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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 .
Trc y, hu ht nhng ngi lm ton ca Vit Nam thng 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 tng 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, trc mi chng chng ti trnh by tm tt l thuyt gip bn cnh li cc kin thc c bn cn thit khi gii bi tp trong chng tng 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 gianmetric 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 Hng (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 Dng 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, Trng 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, KhoaTon C Tin hc, trng i hc Khoa hc T nhin, i hc Quc gia
H 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 dng Z - tp cc s nguyn N - tp cc s nguyn dng 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
sgnx =
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 ngca n, nu n khng b chn trn th ta quy c rng supA = +1.
Nu A R khc rng v b chn di th ta k hiu infA l cn di ngca n, nu n khng b chn di th ta quy c rng infA = 1.
Dy fang cc s thc c gi l n iu tng (tng ng n iu gim)nu an+1 an (tng 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 di ng v cn trn ng cady , k hiu ln lt 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!1
an =
8>:1 nu fang khng b chn di;+1 nu fang b chn di v S = ;;inf S nu fang b chn di 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 nc ta t trc n nay, cc hm tang vctang cng nh cc hm ngc 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 nc chu u,chng c k hiu tng t l tanx, cotx, arctanx, arccotx. 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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Chng 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 (tng ng, tng thc s, gim, gim thcs) trn tp khc rng A 2 R nu x1 < x2; x1; x2 2 A ko theo f(x1) f(x2)(tng ng f(x1) < f(x2), f(x1) f(x2), f(x1) > f(x2) ). Hm tng hay gim(tng ng, tng thc s hay gim thc s) gi l hm n iu (tng 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) lim
x!0[x]
x;
(e) limx!1
x(px2 + 1 3
px3 + 1); (f) lim
x!0cos(
2cosx)
sin(sinx):
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(sinx) = l,
3
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4 Chng 1. Gii hn v tnh lin tc
(b) limx!0
f(x) = l th limx!0
f(jxj) = l. iu ngc li c ng khng ?
1.1.3. Gi s hm f : (a; a) n f0g ! (0;+1) tho mn limx!0(f(x) + 1
f(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 1av a; b > 1 th lim
x!0+f(x) = f(0).
1.1.6. Tnh
(a) limx!0(x2(1 + 2 + 3 + + [ 1jxj ]));
(b) limx!0+
(x([ 1x] + [ 2
x] + + [k
x])); k 2 N.
1.1.7. Tnh limx!1
[P (x)]P (jxj) , y P (x) l a thc vi h s dng.
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, th
limx!0
f(x) = 0.
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5
1.1.10. Cho trc s thc , gi s limx!1
f(ax)x
= g(a) vi mi s dng 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!1
lnxx= 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 1x
; a > 0;
(c) limx!0
(1 + x) 1x
; 2 R:
1.1.18. Tm
(a) limx!1
(lnx)1x ; (b) lim
x!0+xsinx;
(c) limx!0(cosx)
1sin2 x ; (d) lim
x!1(ex 1) 1x ;
(e) limx!0(sin x)
1lnx :
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6 Chng 1. Gii hn v tnh lin tc
1.1.19. Tm cc gii hn sau:
(a) limx!0
sin 2x+ 2arctg 3x+ 3x2
ln(1 + 3x+ sin2 x) + xex; (b) lim
x!0ln cosx
tg x2;
(c) limx!0+
p1 ex p1 cosxp
sinx; (d) lim
x!0(1 + x2)cotg x:
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 2 R , cc s dng m;M saocho m f(x)
xM vi nhng gi tr dng ca x trong ln cn ca 0. Chng
minh rng nu limx!0+
g(x) lnx = ; th limx!0+
f(x)g(x) = e . Trng 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+
2 sin
px+
px sin 1
x
x,
(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 dng 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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7
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 di 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 + 1limx!+1
f(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 nulimx!+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
fx1x 1
x
= 0, th lim
x!0f(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 Chng 1. Gii hn v tnh lin tc
(b) f(x0 ) = limx!x0
f(x) = supx 0 v p l s nguyn dng c nh. K hiu fn l php lp
th n ca f . Chng minh rng nu mp l s nguyn dng nh nht sao cho
fmp(0) > 0, th
p
mp lim
n!1
fn(0)
n lim
n!1fn(0)
n pmp
+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; q 2 N, vp; q nguyn 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; q 2 N, v
p; q nguyn 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 jf j 2 C([a; b]). Ch ra bng vd rng iu ngc li khng ng.
1.2.5. Xc nh tt c cc an v bn sao cho hm xc nh bi
f(x) =
(an + sinx nu x 2 [2n; 2n+ 1]; n 2 Z ,bn + cosx nu x 2 (2n 1; 2n); n 2 Z ,
lin tc trn R.
1.2.6. Cho f(x) = [x2] sinx 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 Chng 1. 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!1
2npcos2n x+ sin2n x; x 2 R:
1.2.9. Chng minh rng nu f : R! R lin tc v tun hon th n c gitr 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] nhng khng c gi tr nh nht,
cng khng c gi tr ln nht.
(b) Cho v d v hm b chn trn [0; 1] nhng 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 2 C([a; b]) v vi x 2 [a; b], t h(x) = minff(x); g(x)g vH(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 trc, 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 Chng 1. 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 thngc 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 dng 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 trc vn cn ng
khi tnh lin tc ca f trn R c thay th bi tnh lin tc ti mtim.
(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 lt vi chuk c bn dng 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!1
1
n
nXk=1
(1)kf(kn) = 0:
1.2.31. Cho f lin tc trn [0; 1]. Chng minh rng
limn!1
1
2n
nXk=0
(1)kn
k
f(k
n) = 0:
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14 Chng 1. 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 mis dng x v mi s t nhin n. Chng minh rng lim
x!1f(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
nhng 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 minhrng 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 chotrc 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 phng trnh (1 x) cosx = sinx c t nht mtnghim trong (0; 1).
