. Vi m khc khng, chng minh rng nu hm lin tc f : R ! Rtho mn ph-ng trnh

f

2x f(x)

m

= mx;

th f(x) = m(x

c):

1.6.23. Chng minh rng cc nghim duy nht ca ph-ng trnh hm

f(x + y) + f(y x) = 2f(x)f(y)

lin tc trn R v khng ng nht bng khng l f(x) = cos(ax) v f(x) =

cosh(ax) vi a thc.

1.6.24. Xc nh mi nghim lin tc trn (1; 1) ca

f x + y

1 + xy

= f(x) + f(y):

1.6.25. Tm mi a thc P sao cho

P(2x x2) = (P(x))2:


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31

1.6.26. Cho m; n

2 l cc s nguyn. Tm tt c cc hm f : [0;

1)

!R

lin tc ti t nht mt im trong [0; 1) v sao cho

f

1

n

nXi=1

xmi

!=

1

n

nXi=1

(f(xi))m vi xi 0; i = 1; 2; : : : ; n:

1.6.27. Tm tt c cc hm khng ng nht bng khng f : R ! R thomn ph-ng trnh

f(xy) = f(x)f(y) v f(x + z) = f(x) + f(z)

vi z6= 0 no .1.6.28. Tm tt c cc hm f : R n f0g ! R sao cho

f(x) = f

1

x

; x 6= 0:

1.6.29. Tm tt c cc hm f : R n f0g ! R tho mn ph-ng trnh hm

f(x) + f(x2) = f

1

x

+ f

1

x2

; x 6= 0

1.6.30. Chng minh rng cc hm f ; g ; : R! R tho mn ph-ng trnhf(x) g(y)

x y =

x + y

2

; y 6= x;

nu v ch nu tn ti a; b v c sao cho

f(x) = g(x) = ax2 + bx + c; (x) = 2ax + b:

1.6.31. Chng minh rng tn ti hm f : R! Q tho mn ba iu kin sauy :

(a) f(x + y) = f(x) + f(y) vi x; y 2 R;

(b) f(x) = x vi x 2 Q;

(c) f khng lin tc trn R:


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1.7 Hm lin tc trong khng gian metric

Trong mc ny, X v Y ln l-t k hiu l cc khng gian metric (X; d1)

v (Y; d2). n gin, ta ni rng X l khng gian metric thay cho (X; d1)

l khng gian metric. R v Rn lun gi s -c trang b metric Euclide, nu

khng pht biu ng-c li.

1.7.1. Gi (X; d1) v (Y; d2) l cc khng gian metric v f : X! Y. Chngminh rng cc iu kin sau y t-ng -ng.

(a) Hm f lin tc.(b) Vi mi tp ng F Y, tp f1(F) ng trong X:

(c) Vi mi tp m G Y, tp f1(G) m trong X:

(d) Vi mi tp con A ca X, f(A f(A)):

(e) Vi mi tp con B ca Y, f1(B) f1(B):1.7.2. Gi (X; d1) v (Y; d2) l cc khng gian metric v f : X! Y lin tc.

Chng minh rng nghch nh f1

(B) ca tp Borel B trong (Y; d2) l tpBorel trong (X; d1):

1.7.3. Cho v d hm lin tc f : X! Y sao cho nh f(F) (t-ng ng, f(G))khng ng (t-ng ng, m) trong Y vi F ng (t-ng ng, G m) trong X.

1.7.4. Gi (X; d1) v (Y; d2) l cc khng gian metric v f : X! Y lin tc.Chng minh rng nh ca tp compact F trong X l tp compact trong Y.

1.7.5. Cho f xc nh trn hp cc tp ng F1;F2; : : : ;Fm. Chng minh

rng nu gii hn ca f trn mi Fi; i = 1; 2; : : : ; m, l lin tc, th f lintc trn F1 [F2 [ : : : [Fm. Ch ra v d rng pht biu trn khng ngtrong tr-ng hp v hn Fi.

1.7.6. Cho f xc nh trn hp cc tp m Gt; t 2 T. Chng minh rng nuvi mi t 2 T, gii hn fjGt l lin tc, th f lin tc trn

St2TGt.


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33

1.7.7. Cho (X; d1) v (Y; d2) l cc khng gian metric. Chng minh rng

f : X ! Y lin tc nu v ch nu vi mi A trong X, hm fjA lin tc.1.7.8. Gi s f l song nh lin tc t khng gian metric compact X ln

khng gian metric Y. Chng minh rng hm ng-c f1 lin tc trn Y.

Cng chng minh rng gi thit compact khng th b b qua.

1.7.9. Gi f l nh x lin tc t khng gian metric compact X vo khng

gian metric Y. Chng minh rng f lin tc u trn X.

1.7.10. Gi (X; d) l khng gian metric v A l tp con khc rng ca X.

Chng minh rng hm f : X ! [0; 1) xc nh bif(x) = dist(x;A) = inffd(x; y) : y 2 Ag

lin tc u trn X.

1.7.11. Gi s f l nh x lin tc ca khng gian metric lin thng X vo

khng gian metric Y. Chng minh rng f(X) lin thng trong Y.

1.7.12. Cho f : A! Y; ; 6= A X. Vi x 2 A nh ngha

of(x; ) = diam(f(A \B(x; ))):Giao ca f ti x -c xc nh bi

of(x) = lim!0+

of(x; ):

Chng minh rng f lin tc ti x0 2 A nu v ch nu of(x0) = 0 (so snhvi 1.4.19 v 1.4.20).

1.7.13. Gi s f : A ! Y; ; 6= A X v vi x 2 A, gi of(x) l giao caf ti x oc xc nh nhtrong bi ton tr-c. Chng minh rng vi mi

" > 0, tp fx 2 A : of(x) "g l ng trong X.1.7.14. Chng minh rng tp im lin tc ca f : X ! Y l giao m -c

cc tp m, ni cch khc, l G trong (X; d1). Cng chng minh rng tpim gin on ca f l hp m -c cc tp ng, ni cch khc, l Ftrong (X; d1).


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1.7.15. Cho v d hm f : R

!R c tp im gin on l Q.

1.7.16. Chng minh rng vi mi tp con F ca R l tp im gin onca hm f : R! R.

1.7.17. Cho A l tp con F ca khng gian metric X. C tn ti hay khnghm f : X ! R m tp im gin on l A ?

1.7.18. Gi A l hm c tr-ng ca A X. Chng minh rng fx 2 X :oA(x) > 0g = @A, y f(x) l giao ca f ti x -c xc nh nh- trong1.7.12. Suy ra rng

A lin tc trn X nu v ch nu A va m, va ng

trong X.

1.7.19. Gi s g1 v g2 l cc nh x lin tc ca khng gian metric (X; d1)

vo khng gian metric (X; d2), v tp A c phn trong rng, tr mt trong

X.Chng minh rng nu

f(x) =

(g1(x) vi x 2 A,g2(x) vi x 2 X nA,

thof(x) = d2(g1(x); g2(x)); x 2 X:

y of(x) l giao ca f ti x -c xc nh nhtrong 1.7.12.

1.7.20. Ta ni rng hm thc f xc nh trn khng gian metric X l thuc

lp Baire th nht nu f l gii hn im ca dy hm lin tc trn X.

Chng minh rng nu f thuc lp Baire th nht, th tp cc im gin

on ca f l tp thuc phm tr th nht; tc l, n l hp ca mt s m

-c cc tp khng u tr mt.

1.7.21. Chng minh rng nu X l khng gian metric y v f thuc lp

Baire th nht trn X, th tp cc im lin tc ca f tr mt trong X.

1.7.22. Gi f : (0; 1) ! R lin tc sao cho vi mi s d-ng x, dy ffxnghi t ti khng. T c suy ra lim

x!0+f(x) = 0 khng ? (so snh vi 1.1.33.)


