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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.
vii
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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,
3
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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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5
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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6 Ch-ng1. Gii hn v tnh lin tc
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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8 Ch-ng1. Gii hn v tnh lin tc
(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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10 Ch-ng1. Gii hn v tnh lin tc
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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12 Ch-ng1. Gii hn v tnh lin tc
(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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(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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14 Ch-ng1. Gii hn v tnh lin tc
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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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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16 Ch-ng1. Gii hn v tnh lin tc
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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18 Ch-ng1. Gii hn v tnh lin tc
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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20 Ch-ng1. Gii hn v tnh lin tc
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.
1.4.14. Chng minh rng
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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22 Ch-ng1. Gii hn v tnh lin tc
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 nh- trong 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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24 Ch-ng1. Gii hn v tnh lin tc
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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26 Ch-ng1. Gii hn v tnh lin tc
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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28 Ch-ng1. Gii hn v tnh lin tc
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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30 Ch-ng1. Gii hn v tnh lin tc
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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32 Ch-ng1. Gii hn v tnh lin tc
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 nh- trong 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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34 Ch-ng1. Gii hn v tnh lin tc
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 nh- trong 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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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
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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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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)!:
2.1.31. Chng minh rng
(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. Cc nh l gi tr trung bnh 47
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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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. Cc nh l gi tr trung bnh 51
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. Cng thc Taylor v quy tc LHpital 55
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. Cng thc Taylor v quy tc LHpital 57
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. Cng thc Taylor v quy tc LHpital 59
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:
2.4.5. Chng minh rng
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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2.5. Cc ng dng ca o hm 69
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.
2.5.13. Chng minh rng
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)
2.5.28. Chng minh rng
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:
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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:
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2.5. Cc ng dng ca o hm 73
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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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.
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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
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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