(b) Vi a thc khc khng P , chng minh rng phng 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) =nYk=0
(x+ ak) + 2nYk=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 Chng 1. Gii hn v tnh lin tc
1.3.12. Vi n 2 N, gi f 2 C([0; n]) sao cho f(0) = f(n). Chng minh rngtn 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(x0k), 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 ?
1.3.14. 6 Vi n 2 N, gi f 2 C([0; n]) sao cho f(0) = f(n). Chng minh rngphng 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 vinhau; tc l, f(g(x)) = g(f(x)) vi mi x 2 R. Chng minh rng nu phngtrnh f2(x) = g2(x) c nghim, th phng 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) f 2(x) = x; x 2 R, nu f gim thc s.
1.3.18. Gi s f : R ! R tho mn iu kin f(f(x)) = f 2(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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17
1.3.20. Chng minh rng nu f : R ! R c tnh cht gi tr trung gian vf1(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 trc, 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 phng trnh
f(x) = 0 c nghim trong khong m (0; 1).
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18 Chng 1. Gii hn v tnh lin tc
1.3.28. Chng minh rng mi song nh f : R! [0;1) c v hn im ginon.
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 trng 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 vhai 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 supA (tng ng inf A) ls thc m rng nh nht (tng ng, ln nht) m ln hn (tng 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 di (tng nggii hn trn) ca f(x) khi x ! x0 c nh ngha l inf (tng 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 di v gii hn trn ca f(x) khi
x! x0 c k hiu tng ng bi limx!x0
f(x) v limx!x0
f(x).
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19
nh ngha 7. Mt hm gi tr thc gi l na lin tc di (tng ngtrn) ti x0 2 A; x0 l im gii hn ca A, nu lim
x!x0f(x) f(x0) (tng ng
limx!x0
f(x) f(x0)). Nu x0 l im c lp ca A, th ta gi s rng f l nalin tc trn v di 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>0supff(x)g : x 2 A; 0 < jx x0j < :
1.4.2. 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+
supff(x)g : x 2 A; 0 < jx x0j < :
1.4.3. Chng minh rng y0 2 R l gii hn di ca f : A! R ti im giihn x0 ca A nu v ch nu vi mi " > 0, hai iu kin sau y c tho
mn :
(i) tn ti > 0 sao cho f(x) > y0 " vi mi x 2 A v 0 < jx x0j < ;
(ii) vi mi > 0, tn ti x0 2 A sao cho 0 < jx0 x0j < v f(x) < y0 + ":
Thit lp bi ton tng 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 Chng 1. 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 nul = lim
x!x0f(x) (tng ng L = lim
x!x0f(x)), th tn ti dy fxng; xn 2 A; xn 6= x0,
hi t ti x0 sao cho l = limn!1
f(xn) (tng ng L = limn!1
f(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
1
f(x)=
1
limx!x0
f(x)v lim
x!x01
f(x)=
1
limx!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 trng hp cc dng bt nh +11v 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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21
1.4.10. Chng minh rng nu limx!x0
f(x) tn ti, th (tr trng hp cc dng
bt nh +11 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 trng 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!af(x) v L =
limx!af(x), th vi mi 2 [l; L], tn ti dy fxng gm cc im trong (a; b) hi
t ti a sao cho limn!1
f(xn) = .
1.4.12. Tm tt c cc im ti f : R! R xc nh bi
f(x) =
(0 nu x v t,sinx nu x hu t
l 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; q 2 N,
v p; q nguyn t cng nhau,
l na lin tc trn.
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22 Chng 1. 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,qxqx+1
nu x = pq; p 2 Z; q 2 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; q 2 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 di.
1.4.16. Cho f; g : A ! R na lin tc trn (tng ng, di) ti x0 2 A.Chng minh rng
(a) nu a > 0 th af na lin tc di (tng ng, trn) ti x0 2 A. Nua > 0 th af na lin tc trn (tng ng, di) ti x0.
(b) f + g na lin tc di (tng ng, trn) ti x0.
1.4.17. Gi s rng fn : A ! R; n 2 N, na lin tc di (tng ng, trn)ti x0 2 A. Chng minh rng sup
n2Nfn (tng ng, sup
n2Nfn) na lin tc di
(tng ng, trn) ti x0.
1.4.18. Chng minh rng gii hn theo tng im ca mt dy tng (tng
ng, gim) cc hm na lin tc di (tng ng, trn) l na lin tc di
(tng 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!xf(z)g v f2(x) = minff(x); lim
z!xf(z)g:
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23
1.4.20. Gi f1; f2, v of nh trong bi ton trc. Chng minh rng f1 v ofl na lin tc trn, v f2 l na lin tc di.
1.4.21. Chng minh rng f : A ! R l na lin tc di (tng ng,trn) ti x0 2 A, iu kin cn v l vi mi a < f(x0) (tng ng,a > f(x0)), tn ti > 0 sao cho f(x) > a (tng 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 di (tng ng,trn) ti x0 2 A, iu kin cn v l vi mi a 2 R, tp fx 2 A : f(x) > ag(tng ng, fx 2 A : f(x) < ag) l m trong A.