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35

1.7.23. K hiu

Fl h cc hm lin tc trn khng gian metric compact X

sao cho vi mi x 2 X, tn ti Mx tho mn

jf(x)j Mx vi mi f 2 F:

Chng minh rng tn ti hng s d-ng M v tp m khc rng G X socho

jf(x)j M vi mi f 2 Fv vi mi x 2 G:

1.7.24. Gi F1 F2 F3 : : : l dy cc tp con khc rng lng nhau cakhng gian metric y X sao cho limn!1diam Fn = 0. Chng minh rngnu f lin tc trn X, th

f

1\n=1

Fn

!=

1\n=1

f(Fn):

1.7.25. Gi (X; d) l khng gian metric v p l im c nh trong X. Vi

a 2 X, xc nh hm fu bi fu(x) = d1(u; x) d1(p;x); x 2 X. Chng minhrng u 7! fu l nh x bo ton khong cch, ni cch khc, l ng c ca(X; d1) vo khng gian C(X;R) cc hm thc lin tc trn X -c trang bmetric d(f; g) = supff(x) g(x) : x 2 Xg.

1.7.26. Chng minh rng khng gian metric X l compact nu v ch nu

vi mi hm lin tc f : X! R l b chn.

1.7.27. Cho (X; d1) l khng gian metric v vi x 2 X, xc nh (x) =dist(x;X n fxg). Chng minh rng hai iu kin sau y t-ng-ng.

(a) Mi hm f : X!R l lin tc u.

(b) Mi dy fxng cc phn t ca X sao cho

limn!1

(xn) = 0

cha dy con hi t.


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45/398

Ch-ng 2

Php tnh vi phn

2.1 o hm ca hm s thc

2.1.1. Tnh o hm (nu c) ca cc hm sau:

(a) f(x) = xjxj; x 2 R;

(b) f(x) =pjxj; x 2 R;

(c) f(x) = [x]sin2

(x); x 2 R;(d) f(x) = (x [x])sin2(x); x 2 R;

(e) f(x) = ln jxj; x 2 Rnf0g;

(f) f(x) = arccos 1jxj ; jxj > 1:

2.1.2. o hm cc hm s sau:

(a) f(x) = logx 2; x > 0; x

6= 1;

(b) f(x) = logx cos x; x 2

0; 2

nf1g:2.1.3. Nghin cu tnh kh vi ca cc hm s sau:

(a) f(x) =

(arctan x vi jxj 1;4

sgn x + x12

vi jxj > 1;

37


46/398

38 Ch-ng 2. Vi phn

(b) f(x) =(x2ex2 vi jxj 1;1

e vi jxj > 1;

(c) f(x) =

(arctan 1jxj vi x 6= 0;2

vi x = 0:

2.1.4. Chng minh rng hm s

f(x) =

(x2cos

x

vi x 6= 0;

0 vi x = 0:

khng kh vi ti cc im xn = 22n+1 ; n 2 Z, nh-ng kh vi ti 0 l im giihn ca dy fxng.

2.1.5. Xc nh cc gi tr a;b;c;d sao cho hm f kh vi trn R:

(a) f(x) =

8>:

4x x 0;

ax2 + bx + c 0 < x < 1;

3 2x x 1

(b) f(x) =8>:

ax + b x 0;

cx2 + dx 0 < x 1;1 1

xx > 1

(c) f(x) =

8>:

ax + b x 1;

ax2 + c 1 < x 2;dx2+1x

x > 2:

2.1.6. Tnh tng:

n

Xk=0 kekx; x 2 R;(a)2nXk=0

(1)k

2n

k

kn; n 1;(b)

nXk=1

k cos(kx); x 2 R:(c)


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48/398

40 Ch-ng 2. Vi phn

2.1.13. Cho f l hm kh vi ti im a v

fxn

gv

fzn

gl cc dy hi t ti

a sao cho xn 6= a, zn 6= a, xn 6= zn, n 2 N. Hy ch ra hm f sao cho gii hn

limn!1

f(xn) f(zn)xn zn

(a) bng f0(a),

(b) khng tn ti hoc c tn ti nh-ng khc f0(a).

2.1.14. Cho f l hm kh vi ti a v xt hai dy fxng v fzng cng hi t

va sao cho xn < a < zn vi mi n 2 N. Chng minh rnglimn!1

f(xn) f(zn)xn zn = f

0(a):

2.1.15.

(a) Chng minh rng hm f xc nh trong khong (0; 2) theo cng thc

f(x) =

(x2 vi cc gi tr x hu t trong khong (0; 2);

2x 1 vi cc gi tr x v t trong khong (0; 2)

ch kh vi ti duy nht im x = 1 v f0(1) 6= 0. Hm ng-c ca f ckh vi ti im 1 = y = f(1) khng?

(b) Cho

A = fy 2 (0; 3) : y 2 Q; py =2 Qg;B =

x : x =

1

2(y + 4); y 2 A

:

Xt hm

f(x) =

8>:

x2 vi x hu t thuc (0; 2);

2x 1 vi x v t thuc (0; 2);2x 4 vi x 2 B:

Chng minh rng khong (0; 3) cha trong min gi tr ca f v hm

ng-c ca f khng kh vi ti im 1.


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2.1. o hm ca hm s thc 41

2.1.16. Xt hm f xc nh trn R sau

f(x) =

(0 nu x v t hoc bng 0,

aq nu x =pq

; p 2 Z; q2 N v p; q nguyn t cng nhau,

trong dy faqg tho mn iu kin limn!1

nkan = 0 vi k 2. Chng minhrng f kh vi ti mi iu v t c bc i s nh hn hoc bng k, tc l...

2.1.17. Cho P l mt a thc bc n vi n nghim thc khc nhau x1; : : : ; xnv Q l a thc bc khng qu n 1. Chng minh rng

Q(x)P(x)

=nXk=1

Q(xk)P0(xk)(x xk)

vi x 2 Rnfx1; x2; : : : ; xng. Tm tngnXk=1

1

P0(xk); n 2:

2.1.18. S dng kt qu bi tr-c hy kim tra cc ng thc sau:

nXk=0

nk(1)kx + k = n!x(x + 1)(x + 2) (x + n)(a)

vi x 2 Rnfn; (n 1); : : : ; 1; 0g,nX

k=0

n

k

(1)kx + 2k

=n!2n

x(x + 2)(x + 4) (x + 2n)(b)

vi x 2 Rnf2n; 2(n 1); : : : ; 2; 0g.

2.1.19. Cho f kh vi trn R. Hy kho st tnh kh vi ca hm jfj.

2.1.20. Gi s f1; f2; : : : ; f n xc nh trong mt ln cn ca x, khc 0 v kh

vi ti x. Chng minh rngnQk=1

fk

0nQk=1

fk

(x) =

nXk=1

f0k(x)fk(x)

:


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42 Ch-ng 2. Vi phn

2.1.21. Gi s cc hm f1; f2; : : : ; f n; g1; g2; : : : ; gn xc nh trong ln cn ca

x, khc 0 va kh vi ti x. Chng minh rngnYk=1

fkgk

!0(x) =

nYk=1

fkgk

(x)nXk=1

f0k(x)fk(x)

g0k(x)

gk(x)

:

2.1.22. Nghin cu tnh kh vi ca f v jfj vi

f(x) =

(x nu x 2 Q;sin x nu x 2 RnQ:(a)

f(x) =(x 32k nu x 2 Q \ 12k1 ; 12k2 ; k 2;

sin

x 32k

nu x 2 (RnQ) \ 1

2k1 ;1

2k2

; k 2:(b)

2.1.23. Chng minh rng nu o hm mt pha f0(x0) v f0+(x0) tn ti th

f lin tc ti x0.

2.1.24. Chng minh rng nu f : (a; b) ! R t cc i ti c 2 (a; b), tc lf(c) = maxff(x) : x 2 (a; b)g v tn ti cc o hm tri v o hm phif0(c) v f

0+(c), th f

0(x0) 0 v f0+(x0) 0. Hy pht biu bi ton t-ng

ng tr-ng hp f t cc tiu.

2.1.25. Chng minh rng nu f 2 C([a; b]); f(a) = f(b) v f0 tn ti trn(a; b) th

infff0(x) : x 2 (a; b)g 0 supff0(x) : x 2 (a; b)g:

2.1.26. Chng minh rng nu f 2 C([a; b]) v f0 tn ti trn (a; b) th

infff0(x) : x 2 (a; b)g f(b) f(a)

b

a supff0(x) : x 2 (a; b)g:

2.1.27. Chng minh rng nu f0 tn ti v lin tc trn (a; b) th f kh vi

trn (a; b) v f0(x) = f0(x) vi x 2 (a; b).