1.4.23. Chng minh rng f : R! R l na lin tc di 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 di (tng
ng, trn) f : A ! R l gii hn ca dy tng (tng 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 di v f(x) g(x) khp ni trn A, th tn ti hm lin tc h trnA 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 trc, 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 Chng 1. 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) = lnx; (h) f(x) = cosx 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) =px; (b) f(x) = x sinx;
(c) f(x) = sin2 x; (d) f(x) = sinx2;
(e) f(x) = ex; (f) f(x) = esin(x2);
(g) f(x) = sin sinx; (h) f(x) = sin(x sinx);
(i) f(x) = sinpx:
1.5.3. Chng minh rng nu f lin tc u trn (a; b); a; b 2 R, th limx!a+
f(x)
v limx!b
f(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) fg;
(c) x 7! f(x) sinx ?
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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25
(b) Gi s A v B l cc tp ng trong R v gi f : A[B ! R l lin tcu 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 tcu 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 1xtrn (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 sdng M sao cho jf(x)j
xM vi x 1.
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26 Chng 1. Gii hn v tnh lin tc
1.5.14. Gi f : [0;1)! R lin tc u. Chng minh rng tn ti s dngM vi tnh cht sau y :
supu>0fjf(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() = supfjf(x1 f(x2))j : x1; x2 2 A; jx1 x2j < g
v 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 sautng ng.
-
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 trc, 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 Phng trnh hm
1.6.1. Chng minh rng hm duy nht lin tc trn R v tho mn phngtrnh 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 phng 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):
-
28 Chng 1. 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 phng 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 phng 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 phng trnh hm
f(x) + f(qx) = 0:
1.6.8. Tm tt c cc hm f : R! R lin tc ti khng v tho mn phngtrnh hm
f(x) + f
2
3x
= x:
1.6.9. Xc nh mi nghim f : R! R lin tc ti khng ca phng trnhhm
2f(2x) = f(x) + x:
1.6.10. Tm tt c cc hm lin tc f : R! R tho mn phng trnh Jensen
f
x+ y
2
=f(x) + f(y)
2:
1.6.11. Tm tt c cc hm lin tc trn (a; b); a; b 2 R, tho mn phngtrnh Jensen
f
x+ y
2
=f(x) + f(y)
2:
-
29
1.6.12. Xc nh tt c cc nghim lin tc ti 1 ca phng trnh hm
f(2x+ 1) = f(x):
1.6.13. Vi a thc, chng minh rng nu f : R ! R l nghim lin tc caphng trnh
f(x+ y) = f(x) + f(y) + axy;
th f(x) = a2x2 + bx, y b = f(1) a
2.
1.6.14. Xc nh mi nghim lin tc ti 0 ca phng 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 phng trnh
f(x+ y) + f(x y) = 2f(x)f(y); x; y 2 R:
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 phng 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 phng trnh hm
f(x) + f
x 1x
= 1 + x vi x 6= 0; 1:
-
30 Chng 1. Gii hn v tnh lin tc
1.6.20. Dy fxng hi 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 phng trnh
f
2x f(x)
m
= mx;
th f(x) = m(x c):
1.6.23. Chng minh rng cc nghim duy nht ca phng 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:
-
31
1.6.26. Cho m;n 2 l cc s nguyn. Tm tt c cc hm f : [0;1) ! Rlin 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 phng trnh
f(xy) = f(x)f(y) v f(x+ z) = f(x) + f(z)
vi z 6= 0 no .
1.6.28. Tm tt c cc hm f : R n f0g ! R sao cho
f(x) = f1
x
; x 6= 0:
1.6.29. Tm tt c cc hm f : R n f0g ! R tho mn phng 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 phng 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:
-
32 Chng 1. Gii hn v tnh lin tc
1.7 Hm lin tc trong khng gian metric
Trong mc ny, X v Y ln lt 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, nukhng pht biu ngc 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 tng 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 tp
Borel trong (X; d1):
1.7.3. Cho v d hm lin tc f : X! Y sao cho nh f(F) (tng ng, f(G))khng ng (tng ng, m) trong Y vi F ng (tng 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 lin
tc trn F1 [F2 [ : : :[Fm. Ch ra v d rng pht biu trn khng ngtrong trng 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.
-
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 ngc 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 nghaof(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 trc. 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 ccc 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).
-
34 Chng 1. 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 trng 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,
th
of (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 dng x, dy ff xn
ghi t ti khng. T c suy ra lim
x!0+f(x) = 0 khng ? (so snh vi 1.1.33.)
-
35
1.7.23. K hiu F l h cc hm lin tc trn khng gian metric compact Xsao 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 dng M v tp m khc rng G X socho
jf(x)j M vi mi f 2 F v 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 lim
n!1diam Fn = 0. Chng minh rng
nu 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 tngng.
(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.
-
36 Chng 1. Gii hn v tnh lin tc
1.7.28. Chng minh rng khng gian metric X l compact nu v ch nu
mi hm thc lin tc trn X l lin tc u v vi mi " > 0, tp fx 2 X :(x) > "g, y c xc nh nh trong 1.7.27, l hu hn.1.7.29. Cho v d khng gian metric khng compact sao cho mi hm lin
tc f : X! R l lin tc u trn X.1.7.30. Xt hm nh ngha bi (so snh vi 1.2.3 (a))
f(x) =
8>:0 nu xhu t.0 nu x = 0;1q
nu x = pq; p 2 Z; q 2 N; v p; q nguyn t cng nhau
1.7.31.
1.7.32.
1.7.33.
1.7.34.
1.7.35.
1.7.36.
1.7.37.
1.7.38.
1.7.39.
1.7.40.
1.7.41.
1.7.42.
1.7.43.
1.7.44.