2.1.28. Tn ti hay khng hm f : (1; 2) ! R sao cho f0(x) = x v f0+(x) = 2xvi x 2 (1; 2) ?


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2.1. o hm ca hm s thc 43

2.1.29. Cho f kh vi trn [a; b] tho mn

f(a) = f(b) = 0;(i)

f0(a) = f0+(a) > 0; f0(b) = f0(b) > 0:(ii)

Chng minh rng tn ti c 2 (a; b) sao cho f(c) = 0 v f0(c) 0.2.1.30. Chng minh rng f(x) = arctan x tho mn ph-ng trnh

(1 + x2)f(n)(x) + 2(n 1)f(n1)(x) + (n 2)(n 1)f(n2)(x) = 0

vix 2

R vn 2

. Chng minh rng

f(2m)(0) = 0; f(2m+1)(0) = (1)m(2m)!:


(ex sin x)(n) = 2n=2ex sin

x + n

4

; x 2 R; n 1;(a)

(xn ln x)(n) = n!

ln x + 1 +

1

2+ + 1

n

; x > 0; n 1;(b)

ln xx (n)

= (

1)nn!xn1ln x 1 1

2 1

n ; x > 0; n 1;(c) xn1e1=x

(n)= (1)n e

1=x

xn+1; x 6= 0; n 1:(d)

2.1.32. Chng minh cc ng nht thc sau:nXk=0

n

k

sin

x + k

2

= 2n=2 sin

x + n

4

; x 2 R; n 1(a)

n

Xk=1(1)k+1 1

k

n

k

= 1 +

1

2+ + 1

n; n 1(b)

2.1.33. Cho f(x) = px2 1 vi x > 1. Chng minh rng f(n)(x) > 0 nu nl v f(n) < 0 vi n chn.

2.1.34. Cho f2n = ln(1 + x2n); n 2 N. Chng minh rng

f(2n)2n (1) = 0:


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44 Ch-ng 2. Vi phn

2.1.35. Cho P l mt a thc bc n, chng minh rngnXk=0

P(k)(0)

(k + 1)!xk+1 =

nXk=0

(1)k P(k)(x)

(k + 1)!xk+1:

2.1.36. Cho 1; 2; : : : ; n l cc gi tr tho mn iu kin

k1 + k2 + : : : +

kn > 0; 8k 2 N:

Kho hm

f(x) =1

(1 1x)(1 2x) (1 nx)s -c xc nh trong ln cn 0. Chng minh rng vi k 2 N ta c f(k)(0) > 0.2.1.37. Cho f l hm kh vi n cp n trn (0; +1). Chng minh rng vi

x > 0,1

xn+1f(n)

1

x

= (1)n

xn1f

1

x

(n):

2.1.38. Cho I;J l hai khong m v f : J! R, g : I! J l cc hm kh viv hn trn J v I. Chng minh cng thc Fa di Bruno cho o hm cp n

ca h = f g sau:

h(n)(t) =X n!

k1! kn!f(k)(g(t))

g(1)(t)1!

k1

g(n)(t)1!

kn;

trong k = k1 + k2 + + kn v tng ly trn tt c cc gi tr k1; k2; : : : ; knsao cho k1 + 2k2 + + nkn = n.2.1.39. Chng minh rng cc hm s sau :

f(x) =

(e1=x

2nu x 6= 0;

0 nu x = 0;(a)

g(x) =(e1=x nu x > 0;

0 nu x 0;(b)

h(x) =

(e

1xa+

1xb nu x 2 (a; b);

0 nu x =2 (a; b);(c)

cng thuc C1(R).


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2.2. Cc nh l gi tr trung bnh 45

2.1.40. Cho f kh vi trn (a; b) sao cho vi x

2(a; b) ta c f0(x) = g(f(x)),

trong g 2 C1(a; b). Chng minh rng f 2 C1(a; b).

2.1.41. Cho f l hm kh vi cp hai trn (a; b) v vi cc s ; ; thc tho

mn 2 + 2 > 0 ta c

f00(x) + f0(x) + f(x) = 0; x 2 (a; b):

Chng minh rng f 2 C1(a; b).

2.2 Cc nh l gi tr trung bnh

2.2.1. Chng minh rng nu f lin tc trong khong ng [a; b], kh vi trn

khong m (a; b) v f(a) = f(b) = 0 th vi 2 R, tn ti x 2 (a; b) sao cho

f(x) + f0(x) = 0:

2.2.2. Cho f v g l cc hm lin tc trn [a; b], kh vi trn khong m (a; b)

v gi s f(a) = f(b) = 0. Chng minh rng tn ti x2

(a; b) sao cho

g0(x)f(x) + f0(x) = 0:

2.2.3. Cho f l hm lin tc trn [a; b]; a > 0 v kh vi trn khong m (a; b).

Chng minh rng nuf(a)

a=

f(b)

b;

th tn ti x0 2 (a; b) sao cho x0f0(x0) = f(x0):

2.2.4. Gi s f lin tc trn [a; b] v kh vi trn (a; b). Chng minh rngnu f2(b) f2(a) = b2 a2 th ph-ng trnh

f0(x)f(x) = x

c t nht mt nghim trong (a; b).


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46 Ch-ng 2. Vi phn

2.2.5. Gi s f v g lin tc, khc 0 trong [a; b] v kh vi trn (a; b). Chng

minh rng nu f(a)g(b) = f(b)g(a) th tn ti x0 2 (a; b) sao chof0(x0)f(x0)

=g0(x0)g(x0)

:

2.2.6. Gi s a0; a1; : : : ; an l cc s thc tho mn

a0n + 1

+a1n

+ + an12

+ an = 0:

Chng minh rng a thc P(x) = a0xn

+ a1xn

1

+ + an c t nht mtnghim trong (0; 1).

2.2.7. Xt cc s thc a0; a1; : : : ; an tho mn

a01

+2a1

1+

22a23

+ 2n1an1

n+

2nann + 1

= 0:

Chng minh rng hm s

f(x) = an lnn

x + + a2 ln2

x + a1 ln x + a0

c t nht mt nghim trong (1; e2).

2.2.8. Chng minh rng nu mi nghim ca a thc P c bc n 2 u lthc th mi nghim ca a thc P0 cng u l thc.

2.2.9. Cho f kh vi lin tc trn [a; b] v kh vi cp hai trn (a; b), gi

s f(a) = f0(a) = f(b) = 0. Chng minh rng tn ti x1 2 (a; b) sao cho

f00(x1) = 0.

2.2.10. Cho f kh vi lin tc trn [a; b] v kh vi cp hai trn (a; b), gi s

f(a) = f(b) v f0(a) = f0(b) = 0. Chng minh rng tn ti hai s x1; x2 2(a; b); x1 6= x2 sao cho

f00(x1) = f00(x2):


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2.2.11. Chng minh rng cc ph-ng trnh sau:

x13 + 7x3 5 = 0;(a)3x + 4x = 5x(b)

c ng mt nghim thc .

2.2.12. Chng minh rng vi cc sa1; a2; : : : ; an khc0vviccs1; 2; : : : ; ntho mn i 6= j; i 6= j, ph-ng trnh

a1x1 + a2x

2 +

+ anx

n = 0

c nhiu nht l n 1 nghim trong (0; +1).2.2.13. Chng minh rng vi cc gi thit ca bi trn, ph-ng trnh

a1e1x + a2e

2x + + anenx = 0

c nhiu nht l n 1 nghim trong (0; +1).2.2.14. Cho cc hm f ; g ; h lin tc trn [a; b] v kh vi trn (a; b), ta nh

ngha hm

F(x) = det

f(x) g(x) h(x)f(a) g(a) h(a)

f(b) g(b) h(b)

; x 2 [a; b]:

Chng minh rng tn ti x0 2 (a; b) sao cho F0(x0) = 0. S dng kt qu vanhn -c pht biu nh l gi tr trung bnh v nh l gi tr trung bnh

tng qut.

2.2.15. Cho f lin tc trn [0; 2] v kh vi cp hai trn (0; 2). Chng minh

rng nu f(0) = 0; f(1) = 1 v f(2) = 2 th tn ti x0 2 (0; 2) sao chof00(x0) = 0.

2.2.16. Gi s f lin tc trn [a; b] v kh vi trn (a; b). Chng minh rng

nu f khng l mt hm tuyn tnh th tn ti x1 v x2 thuc (a; b) sao cho

f0(x1) 2 vi c 2 (0; 1).2.2.18. Cho f lin tc trn [a; b]; a > 0, kh vi trn (a; b). Chng minh rng

tn ti x0 2 (a; b) sao chobf(a) af(b)

b a = f(x1) x1f0(x1):

2.2.19. Chng minh rng cc hm s x 7! ln(1 + x), x 7! ln(1 + x2) vx

7!arctan x lin tc u trn [0; +

1).

2.2.20. Gi s f kh vi cp hai trn (a; b) v tn ti M 0 sao cho jf00(x)j M vi mi x 2 (a; b). Chng minh rng f lin tc u trn (a; b).2.2.21. Gi s f : [a; b] ! R, b a 4 kh vi trn khong m (a; b). Chng

minh rng tn ti x0 2 (a; b) sao chof0(x0) < 1 + f2(x0):

2.2.22. Chng minh rng nu f kh vi trn (a; b) v nu

limx!a+ f(x) = +1; limx!b f(x) = 1;(i)f0(x) + f2(x) + 1 0; vi x 2 (a; b);(ii)

th b a .2.2.23. Cho f lin tc trn [a; b] v kh vi trn (a; b). Chng minh rng nu

limx!b

f0(x) = A th f0(b) = A.

2.2.24. Gi s f kh vi trn (0; 1) v f0(x) = O(x) khi x ! 1. Chng minhrng f(x) = O(x2) khi x ! 1.

2.2.25. Cho f1; f2; : : : ; f n v g1; g2; : : : ; gn l cc hm lin tc trn [a; b] vkh vi trn (a; b). Gi s rng gk(a) 6= gk(b) vi mi k = 1; 2; : : : ; n. Chngminh rng tn ti c 2 (a; b) sao cho

nXk=1

f0k(c) =nXk=1

g0k(c)fk(b) fk(a)gk(b) gk(a) :


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58/398

50 Ch-ng 2. Vi phn

(b) nu limx!+1

(f(x)

2p

xf0(x)) = 0 th limx!+1

f(x) = 0.

2.2.33. Chng minh rng nu f 2 C2([a; b]) c t nht ba nghim trong [a; b]th ph-ng trnh f(x) + f00(x) = 2f0(x) c t nht mt nghim trong [a; b].

2.2.34. Chng minh rng nu a thcP bc n c n nghim phn bit ln

hn 1 th a thc

Q(x) = (x2 + 1)P(x)P0(x) + xP2(x) + (P0(x))2

c t nht 2n 1 nghim phn bit.2.2.35. Gi s rng a thc P(x) = amxm+am1xm1+ +a1x+a0 vi am > 0

c m nghim thc phn bit. Chng minh rng a thc Q(x) = (P(x))2P0(x)c

(1) ng m + 1 nghim thc phn bit nu m l,

(2) ng m nghim thc phn bit nu m chn.

2.2.36. Gi s a thc P(x) bc n 3 c cc nghim u thc, vit

P(x) = (x a1)(x a2) (x an);trong ai ai+1; i = 1; 2; : : : ; n 1 v

P0(x) = n(x c1)(x c2) (x cn1);

trong ai ci ai+1; i = 1; 2; : : : ; n 1. Chng minh rng nu

Q(x) = (x a1)(x a2) (x an1);Q0(x) = (n 1)(x d1)(x d2) (x dn2);

th di ci vi i = 1; 2; : : : ; n 2. Hn na chng minh rng nu

R(x) = (x a2)(x a3) (x an);R0(x) = (n 1)(x e1)(x e2) (x en2);

th ei ci+1 vi i = 1; 2; : : : ; n 2.


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2.2.37. S dng gi thit ca bi trn hy chng minh rng

(1) nu S(x) = (x a1 ")(x a2) : : : (x an), trong " > 0 tho mna1+" an1 v nu S0(x) = n(xf1)(xf2) : : : (xfn1) th fn1 cn1,

(2) nu T(x) = (x a1)(x a2) : : : (x an + "), vi " > 0 tho mn an " a2v nu T0(x) = n(x g1)(x g2) : : : (x gn1) th g1 c1.

2.2.38. S dng gi thit ca bi 2.2.36 hy chng minh rng

ai +ai+1 ain

i + 1

ci ai+1 ai+1 aii + 1

; i = 1; 2; : : : ; n 1:

2.2.39. Chng minh rng nu f kh vi trn [0; 1] v

(i) f(0) = 0,

(ii) tn ti K > 0 sao cho jf0(x)j Kjf(x)j vi x 2 [0; 1],th f(x) 0.2.2.40. Cho f l mt hm kh vi v hn trn khong (1; 1), J (1; 1)

l mt khong c di . Gi s J -c chia thnh ba khong lin tip

J1; J2; J3 c di t-ng ng l 1; 2; 3, tc l ta c J1 [ J2 [ J3 = J v1 + 2 + 3 = . Chng minh rng nu

mk(J) = infjf(k)(x)j : x 2 J ; k 2 N;

th

mk(J) 1

2(mk1(J1) + mk1(J3)):

2.2.41. Chng minh rng vi gi thit ca bi tr-c, nu jf(x)j 1 vix

2(

1; 1) th

mk(J) 2k(k+1)

2 kk

k; k 2 N:

2.2.42. Gi s rng a thc P(x) = anxn + an1xn1 + + a1x + a0 c nnghim thc phn bit. Chng minh rng nu tn ti p; 1 p n 1 saocho ap = 0 v ai 6= 0 vi mi i 6= p th ap1ap+1 < 0.


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52 Ch-ng 2. Vi phn

2.3 Cng thc Taylor v quy tc LHpital

2.3.1. Gi s f : [a; b] ! R kh vi cp n 1 trn [a; b]. Nu f(n)(x0) tn tith vi mi x 2 [a; b],

f(x) = f(x0) +f0(x0)

1!(x x0) + f

00(x0)2!

(x x0)2

+ + f(n)(x0)

n!(x x0)n + o((x x0)n):

(Cng thc ny -c gi l cng thc Taylor vi phn d- dng Peano).

2.3.2. Gi s f : [a; b] ! R kh vi lin tc cp n trn [a; b] v gi s rngf(n+1) tn ti trong khong m (a; b). Chng minh rng vi mi x; x0 2 [a; b]v mi p > 0 tn ti 2 (0; 1) sao cho ,

f(x) = f(x0) +f0(x0)

1!(x x0) + f

00(x0)2!

(x x0)2

+ + f(n)(x0)

n!(x x0)n + rn(x);

trong

rn(x) =f(n+1)(x0 + (x x0))

n!p(1 )n+1p(x x0)n+1

-c gi l phn d- dng Schlomilch-Roche.

2.3.3. S dng kt qu trn hy chng minh cc dng phn d- sau:

rn(x) = f(n+1)

(x0 + (x x0))(n + 1)!

(x x0)n+1(a)(dng Lagrange),

rn(x) =f(n+1)(x0 + (x x0))

n!(1 )n(x x0)n+1(b)

(dng Cauchy).


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2.3. Cng thc Taylor v quy tc LHpital 53

2.3.4. Cho f : [a; b]

!R l hm kh vi cp n + 1 trn [a; b], x; x0

2[a; b].

Chng minh cng thc Taylor vi phn d- dng tch phn sau:

f(x) = f(x0) +f0(x0)

1!(x x0) + f

00(x0)2!

(x x0)2

+ + fn(x0)

n!(x x0)n + 1

n!

Zxx0

f(n+1)(t)(x t)ndt:

2.3.5. Cho f : [a; b] ! R l hm kh vi cp n + 1 trn [a; b], x; x0 2 [a; b].Chng minh cng thc Taylor sau:

f(x) = f(x0) +f0(x0)

1! (x x0) +f00(x0)

2! (x x0)2

+ + fn(x0)

n!(x x0)n + Rn+1(x);

trong

Rn+1(x) =

Zxx0

Ztn+1x0

Ztnx0

Zt2x0

f(n+1)(t1)dt1 dtndtn+1:

2.3.6. Chng minh cng thc xp x sau

p1 + x 1 +1

2 1

8x2

cho sai s kt qu khng v-t qu 12jxj3 khi jxj < 1

2.

2.3.7. Chng minh cc bt ng thc sau:

(1 + x) > 1 + x vi > 1 hoc < 0;(a)

(1 + x) < 1 + x vi 0 < < 1;(b)

gi thit rng x >

1; x

6= 0.