Chng minh. Khng c chi a2 = 1
Chng minh. Li gii tip theo
-
Chng 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 cosx; x 20;
2
nf1g:2.1.3. Nghin cu tnh kh vi ca cc hm s sau:
(a) f(x) =
(arctanx vi jxj 1;4sgnx+ x1
2vi jxj > 1;
37
-
38 Chng 2. Vi phn
(b) f(x) =
(x2ex
2vi jxj 1;
1e
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, nhng 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:
nXk=0
kekx; x 2 R;(a)2nXk=0
(1)k2n
k
kn; n 1;(b)
nXk=1
k cos(kx); x 2 R:(c)
-
2.1. o hm ca hm s thc 39
2.1.7. Chng minh rng nu ja1 sinx+aj2 sin 2x+ +an sinnxj j sinxj vix 2 R th ja1 + 2a2 + + nanj 1.
2.1.8. Gi s rng f v g kh vi ti a, hy xc nh
(a) limx!a
xf(a) af(x)x a ; (b) limx!a
f(x)g(a) f(a)g(x)x a
2.1.9. Gi s rng f(a) > 0 v f kh vi ti a. Hy tnh cc gii hn sau:
(a) limn!1
fa+ 1
n
f(a)
!; (b) lim
x!a
f(x)
f(a)
1lnxln a
; a > 0:
2.1.10. Cho f kh vi ti a. Hy tnh cc gii hn sau:
limx!a
anf(x) xnf(a)x a ; n 2 N;(a)
limx!a
f(x)ex f(a)f(x) cosx f(a) ; a = 0; f
0(0) 6= 0;(b)
limn!1
n
f
a+
1
n
+ f
a+
2
n
+ + f
a+
k
n
kf(a)
; k 2 N;(c)
limn!1
f
a+
1
n2
+ f
a+
2
n2
+ + f
a+
n
n2
nf(a)
:(d)
2.1.11. Vi a > 0 v m; k 2 N hy tnh
limn!1
(n+ 1)m + (n+ 2)m + + (n+ k)m
nm1 kn
;(a)
limn!1
a+ 1
n
n a+ 2
n
n a+ kn
nank
;(b)
limn!1
1 +
a
n2
1 +
2a
n2
1 +
na
n2
:(c)
2.1.12. Gi s rng f(0) = 0 v f kh vi ti im 0. Hy tnh tng
limx!0
1
x
f(x) + f
x2
+ f
x3
+ + f
xk
;
vi k l mt s nguyn dng cho trc.
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40 Chng 2. Vi phn
2.1.13. Cho f l hm kh vi ti im a v fxng v fzng l cc dy hi t tia 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 f 0(a),
(b) khng tn ti hoc c tn ti nhng khc f 0(a).
2.1.14. Cho f l hm kh vi ti a v xt hai dy fxng v fzng cng hi tv a sao cho xn < a < zn vi mi n 2 N. Chng minh rng
limn!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 f 0(1) 6= 0. Hm ngc 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
ngc ca f khng kh vi ti im 1.
-
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; q 2 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)
P 0(xk)(x xk)
vi x 2 Rnfx1; x2; : : : ; xng. Tm tngnXk=1
1
P 0(xk); n 2:
2.1.18. S dng kt qu bi trc hy kim tra cc ng thc sau:
nXk=0
n
k
(1)kx+ k
=n!
x(x+ 1)(x+ 2) (x+ n)(a)
vi x 2 Rnfn;(n 1); : : : ;1; 0g,nXk=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 jf j.
2.1.20. Gi s f1; f2; : : : ; fn 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
f 0k(x)fk(x)
:
-
42 Chng 2. Vi phn
2.1.21. Gi s cc hm f1; f2; : : : ; fn; 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
f 0k(x)fk(x)
g0k(x)
gk(x)
:
2.1.22. Nghin cu tnh kh vi ca f v jf j vi
f(x) =
(x nu x 2 Q;sinx nu x 2 RnQ:(a)
f(x) =
(x 3
2knu x 2 Q \ 1
2k1 ;1
2k2; k 2;
sinx 3
2k
nu x 2 (RnQ) \ 1
2k1 ;1
2k2; k 2:(b)
2.1.23. Chng minh rng nu o hm mt pha f 0(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 phif 0(c) v f
0+(c), th f
0(x0) 0 v f 0+(x0) 0. Hy pht biu bi ton tng
ng trng hp f t cc tiu.
2.1.25. Chng minh rng nu f 2 C([a; b]); f(a) = f(b) v f 0 tn ti trn(a; b) th
infff 0(x) : x 2 (a; b)g 0 supff 0(x) : x 2 (a; b)g:
2.1.26. Chng minh rng nu f 2 C([a; b]) v f 0 tn ti trn (a; b) th
infff 0(x) : x 2 (a; b)g f(b) f(a)b a supff
0(x) : x 2 (a; b)g:
2.1.27. Chng minh rng nu f 0 tn ti v lin tc trn (a; b) th f kh vi
trn (a; b) v f 0(x) = f 0(x) vi x 2 (a; b).
2.1.28. Tn ti hay khng hm f : (1; 2)! R sao cho f 0(x) = x v f 0+(x) = 2xvi x 2 (1; 2) ?
-
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)
f 0(a) = f 0+(a) > 0; f0(b) = f 0(b) > 0:(ii)
Chng minh rng tn ti c 2 (a; b) sao cho f(c) = 0 v f 0(c) 0.2.1.30. Chng minh rng f(x) = arctanx tho mn phng trnh
(1 + x2)f (n)(x) + 2(n 1)f (n1)(x) + (n 2)(n 1)f (n2)(x) = 0
vi x 2 R v n 2. Chng minh rng
f (2m)(0) = 0; f (2m+1)(0) = (1)m(2m)!:
2.1.31. Chng minh rng
(ex sinx)(n) = 2n=2ex sinx+ n
4
; x 2 R; n 1;(a)
(xn lnx)(n) = n!
lnx+ 1 +
1
2+ + 1
n
; x > 0; n 1;(b)
lnx
x
(n)= (1)nn!xn1
lnx 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
sinx+ k
2
= 2n=2 sin
x+ n
4
; x 2 R; n 1(a)
nXk=1
(1)k+1 1k
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 n
l 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:
-
44 Chng 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 vix > 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).