2.3.8. Cho cc hm f; g 2 C2([0; 1]), g0(x) 6= 0 vi x 2 (0; 1) tho mnf0(0)g00(0) 6= f00(0)g0(0). Vi x 2 (0; 1) xt hm (x) l mt s tho mnnh l gi tr trung bnh tng qut, tc l

f(x) f(0)g(x) g(0) =

f0((x))g0((x))

:


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54 Ch-ng 2. Vi phn

Hy tnh gii hn

limx!0+

(x)x

:

2.3.9. Cho f : R ! R kh vi cp n + 1 trn R. Chng minh rng vi mix 2 R tn ti 2 (0; 1) sao cho

f(x) = f(0) + xf0(x) x2

2f00(x) + + (1)n+1 x

n

n!f(n)(x)(a)

+ (1)n+2 xn+1

(n + 1)!f(n+1)(x);

f x1 + x = f(x) x2

1 + xf0(x) + + (1)n x

2n

(1 + x)n f

(n)

(x)n!(b)

+ (1)n+1 x2n+2

(1 + x)n+1

f(n+1)x+x2

1+x

(1 + n)!

; x 6= 1:

2.3.10. Cho f : R ! R kh vi cp 2n + 1 trn R. Chng minh rng vi mix 2 R tn ti 2 (0; 1) sao cho

f(x) = f(0) +2

1!f0

x

2x

2+

2

3!f(3)

x

2x

23

+ + 2(2n 1)!f(2n1)

x2x

22n1

+2

(2n + 1)!f(2n+1)(x)

x2

2n+1:

2.3.11. S dng kt qu bi trn hy chng minh rng

ln(1 + x) > 2nXk=0

1

2k + 1

x

2 + x

2k+1vi n = 0; 1; : : : v x > 0.

2.3.12. Chng minh rng nu f00(x) tn ti th

limh!0

f(x + h) 2f(x) + f(x h)h2

= f00(x);(a)

limh!0

f(x + 2h) 2f(x + h) + f(x)h2

= f00(x):(b)


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2.3.13. Chng minh rng nu f000(x) tn ti th

limh!0

f(x + 3h) 3f(x + 2h) + 3f(x + h) f(x)h3

= f000(x):

2.3.14. Cho x > 0, hy kim tra cc bt ng thc sau:

ex >nXk=0

xk

k!;(a)

x x2

2+

x3

3 x

4

4< ln(1 + x) < x x

2

2+

x3

3;(b)

1 + 12x 18x2 < p1 + x < 1 + 12x 18x2 + 116x3:(c)2.3.15. Chng minh rng nu tn ti f(n+1)(x) khc 0 v (x) l gi tr -c

xc nh qua cng thc Taylor

f(x + h) = f(x) + hf0(x) + + hn1

(n 1)!f(n1)(x) +

hn

n!f(n)(x + (h)h);

th

limh

!0

(h) =1

n + 1:

2.3.16. Gi s f kh vi trn [0; 1] v f(0) = f(1) = 0. Hn na tn ti f00

trong (0; 1) gii ni, tc l jf00(x)j A; vi mi x 2 (0; 1), Chng minh rng

jf0(x)j A2

; vi x 2 [0; 1]

2.3.17. Gi s f : [c; c] ! R kh vi cp hai trn [c; c] v t Mk =supff(k)(x) : x 2 [c; c]g vi k = 0; 1; 2. Chng minh rng

jf0(x)

j

M0

c+ (x2 + c2)

M2

2cvi x

2[

c; c];(a)

M1 2p

M0M2 vi c r

M0M2

:(b)

2.3.18. Cho f kh vi cp hai trn (a; 1), a 2 R, t

Mk = supff(k)(x) : x 2 (0; 1g < 1; k = 0; 1; 2:


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56 Ch-ng 2. Vi phn

Chng minh rng

M1 2p

M0M2:

Hy ch ra tr-ng hp hm f lm cho bt ng thc tr thnh ng thc.

2.3.19. Cho f kh vi cp hai trn R, t

Mk = supff(k)(x) : x 2 (0; 1)g < 1; k = 0; 1; 2:

Chng minh rng

M1 2pM0M2:2.3.20. Cho f kh vi cp hai trn R, t

Mk = supff(k)(x) : x 2 (0; 1)g < 1; k = 0; 1; 2; : : : ; p; p 2:

Chng minh rng

Mk 2k(pk)=2M1(k=p)0 M

k=p2 ; k = 1; 2; : : : ; p 1:

2.3.21. Gi s f00 tn ti v gii ni trong (0;

1). Chng minh rng nu

limx!1 f(x) = 0 th limx!1 f

0(x) = 0.

2.3.22. Gi s f kh vi lin tc cp hai trn (0; 1), tho mn

limx!+1

xf(x) = 0 v limx!+1

xf00(x) = 0:

Chng minh rng limx!+1

xf0(x) = 0:


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2.3.23. Gi s f kh vi lin tc cp hai trn (0; 1) v tho mn

(i) limx!1

f(x) = 0;

(ii) tn ti M > 0 sao cho (1 x2)jf00(x)j M vi x 2 (0; 1).

Chng minh rng limx!1

(1 x)f0(x) = 0:

2.3.24. Cho f kh vi trn [a; b] v gi s rng f0(a) = f0(b) = 0. Chng minh

rng nu f00 tn ti trong (a; b) th tn ti c 2 (a; b) sao cho

jf00(c)j 4

(b a)2 jf(b) f(a)j:

2.3.25. Gi s f[1; 1] ! R kh vi cp ba v bit rng f(1) = f(0) =0; f(1) = 1 v f0(0) = 0. Chng minh rng tn ti c 2 (1; 1) sao chof000(c) 3.

2.3.26. Cho f kh vi lin tc cp n trn [a; b] v t

Q(t) =f(x) f(t)

x

t

; x; t 2 [a; b]; x 6= t:

Chng minh cng thc Taylor d-i dng sau:

f(x) = f(x0) +f0(x0)

1!(x x0) + + f

(n)(x0)

n!(x x0)n + rn(x);

vi

rn(x) =Q(n)(x0)

n!(x x0)n+1:

2.3.27. Gi s rng f : (1; 1) ! R kh vi ti 0, cc dy fxng, fyng thomn

1 < xn < yn < 1; n

2N sao cho lim

n!1

xn = limn!1

yn = 0. Xt th-ng

Dn =f(yn) f(xn)

yn xn :

Chng minh rng

(a) nu xn < 0 < yn th limn!1

Dn = f0(0).


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58 Ch-ng 2. Vi phn

(b) nu 0 < xn < yn v dy n ynynxno gii ni th limn!1Dn = f0(0).(c) nu f0 tn ti trong (1; 1) v lin tc ti 0 th lim

n!1Dn = f

0(0).

(Hy so snh vi 2.1.13 v 2.1.14.)

2.3.28. Cho m 2 N , xt a thc P sau

P(x) =m+1Xk=1

m + 1

k

(1)k(x k)m; x 2 R:

Chng minh rng P(x) 0.2.3.29. Gi s rng f(n+2) lin tc trn [0; 1]. Chng minh rng tn ti

2 (0; 1) sao cho

f(x) = f(0) +f0(0)

1!x + + f

(n1)(0)(n 1)! x

n1 +f(n)

x

n+1

n!

xn

+n

2(n + 1)f(n+2)(x)

xn+2

(n + 2)!:

2.3.30. Gi s rng f(n+p) tn ti trong [a; b] v lin tc ti x0 2 [a; b]. Chngminh rng nu f(n+j)(x0) = 0 vi j = 1; 2; : : : ; p 1, f(n+p)(x0) 6= 0 v

f(x) = f(x0) +f0(x0)

1!(x x0) + + f

(n1)(x0)(n 1)! (x x0)

n1

+f(n)(x0 + (x)(x x0))

n!(x x0)n:

th

limx!x0

(x) = n + p

n 1=p

:

2.3.31. Cho f l hm kh vi lin tc cp hai trn (1; 1) v f(0) = 0. Hytnh gii hn

limx!0+

h1px

iXk=1

f(kx):


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2.3.32. Cho f kh vi v hn trn (a; b). Chng minh rng nu f bng 0 ti

v hn im trong khong ng [c; d] (a; b) v

supfjf(n)(x)j : x 2 (a; b)g = O(n!) khi n ! 1

th f bng khng trn mt khong m nm trong (a; b).