-
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 f 0(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
f 00(x) + f 0(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) + f 0(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 x 2 (a; b) sao cho
g0(x)f(x) + f 0(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 x0f 0(x0) = f(x0):
2.2.4. Gi s f lin tc trn [a; b] v kh vi trn (a; b). Chng minh rng
nu f2(b) f 2(a) = b2 a2 th phng trnh
f 0(x)f(x) = x
c t nht mt nghim trong (a; b).
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46 Chng 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 cho
f 0(x0)f(x0)
=g0(x0)g(x0)
:
2.2.6. Gi s a0; a1; : : : ; an l cc s thc tho mn
a0n+ 1
+a1n+ + an1
2+ an = 0:
Chng minh rng a thc P (x) = a0xn + a1xn1 + + an c t nht mtnghim trong (0; 1).
2.2.7. Xt cc s thc a0; a1; : : : ; an tho mn
a01+2a11+22a23 + 2
n1an1n
+2nann+ 1
= 0:
Chng minh rng hm s
f(x) = an lnn x+ + a2 ln2 x+ a1 lnx+ 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 P 0 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) = f 0(a) = f(b) = 0. Chng minh rng tn ti x1 2 (a; b) sao chof 00(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 f 0(a) = f 0(b) = 0. Chng minh rng tn ti hai s x1; x2 2(a; b); x1 6= x2 sao cho
f 00(x1) = f 00(x2):
-
2.2. Cc nh l gi tr trung bnh 47
2.2.11. Chng minh rng cc phng trnh sau:
x13 + 7x3 5 = 0;(a)3x + 4x = 5x(b)
c ng mt nghim thc .
2.2.12. Chngminh rng vi cc s a1; a2; : : : ; an khc 0 v vi cc s 1; 2; : : : ; ntho mn i 6= j; i 6= j, phng trnh
a1x1 + a2x
2 + + anxn = 0
c nhiu nht l n 1 nghim trong (0;+1).2.2.13. Chng minh rng vi cc gi thit ca bi trn, phng 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 F 0(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 chof 00(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
f 0(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! arctanx 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 jf 00(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). Chngminh rng tn ti x0 2 (a; b) sao cho
f 0(x0) < 1 + f 2(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)f 0(x) + f 2(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
f 0(x) = A th f 0(b) = A.
2.2.24. Gi s f kh vi trn (0;1) v f 0(x) = O(x) khi x!1. Chng minhrng f(x) = O(x2) khi x!1.2.2.25. Cho f1; f2; : : : ; fn v g1; g2; : : : ; gn l cc hm lin tc trn [a; b] v
kh 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
f 0k(c) =nXk=1
g0k(c)fk(b) fk(a)gk(b) gk(a) :
-
2.2. Cc nh l gi tr trung bnh 49
2.2.26. Cho hm f kh vi trn khong m I v gi s [a; b] I. Ta ni rngf kh vi u trn [a; b] nu vi mi " > 0, tn ti > 0 sao cho
f(x+ h) f(x)h
f 0(x)< "
vi mi x 2 [a; b] v jhj < , x + h 2 I. Chng minh rng f kh vi u trn[a; b] khi v ch khi f 0 lin tc trn [a; b].
2.2.27. Cho f lin tc trn [a; b], g kh vi trn [a; b] v g(a) = 0. Chng minh
rng nu tn ti 6= 0 sao cho
jg(x)f(x) + g0(x)j jg(x)j; vi x 2 [a; b];
th g(x) 0 trn [a; b].
2.2.28. Cho f kh vi trn (0;+1). Chng minh rng nu limx!+1
f(x)x= 0 th
limx!+1
jf 0(x)j = 0:
2.2.29. Tm tt c cc hm f : R! R l tho mn phng trnh hmf(x+ h) f(x)
h= f 0
x+
1
2h
vi x; h 2 R; h 6= 0:
(HD. Chng minh rng phng trnh ch c duy nht nghim l mt a thc
bc hai bt k).
2.2.30. Cho cc s dng p; q tho mn p + q = 1, hy tm tt c cc hm
f : R! R tho mn phng trnhf(x) f(y)x y = f
0(px+ qy) vi x; y 2 R; x 6= y:
2.2.31. Chng minh rng nu f kh vi trn khong m I th f 0 nhn mi
gi tr trung gian trong I.
2.2.32. Cho f kh vi trn (0;1). Chng minh rng
(a) nu limx!+1
(f(x) f 0(x)) = 0 th limx!+1
f(x) = 0,
-
50 Chng 2. Vi phn
(b) nu limx!+1
(f(x) 2pxf 0(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 phng trnh f(x) + f 00(x) = 2f 0(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)P 0(x) + xP 2(x) + (P 0(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 > 0cm nghim thc phn bit. Chng minh rng a thcQ(x) = (P (x))2P 0(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
P 0(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.
-
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 S 0(x) = n(xf1)(xf2) : : : (xfn1) th fn1 cn1,
(2) nu T (x) = (xa1)(xa2) : : : (xan+ "), vi " > 0 tho mn an " a2v nu T 0(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 jf 0(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 tng 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) 12(mk1(J1) +mk1(J3)):
2.2.41. Chng minh rng vi gi thit ca bi trc, 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.
-
52 Chng 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) +f 0(x0)1!
(x x0) + f00(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) +f 0(x0)1!
(x x0) + f00(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).
-
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) +f 0(x0)1!
(x x0) + f00(x0)2!
(x x0)2
+ + fn(x0)
n!(x x0)n + 1
n!
Z xx0
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) +f 0(x0)1!
(x x0) + f00(x0)2!
(x x0)2
+ + fn(x0)
n!(x x0)n +Rn+1(x);
trong
Rn+1(x) =
Z xx0
Z tn+1x0
Z tnx0
Z t2x0
f (n+1)(t1)dt1 dtndtn+1:
2.3.6. Chng minh cng thc xp x sau
p1 + x 1 + 1
2 18x2
cho sai s kt qu khng vt 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 mnf 0(0)g00(0) 6= f 00(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) =
f 0((x))g0((x))
:
-
54 Chng 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) + xf 0(x) x2
2f 00(x) + + (1)n+1x
n
n!f (n)(x)(a)
+ (1)n+2 xn+1
(n+ 1)!f (n+1)(x);
f
x
1 + x
= f(x) x
2
1 + xf 0(x) + + (1)n x
2n
(1 + x)nf (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!f 0x2
x2
+2
3!f (3)
x2
x2
3+ + 2
(2n 1)!f(2n1)
x2
x2
2n1+
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 f 00(x) tn ti th
limh!0
f(x+ h) 2f(x) + f(x h)h2
= f 00(x);(a)
limh!0
f(x+ 2h) 2f(x+ h) + f(x)h2
= f 00(x):(b)
-
2.3. Cng thc Taylor v quy tc LHpital 55
2.3.13. Chng minh rng nu f 000(x) tn ti th
limh!0
f(x+ 3h) 3f(x+ 2h) + 3f(x+ h) f(x)h3
= f 000(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 +1
2x 1
8x2 0 sao cho (1 x2)jf 00(x)j M vi x 2 (0; 1).
Chng minh rng limx!1
(1 x)f 0(x) = 0:
2.3.24. Cho f kh vi trn [a; b] v gi s rng f 0(a) = f 0(b) = 0. Chng minh
rng nu f 00 tn ti trong (a; b) th tn ti c 2 (a; b) sao cho
jf 00(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 f 0(0) = 0. Chng minh rng tn ti c 2 (1; 1) sao chof 000(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 di dng sau:
f(x) = f(x0) +f 0(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 2 N sao cho lim
n!1xn = lim
n!1yn = 0. Xt thng
Dn =f(yn) f(xn)yn xn :
Chng minh rng
(a) nu xn < 0 < yn th limn!1
Dn = f0(0).
-
58 Chng 2. Vi phn
(b) nu 0 < xn < yn v dyn
ynynxn
ogii ni th lim
n!1Dn = f
0(0).
(c) nu f 0 tn ti trong (1; 1) v lin tc ti 0 th limn!1
Dn = f0(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) +f 0(0)1!
x+ + f(n1)(0)(n 1)! x
n1 +f (n)
xn+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) +f 0(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):
-
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!+1x1 +
1
x
x e;
(c) limx!5(6 x) 1x5 ; (d) lim
x!0+
sinx
x
1=x;
(e) limx!0+
sinx
x
1=x2:
2.3.35. Chng minh rng vi f kh vi lin tc cp hai trn R tho mnf(0) = 1, f 0(0) = 0 v f 00(0) = 1 th
limx!+1
f
apx
x= ea
2=2; trong a 2 R:
2.3.36. Vi a > 0 v a 6= 1 hy tnh
limx!+1
ax 1x(a 1)
1=x:
-
60 Chng 2. Vi phn
2.3.37. C th s dng quy tc lHpital trong nhng trng hp sau c
khng ?
limx!1
x sinx2x+ sinx
;(a)
limx!1
2x+ sin 2 + 1
(2x+ sin 2x)(sinx+ 3)2;(b)
limx!0+
2 sin
px+
px sin
1
x
x;(c)
limx!0
1 + xe1=x
2
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
nXk=0
(1)nk
n
k
f(a+ kh)
:
2.3.40. Chng minh quy tc lHpital di 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+
f 0(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.
-
2.4. Hm li 61
(a) Nu limx!+1
(af(x) + f 0(x)) = L; th limx!+1
f(x) = La:
(b) Nu limx!+1
(af(x) + 2pxf 0(x)) = L; th lim
x!+1f(x) = L
a:
Cc kt qu trn c cn ng i vi trng hp a m khng ?
2.3.42. Gi s f kh vi cp ba trn (0;1) sao cho f(x) > 0, f 0(x) > 0,f 00(x) > 0 vi mi x > 0. Chng minh rng nu
limx!1
f 0(x)f 000(x)(f 00(x))2
= c; c 6= 1;
th
limx!1
f(x)f 00(x)(f 0(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 aphng 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
f 0 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 f 00(x) 0 vi mi x 2 I.
-
62 Chng 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 dng 1; 2; : : : ; n thomn 1 + 2 + + n = 1.2.4.4. Cho x; y > 0 v p; q > 0 tho mn 1
p+ 1
q= 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 dng x1; x2; : : : ; xn. Chng minh rng1
n
nXk=1
xk
! 1n
nXk=1
xk :
2.4.9. Cho x1; x2; : : : ; xn 2 (0; 1) v cc s dng 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
1nPk=1
pkxk
nYk=1
1 + xk1 xk
pk:(b)
-
2.4. Hm li 63
2.4.10. Cho x = 1n
nPk=1
xk vi x1; x2; : : : ; xn 2 (0; ). Chng minh rng
nYk=1
sinxk (sinx)n;(a)nYk=1
sinxkxk
sinx
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 (n
2 + 1)a
na1:
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)
11x1+ + n
xn
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.