2.3.33. Gi s rng

(i) f kh vi v hn trn R,

(ii) tn ti L > 0 sao cho jf(n)(x)j L vi mi x 2 R v mi n 2 N,

(iii) f1n

= 0 vi n 2 N: Chng minh rng f(x) 0 trn R:

2.3.34. S dng quy tc lHpital tnh cc gii hn sau:

(a) limx!1

arctan x21x2+1

x

1

; (b) limx!+1

x1 +1

xx

e ;(c) lim

x!5(6 x) 1x5 ; (d) lim

x!0+

sin x

x

1=x;

(e) limx!0+

sin x

x

1=x2:

2.3.35. Chng minh rng vi f kh vi lin tc cp hai trn R tho mn

f(0) = 1, f0(0) = 0 v f00(0) = 1 th

limx!+1f apxx

= ea2=2

; trong a 2 R:

2.3.36. Vi a > 0 v a 6= 1 hy tnh

limx!+1

ax 1

x(a 1)1=x

:


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60 Ch-ng 2. Vi phn

2.3.37. C th s dng quy tc lHpital trong nhng tr-ng hp sau -c

khng ?

limx!1

x sin x2x + sin x

;(a)

limx!1

2x + sin 2 + 1

(2x + sin 2x)(sin x + 3)2;(b)

limx!0+

2sin

px +

px sin

1

x

x;(c)

limx!

01 + xe1=x2

sin1

x4

e1=x2

:(d)

2.3.38. Hm

f(x) =

(1

x ln 2 1

2x1 nu x 6= 0;12

nu x = 0

c kh vi ti im 0 khng ?

2.3.39. Gi s f kh vi lin tc cp n trn R, a 2 R. Chng minh ng thcsau:

f(n)(a) = limh!

0

1

hn

n

Xk=0 (1)nk

n

kf(a + kh) :2.3.40. Chng minh quy tc lHpital d-i dng sau:

Gi s f; g : (a; b) ! R , 1 < a < b < +1 l cc hm kh vi trn (a; b),ng thi tho mn iu kin

(i) g0(x) 6= 0 vi x 2 (a; b),

(ii) limx!a+

g(x) = +1(1),

(iii) limx!a+

f0(x)

g0(x)= L;

1 L +

1:

Khi

limx!a+

f(x)

g(x)= L:

2.3.41. S dng quy tc lHpital va nu trn hy chng minh kt qu

tng qut ca 2.2.32 : Cho f kh vi trn (0; 1) v a > 0.


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2.4. Hm li 61

(a) Nu limx!+1

(af(x) + f0(x)) = L; th limx!+1

f(x) = La

:

(b) Nu limx!+1

(af(x) + 2p

xf0(x)) = L; th limx!+1

f(x) = La :

Cc kt qu trn c cn ng i vi tr-ng hp a m khng ?

2.3.42. Gi s f kh vi cp ba trn (0; 1) sao cho f(x) > 0, f0(x) > 0,f00(x) > 0 vi mi x > 0. Chng minh rng nu

limx!1

f0(x)f000(x)(f00(x))2

= c; c 6= 1;

thlimx!1

f(x)f00(x)(f0(x))2

=1

2 c:

2.3.43. Gi s rng f l hm kh vi v hn trn (1; 1) v f(0) = 0. Chngminh rng nu g -c xc nh trn (0; 1)nf0g theo cng thc g(x) = f(x)

xth

tn ti mt m rng ca g kh vi v hn trn (1; 1).

2.4 Hm li

nh ngha 1. Mt hm f -c gi l li trong khong I R nu

f(x1 + (1 )x2) f(x1) + (1 )f(x2)(1)

trong x1; x2 2 I v 2 (0; 1): Mt hm li f -c gi l li cht trong Inu bt ng thc (1) l cht vi x1 6= x2. f l hm lm nu f l hm li.nh ngha 2. Hm f(x) -c gi l tho mn iu kin Lipschitz a

ph-ng trn mt khong m I vi hng s Lipschitz L > 0 nu vi mi

x; y 2 I, x 6= y th jf(x) f(y)j Ljx yj:2.4.1. Chng minh rng f kh vi trn mt khong m I l li khi v ch khi

f0 tng trong I.

2.4.2. Chng minh rng f kh vi cp hai trn mt khong m I l li khi

v ch khi f00(x) 0 vi mi x 2 I.


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62 Ch-ng 2. Vi phn

2.4.3. Chng minh rng nu f li trong khong m I th bt ng thc

Jensen

f(1x1 + 2x2 + + nxn) 1f(x1) + 2f(x2) + + nf(xn)ng vi mi x1; x2; : : : ; xn 2 I v mi b s thc d-ng 1; 2; : : : ; n thomn 1 + 2 + + n = 1.2.4.4. Cho x; y > 0 v p;q > 0 tho mn 1p +

1q = 1. Chng minh bt ng

thc

xy xp

p+

xq

q:


1

n

nXk=1

xk nvuut nY

k=1

xk vi x1; x2; : : : ; xn > 0:

2.4.6. Chng minh rng vi a 6= b ta c bt ng thceb dab a 1 v cc s d-ng x1; x2; : : : ; xn. Chng minh rng

1

n

nXk=1

xk

!

1

n

nXk=1

xk :

2.4.9. Cho x1; x2; : : : ; xn 2 (0; 1) v cc s d-ng p1; p2; : : : ; pn tho mnnPk=1

pk = 1. Chng minh rng

1 + nXk=1

pkxk!1

nYk=1

1 + xkxk

pk ;(a)1 +

nPk=1

pkxk

1 nPk=1

pkxk

nYk=1

1 + xk1 xk

pk:(b)


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2.4. Hm li 63

2.4.10. Cho x = 1n

nPk=1 xk vi x1; x2; : : : ; xn 2 (0; ). Chng minh rngnYk=1

sin xk (sin x)n;(a)

nYk=1

sin xkxk

sin x

x

n:(b)

2.4.11. Chng minh rng vi a > 0 v x1; x2; : : : ; xn 2 (0; 1) tho mn x1 +x2 + + xn = 1 th

nXk=1

xk + 1

xk

a (n2 + 1)ana1

:

2.4.12. Cho n 2, hy kim tra khng nh sau:nYk=1

2k 12k1

2 2

n+

1

n 2n1n

:

2.4.13. Chng minh cc bt ng thc sau:

n2

x1 + x2 + + xn 1

x1 +

1

x2 + +1

x1 ; x1; x2; : : : ; xn > 0;(a)1

1x1

+ + nxn x11 xnn 1x1 + + nxn(b)

vi k; xk > 0; k = 1; 2; : : : ; n tho mnnPk=1

k = 1.

x11 xnn + y11 ynn (x1 + y1)1 (xn + yn)n(c)vi yk; xk 0; k > 0; k = 1; 2; : : : ; n sao cho

nPk=1

k = 1.

mXj=1

nYi=1

xii;j

nYi=1

mXj=1

xi;j!i

(d)

vi ; xi;j 0; k > 0; i ; j = 1; 2; : : : ; n sao chonPk=1

k = 1.

2.4.14. Chng minh rng nu f : R! R li v b chn trn th l hm hngtrn R.


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64 Ch-ng 2. Vi phn

2.4.15. Liu mt hm li gii ni trn (a;

1) hoc trn (

1; a) c lun l

hm hng khng ?

2.4.16. Gi s rng f : (a; b) ! R li trn (a; b) , trong 1 a; b 1.Chng minh rng hoc f n iu trn (a; b) hoc tn ti c 2 (a; b) sao cho

f(c) = minff(x) : x 2 (a; b)g

ng thi f gim trong (a; c] v tng trong [c; b).

2.4.17. Cho f : (a; b)

!R li trn (a; b), trong

1 a; b

1. Chng

minh rng cc gii hn

limx!a+

f(x) v limx!b

f(x)

tn ti, hu hn hoc v hn.

2.4.18. Gi s f : (a; b) ! R li v gii ni trn (a; b) , 1 a; b 1.Chng minh rng f lin tc u trn (a; b). (So snh vi bi 2.4.14).

2.4.19. Gi s f : (a; b)

!R li trn (a; b), trong

1 a; b

1. Chng

minh rng o hm mt pha ca f tn ti v n iu trn (a; b). Hn nao hm phi v tri ca n bng nhau bn ngoi mt tp m -c.

2.4.20. Gi s f kh vi cp hai trn R v f; f0; f00 tng cht trn R. Vi a; b

cho tr-c, a b cho x ! (x); x > 0 xc nh qua nh l gi tr trung bnh,tc l

f(b + x) f(a x)b a + 2x = f

0():

Chng minh rng hm tng trn (0; 1).

2.4.21. S dng kt qu bi 2.4.4 chng minh bt ng thc Holder: Cho

p; q > 1 tho mn 1p

+ 1q

= 1. Chng minh rng

nXi=1

jxiyij

nXi=1

jxijp!1=p nX

i=1

jyijq!1=q

:


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2.4. Hm li 65

2.4.22. S dng bt ng thc Holder chng minh bt ng thc Mikowski

sau: Nu p > 1 thnXi=1

jxi + yijp!1=p

nXi=1

jxijp!1=p

+

nXi=1

jyijp!1=p

:

2.4.23. Chng minh rng nu chui1Pn=1

a4n hi t th1Pn=1

ann4=5

hi t.

2.4.24. Cho xi; yi 0, i = 1; 2; : : : ; n v p > 1. Chng minh bt ng thcsau

((x1 + + xn)p + (y1 + + yn)p)1=p (xp1 + yp1)1=p + + (xpn + ypn)1=p :

2.4.25. Chng minh bt ng thc Minkowski tng qut sau: Cho xi;j 0,i = 1; 2; : : : ; n; j = 1; 2; : : : ; m v p > 1, chng minh rng

nXi=1

mXj=1

xi;j

!p!1=p

mXj=1

nXi=1

xpi;j

!1=p:

2.4.26. Gi s hm lin tc f trn khong I l li trung bnh tc l

f

x + y

2

f(x) + f(y)

2vi x; y 2 I:

Chng minh rng f li trn I.

2.4.27. Chng minh rng iu kin lin tc trong bi 2.4.26 l khng th

b -c. (Hy ch ra phn v d).

2.4.28. Cho f lin tc trn I sao cho

fx + y2

< f(x) + f(y)2

vi x; y 2 I, x 6= y. Chng minh rng f li cht trn I.

2.4.29. Gi s f li trong khong m I. Chng minh rng f tho mn iu

kin Lipschitz a ph-ng trn I.


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66 Ch-ng 2. Vi phn

2.4.30. Cho f : (0;

1)

!R li, t

limx!0+

f(x) = 0:

Chng minh rng hm f 7! f(x)x

tng trn (0; 1).

2.4.31. Ta ni rng hm f d-i cng tnh trn (0; 1) nu vi x1; x2 2 (0; 1),

f(x1; x2) f(x1) + f(x2):

Chng minh rng

(a) nu x 7! f(x)x

gim trn (0; 1) th f d-i cng tnh.

(b) nu f li v d-i cng tnh trn (0; 1) th hm x 7! f(x)x

l hm gim

trn khong .

2.4.32. Gi s f kh vi trn (a; b) v vi mi x; y 2 (a; b), x 6= y, tn ti duynht sao cho

f(y) f(x)y

x

= f0():

Chng minh rng f li cht hoc lm cht trn (a; b).

2.4.33. Cho f : R ! R lin tc v tho mn iu kin vi mi d 2 R, hmgd(x) = f(x + d) f(x) thuc lp C1(R). Chng minh rng f thuc C1(R).

2.4.34. Gi s an : : : a2 a1 v f li trn on [an; a1]. Chng minh

rngn

Xk=1f(ak+1)ak

n

Xk=1f(ak)ak+1;

trong an+1 = a1.

2.4.35. Gi s rng f lm v tng cht trn mt khong (a; b), 1 a; b 1. Chng minh rng nu a < f(x) < x vi x 2 (a; b) v

limx!a+

f0+(x) = 1;


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2.5. Cc ng dng ca o hm 67

th vi x; y

2(a; b) ta c

limn!1

fn+1(x) fn(x)fn+1(y) fn(y) = 1;

trong fn l thnh phn lp th n ca f (xem 1.1.40).

2.5 Cc ng dng ca o hm

2.5.1. S dng nh l gi tr trung bnh tng qut hy chng minh

1 x2

2!< cos x; vi x 6= 0;(a)

x x3

3!< sin x; vi x > 0;(b)

cos x < 1 x2

2!+

x4

4!; vi x 6= 0;(c)

sin x < x x3

3!+

x5

5!; vi x > 0:(d)

2.5.2. Cho n 2 N v x > 0 hy kim tra cc khng nh sau:

x x3

3!+

x5

5! + x

4n3

(4n 3)! x4n1

(4n 1)! < sin x(a)

< x x3

3!+

x5

5! + x

4n3

(4n 3)! x4n1

(4n 1)! +x4n+1

(4n + 1)!;

1 x2

2!+

x4

4! + x

4n4

(4n 4)! x4n2

(4n 2)! < cos x(b)

< 1 x2

2!+

x4

4! + x

4n4

(4n

4)! x

4n2

(4n

2)!+

x4n

(4n)!:

2.5.3. Cho f lin tc trn [a; b] v kh vi trn khong m (a; b). Chng minh

rng nu a 0 th tn ti x1; x2; x3 2 (a; b) sao cho

f0(x1) = (b + a)f0(x2)

2x2= (b2 + ab + a2)

f0(x3)3x23

:


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77/398


2.5.11. S dng nh l gi tr trung bnh chng minh rng vi 0 < p < q

ta c 1 +

x

p

p 0.

2.5.12. Chng minh rng ex 1 + x vi x 2 R. S dng kt qu chngminh bt ng thc lin h gia trung bnh cng v trung bnh nhn.


xy ex + y(ln y 1)

vi x 2 R v y > 0. Chng minh rng du ng thc xy ra khi v ch khiy = ex.

2.5.14. Gi s f : R ! [1; 1] thuc lp C2(R) v (f(0))2 + (f0(0))2 = 4.Chng minh rng tn ti x0 2 R sao cho f(x0) + f0(x0) = 0:

2.5.15. Kim tra cc bt ng thc sau:x +

1

x

arctan x > 1 vi x > 0;(a)

2tan x sinh x > 0 vi 0 < x < 2

;(b)

ln x 0; x 6= e;(c)

x ln x

x2 1 0; x 6= 1:(d)

2.5.16. So snh cc s sau:

(a) e hay e,

(b) 2p2 hay e,

(c) ln 8 hay 2.


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70 Ch-ng 2. Vi phn

2.5.17. Kim tra cc khng nh sau:

ln

1 +x

a

ln

1 +

b

x

0;(a)

1 +x

m

m 1 +

x

m

m< 1; x 2 Rnf0g; m ; n 2 N; m ; n jxj;(b)

ln

1 +p

1 + x2

0:(c)

2.5.18. Cho x > 0 hy kim tra cc bt ng thc sau:

ln(1 + x) 0(a)

ln(1 + cos x) < ln 2 x2

4; vi x 2 (0; ):(b)

2.5.20. Cho x > 0, chng minh cc bt ng thc sau:

(a) ex < 1 + xex; (b) ex 1 x < x2ex;(c) xex=2 < ex 1; (d) ex < (1 + x)1+x;

(e)

x + 1

2

x+1 xx:

2.5.21. Chng minh rng (e + x)ex > (e x)e+x vi x 2 (0; e).

2.5.22. Chng minh rng nu x > 1 th ex1 + ln x + 1 > 0:

2.5.23. Chng minh cc bt ng thc sau:1

2tan x +

2

3sin x > x; vi 0 < x 3sin x; vi x > 0;(b)

cos x 0 :

(x + y) < x + y:

2.5.26. Cho 2 (0; 1) v x 2 [1; 1], chng minh rng

(1 + x) 1 + x

( 1)

8

x2:

2.5.27. Chng minh kt qu tng qut ca bi trn: Cho B 0 v x 2(1; B], chng minh rng:

(1 + x) 1 + x (1 )2(1 + B)2

x2 vi 0 < < 1;(a)

(1 + x) 1 + x (1 )2(1 + B)2

x2 vi 1 < < 2:(b)


sin x 2

x; vi x 2 h0; 2

i;(a)

sin x 2

x +x

3(2 4x2); vi x 2

h0;

2

i:(b)

2.5.29. Chng minh rng vi x 2 (0; 1) ta c

x(1 x) < sin x 4x(1 x):

2.5.30. Chng minh rng vi x d-ng v n nguyn d-ng ta c

ex nXk=0

xk

k!< x

n(ex 1):

2.5.31. Cho n nguyn d-ng. Hy tm cc cc tr a ph-ng ca hm

f(x) =

1 + x +

x2

2!+ + x

n

n!

ex:


80/398

72 Ch-ng 2. Vi phn

2.5.32. Cho m v n nguyn d-ng, tm cc cc tr a ph-ng ca

f(x) = xm(1 x)n:

2.5.33. Cho m; n nguyn d-ng, tm gi tr ln nht ca

f(x) = sin2m x cos2n x:

2.5.34. Tm cc cc tr a ph-ng ca hm f(x) = x1=3(1 x)2=3:

2.5.35. Tm gi tr ln nht v gi tr nh nht ca hm

f(x) = x arcsin x +p

1 x2

trn [1; 1].