-
64 Chng 2. Vi phn
2.4.15. Liu mt hm li gii ni trn (a;1) hoc trn (1; a) c lun lhm 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. Chngminh 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. Chngminh rng o hm mt pha ca f tn ti v n iu trn (a; b). Hn na
o 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; f 0; f 00 tng cht trn R. Vi a; bcho trc, 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+ 1
q= 1. Chng minh rng
nXi=1
jxiyij
nXi=1
jxijp!1=p nX
i=1
jyijq!1=q
:
-
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
f
x+ y
2
0;(b)
cosx < 1 x2
2!+x4
4!; vi x 6= 0;(c)
sinx < 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)! < sinx(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)! < cosx(b)
< 1 x2
2!+x4
4! + x
4n4
(4n 4)! x4n2
(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
f 0(x1) = (b+ a)f 0(x2)2x2
= (b2 + ab+ a2)f 0(x3)3x23
:
-
68 Chng 2. Vi phn
2.5.4. Chng minh kt qu tng qut ca 2.2.32: Cho f l mt hm bin
phc trn (0;1) v l mt s phc c phn thc dng. Chng minh rngnu f kh vi v lim
x!+1(f(x) + f 0(x)) = 0 th lim
x!+1f(x) = 0.
2.5.5. Cho f kh vi cp hai trn (0;1). Chng minh rng nu limx!+1
(f(x) +
f 0(x) + f 00(x)) = L th limx!+1
f(x) = L.
2.5.6. Cho f kh vi cp ba trn (0;1). Liu t s tn ti ca gii hn
limx!+1
(f(x) + f 0(x) + f 00(x) + f 000(x))
c suy ra s tn ti ca gii hn limx!+1
f(x) khng ?
2.5.7.
(a) Gi s f kh vi lin tc trn (0;1) v cho f(0) = 1. Chng minh rngnu jf(x)j ex vi x 0 th tn ti x0 > 0 sao cho f 0(x0) = ex0 .
(b) Gi s f kh vi lin tc trn (1;1) v cho f(1) = 1. Chng minh rngnu jf(x)j 1
xvi x 1 th tn ti x0 > 1 sao cho f 0(x0) = 1x20 .
2.5.8. Gi s rng f v g kh vi trn [0; a] tho mn f(0) = g(0) = 0, v
g(x) > 0, g0(x) > 0 vi mi x 2 (0; a]. Chng minh rng nu f 0g0 tng trong
(0; a] th fgcng tng trong (0; a].
2.5.9. Chng minh rng cc phng trnh
sin(cosx) = x v cos(sinx) = x
c duy nht nghim trong [0; =2]. Hn na chng minh rng nu x1 v x2ln lt l nghim ca hai phng trnh trn th x1 < x2:
2.5.10. Chng minh rng nu f kh vi trn [a; b], f(a) = 0 v tn ti hng
s C 0 sao cho jf 0(x)j Cjf(x)j vi mi x 2 [a; b] th f(x) 0.
-
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 + (f 0(0))2 = 4.Chng minh rng tn ti x0 2 R sao cho f(x0) + f 0(x0) = 0:
2.5.15. Kim tra cc bt ng thc sau:x+
1
x
arctanx > 1 vi x > 0;(a)
2 tanx sinhx > 0 vi 0 < x < 2;(b)
lnx 0; x 6= e;(c)
x lnx
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.
-
70 Chng 2. Vi phn
2.5.17. Kim tra cc khng nh sau:
ln1 +
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)
ln1 +
p1 + x2
0:(c)
2.5.18. Cho x > 0 hy kim tra cc bt ng thc sau:
ln(1 + x) 0(a)
ln(1 + cosx) < 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 + lnx + 1 > 0:
2.5.23. Chng minh cc bt ng thc sau:
1
2tanx+
2
3sinx > x; vi 0 < x 3 sinx; 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
sinx 2x; vi x 2
h0;
2
i;(a)
sinx 2x+
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) < sinx 4x(1 x):
2.5.30. Chng minh rng vi x dng v n nguyn dng ta c
ex nXk=0
xk
k! 0, chng minh bt ng thc sau
sinhxpsinh2 x+ cosh2 x
< tanh x < x < sinhx 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] trong
bi 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.
-
76 Chng 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)
x1 6=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 m A R c gi l khvi theo ngha Schwarz ti a 2 A nu gii hn
limh!0
f(a+ h) f(a h)2h
= f s(a)
tn ti hu hn, f s(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 (tng ng di) ca f ti a c xcnh bng cch thay th lim trong nh ngha 1 bng lim (tng ng lim ),
k hiu l Df(a) (tng ng Df(a)). o hm Schwarz trn v di ca f
ti a c xc nh bng cch thay th tng 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) = f 0(a). Hy ch ra phn v d chng t iu ngc 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 2 A l mt im gii hn ca Ath
limx!Ax2A
f(x) = limx!Ax2A1
f 0(x) = f (a) = f 0(a):
2.6.3. Chng minh rng mi hm kh vi lin tc ti a th kh vi mnh ti
a.
-
2.6. Kh vi mnh v kh vi theo ngha Schwarz 77
2.6.4. T tnh kh vi mnh ca f ti a c suy ra c tnh lin tc ca f 0
ti im khng ?
2.6.5. Cho tp m G A. Chng minh rng f kh vi mnh trn G khi vch khi o hm f 0 lin tc trn G.
2.6.6. Chng minh rng nu f kh vi trn R th n kh vi mnh trong mttp thng d, tc l trong tp RnB trong B l mt tp thuc phm trth nht trn R. (xem 1.7.20)
2.6.7. Gi s f lin tc trn [a; b] v tn ti o hm Schwarz fs trong mt
khong m (a; b). Chng minh rng nu f(b) > f(a) th tn ti c 2 (a; b) saocho fs(c) 0.