2.5.36. Tm gi tr ln nht trn R ca

f(x) =1

1 + jxj +1

1 + j1 xj :

2.5.37. Cho cc s khng m a1; a2; : : : ; an. Chng minh cc bt ng thc

sau:

1

n

nXk=1

akeak

1

e;(a)

1

n

nXk=1

a2keak

4

e2;(b)

n

Yk=1

ak 3en

exp(13n

Xk=1

ak) :(c)2.5.38. Tm cc cc tr a ph-ng ca hm

f(x) =

(e1=jxj

p2 + sin 1

x

vi x 6= 0;

0 vi x = 0:


81/398


2.5.39. Cho

f(x) =(x4 2 + sin 1x vi x 6= 0;

0 vi x = 0:

Chng minh rng f kh vi trn R v ti 0 f t gi tr ln nht tuyt i

nh-ng f khng n iu trong bt k khong ("; 0) hay (0; ") no.

2.5.40. Cho x > 0, chng minh bt ng thc sau

sinh x

psinh2 x + cosh2 x< tanh x < x < sinh x 0 tho mn y + z < 1 ta c f(y + z) 0

:

2.5.56. Cho f kh vi v hn trn (0; 1), gi s rng vi mi x 2 [0; 1] tn tin(x) sao cho f(n(x))(x) = 0: Chng minh rng trn on [0; 1] f s ng nht

vi mt a thc.

2.5.57. Ch ra v d chng t rng gi thit kh vi v hn trn [0; 1] trongbi tp trn l cn thit. Chng minh rng nu

limn!1

f(n)(x) = 0

vi mi x 2 [0; 1] th ta khng th suy ra kt lun trong bi 2.5.56.


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76 Ch-ng 2. Vi phn

2.6 Kh vi mnh v kh vi theo ngha Schwarz

nh ngha 1. Mt hm thc xc nh trn tp m A R -c gi l khvi mnh ti im a 2 A nu

lim(x1;x2)!(a;a)

x16=x2

f(x1) f(x2)x1 x2 = f

(a)

tn ti hu hn. f(a) -c gi l o hm mnh ca f ti a.

nh ngha 2. Mt hm thc f xc nh trn tp mA

R -

c gi l khvi theo ngha Schwarz ti a 2 A nu gii hn

limh!0

f(a + h) f(a h)2h

= fs(a)

tn ti hu hn, fs(a) -c gi l o hm theo ngha Schwarz hay ni gn

li l o hm Schwarz ca f ti im a.

nh ngha 3. o hm mnh trn (t-ng ng d-i) ca f ti a -c xc

nh bng cch thay th lim trong nh ngha 1 bng lim (t-ng ng lim ),

k hiu l Df(a) (t-ng ng Df(a)). o hm Schwarz trn v d-i ca f

ti a -c xc nh bng cch thay th t-ng t. Ta k hiu chng l Dsf(a)

v Dsf(a).

2.6.1. Chng minh rng nu f kh vi mnh ti a th n kh vi ti a v

f(a) = f0(a). Hy ch ra phn v d chng t iu ng-c li khng ng.

2.6.2. Cho f : A! R v k hiu A1, A l tp cc im m ti f kh viv kh vi mnh. Chng minh rng nu a

2A l mt im gii hn ca A

th

limx!Ax2A

f(x) = limx!Ax2A1

f0(x) = f(a) = f0(a):

2.6.3. Chng minh rng mi hm kh vi lin tc ti a th kh vi mnh ti

a.


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86/398

78 Ch-ng 2. Vi phn

2.6.14. Cho f kh vi Schwarz trn (a; b) , xt x02

(a; b) l cc tr a ph-ng

ca f, hi o hm Schwarz ca f c bng 0 ti x0 khng ?

2.6.15. Ta ni hm f : R ! R c tnh cht Baire nu tn ti mt tp thngd- S R f lin tc trn . Chng minh rng nu f c tnh cht Baireth tn ti mt tp thng d- B sao cho vi mi x 2 B,

Dsf(x) = Df(x) v Dsf(x) = Df(x):

2.6.16. Chng minh rng nu f c tnh cht Baire v kh vi Schwarz trn

R th f kh vi mnh trn mt tp thng d-.

2.6.17. Cho f kh vi Schwarz trn mt khong m I v xt [a; b] I, ta nirng f kh vi Schwarz u trn [a; b] nu vi mi " > 0 tn ti > 0 sao cho

vi jhj < , f(x + h) f(x h)2h fs(x)

< ";

vi x 2 [a; b] v x + h; x h 2 I. Gi s f kh vi Schwarz trn I v [a; b] I.Chng minh rng nu tn ti x0 2 (a; b) sao cho lim

h

!0jf(x0 + h)j = +1 v tn

ti x1 sao cho f b chn trong [x1; x0), th f khng kh vi Schwarz u trn[a; b].

2.6.18. Gi s f lin tc trn I cha [a; b]. Chng minh rng f kh vi

Schwarz u trn [a; b] khi v ch khi fs lin tc trn [a; b].

2.6.19. Hy ch ra phn v d chng t rng gi thit lin tc ca hm

f bi tp trn l cn thit.

2.6.20. Chng minh rng mt hm b chn a ph-ng trn khong m I f

s kh vi Schwarz u trn mi on [a; b] I khi v ch khi f0 lin tc trnI.


87/398

Ch-ng 3

Dy v chui hm

3.1 Dy hm v s hi t u

Chng ta nhc li nh ngha sau.

nh ngha. Chng ta ni rng dy hm ffng hi t u v hm f trn Anu vi mi s" > 0 c mt sn0 2 N sao cho vi mi n n0 bt ng thcjfn(x) f(x)j < " tho mn vi mi x 2 A. Chng ta k hiu l fn

A

f.

3.1.1. Chng minh rng dy hm ffng xc nh trn A l hi t u trnB A v hm f : B ! R nu v ch nu dy sfdng , vi

dn = supfjfn(x) f(x)j : x 2 Bg; n 2 N;

hi t v 0.

3.1.2. Gi s fn A

f v gn A

g. Chng minh rng fn + gn A

f + g. Khng

nh fn gn A

f g c ng khng?

3.1.3. Gi s fn A

f , gn A

g, v tn ti s M > 0 sao cho jf(x)j < M vjg(x)j < M vi mi x 2 A. Chng minh rng fn gn

A

f g.

3.1.4. Cho fang l dy s thc hi t, v ffng l dy hm tho mn

supfjfn(x) fm(x)j : x 2 Ag jan amj; n; m 2 N:

79


88/398

80 Ch-ng 3. Dy v chui hm

Chng minh rng dy hm

ffn

ghi t u trn A.

3.1.5. Chng minh rng hm gii hn ca mt dy hm b chn hi t u

trn A l mt hm b chn. Khng nh ny c ng trong tr-ng hp hi

t im khng?

3.1.6. Chng minh rng dy hm ffng, vi

fn(x) =

(xn

nu n chn,1n

nu n l.

hi t im nh-

ng khng hi t u trn R. H

y tm d

y con hi t u.3.1.7. Chng minh tiu chun Cauchy cho s hi t u.

Dy hm ffng, x c nh trnA, hi t u trnA nu v ch nu vi mi " > 0tn ti sn0 2 N sao cho vi mi m > n0 bt ng thc jfn+m(x) fm(x)j < "tho mn vi mi n 2 N v vi mi x 2 A.3.1.8. Xt s hi t u trn on [0; 1] ca cc dy hm cho b

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