2.6.8. Gi s f lin tc trn [a; b] v f(a) = f(b) = 0. Chng minh rng nu
f kh vi Schwarz trn mt khong m (a; b) th tn ti x1; x2 2 (a; b) sao chof s(x1) 0 v f s(x2) 0.
2.6.9. Cho f lin tc trn [a; b] v kh vi Schwarz trn khong m (a; b).
Chng minh rng tn ti x1; x2 2 (a; b) sao cho
f s(x2) f(b) f(a)b a f
s(x1):
2.6.10. Gi s rng f lin tc v kh vi Schwarz trn (a; b). Chng minh
rng nu o hm Schwarz fs gii ni trn (a; b) th f tho mn iu kin
Lipschitz trong khong ny.
2.6.11. Gi s f v fs lin tc trn (a; b). Chng minh rng f kh vi v
f 0(x) = fs(x) vi mi x 2 (a; b).
2.6.12. Gi s rng f lin tc v kh vi Schwarz trn mt khong m I.
Chng minh rng nu fs 0 ti im x 2 I th f tng trn I.
2.6.13. Gi s rng f lin tc v kh vi Schwarz trn mt khong m I.
Chng minh rng nu fs(x) = 0 ti x 2 I th f l hm hng trn I.
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78 Chng 2. Vi phn
2.6.14. Cho f kh vi Schwarz trn (a; b) , xt x0 2 (a; b) l cc tr a phngca 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 f s(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 f s 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 phng trn khong m I f
s kh vi Schwarz u trn mi on [a; b] I khi v ch khi f 0 lin tc trnI.
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Chng 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 s n0 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
Af .
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 s fdng , vi
dn = supfjfn(x) f(x)j : x 2 Bg; n 2 N;
hi t v 0.
3.1.2. Gi s fn Af v gn
Ag. Chng minh rng fn + gn
Af + g. Khng
nh fn gn Af g c ng khng?
3.1.3. Gi s fn Af , gn
Ag, v tn ti s M > 0 sao cho jf(x)j < M v
jg(x)j < M vi mi x 2 A. Chng minh rng fn gn Af 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 Chng 3. Dy v chui hm
Chng minh rng dy hm ffng hi 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 trng 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 nhng khng hi t u trn R. Hy tm dy con hi t u.
3.1.7. Chng minh tiu chun Cauchy cho s hi t u.
Dy hm ffng, xc nh trnA, hi t u trnA nu v ch nu vi mi " > 0tn ti s n0 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 bi cc cng
thc sau
fn(x) =1
1 + (nx 1)2 ;(a)
fn(x) =x2
x2 + (nx 1)2 ;(b)fn(x) = x
n(1 x);(c)fn(x) = nx
n(1 x);(d)fn(x) = n
3xn(1 x)4;(e)fn(x) =
nx2
1 + nx;(f)
fn(x) =1
1 + xn:(g)
3.1.9. Xt s hi t u trn A v B ca cc dy hm khi
fn(x) = cosn x(1 cosn x); A = [0; =2];B = [=4; =2];(a)
fn(x) = cosn x sin2n x; A = R;B = [0; =4]:(b)
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3.1. Dy hm v s hi t u 81
3.1.10. Xc nh dy hm ffng hi t u trn A hay khng vi
fn(x) = arctan2x
x2 + n3; A = R;(a)
fn(x) = n ln
1 +
x2
n
; A = R;(b)
fn(x) = n ln1 + nx
nx; A = (0;1);(c)
fn(x) =2np1 + x2n; A = R;(d)
fn(x) =np2n + jxjn; A = R;(e)
fn(x) =pn+ 1 sinn x cos x; A = R;(f)
fn(x) = n(npx 1); A = [1; a]; a > 1:(g)
3.1.11. Vi hm f xc nh trn on [a; b], t fn(x) =[nf(x)]n; x 2 [a; b]; n 2 N.
Chng minh rng fn [a;b]
f .
3.1.12. Kim tra rng dy hm ffng, vi
fn(x) = n sinp42n2 + x2;
hi t u trn on [0; a], a > 0. Dy hm ffng c hi t u trn R khng?
3.1.13. Chng minh rng dy a thc fPng xc nh bi cng thc truy hi
P0(x) = 0; Pn+1(x) = Pn(x) +1
2(x P 2n(x)); n = 0; 1; 2; : : : ;
hi t u trn on [0; 1] n hm f(x) =px.
Suy ra rng c dy a thc hi t u trn on [1; 1] n hm x 7! jxj.
3.1.14. Gi s hm f : R! R kh vi v hm f 0 lin tc u trn R . Kimtra rng
n
f
x+
1
n
f(x)
! f 0(x)
u trn R. Bng v d ch ra rng gi thit lin tc u ca hm f 0 lkhng th b qua c.
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82 Chng 3. Dy v chui hm
3.1.15. Cho ffng l dy hm lin tc u hi t u trn R. Chng minhrng hm gii hn cng l hm lin tc u trn R.
3.1.16. Chng minh nh l Dini: Cho ffng l dy hm lin tc trn tpcompact K hi t im v hm f cng l hm lin tc trn K. Khi nu
fn+1(x) fn(x) vi x 2 K v n 2 N th dy hm ffng hi t v hm f utrn K.
Bng v d hy ch ra rng mi iu kin trong nh l Dini (tnh compact,
tnh lin tc ca hm gii hn, tnh lin tc v n iu ca dy hm ffng)l cn thit.
3.1.17. Dy hm ffng xc nh trn tp A c ni l lin tc ng bc trnA nu vi mi " > 0 tn ti s > 0 sao cho jfn(x) fn(x0)j < " mi khijx x0j < ; x; x0 2 A, v n 2 N. Chng minh rng nu ffng l dy hm hit u ca dy hm lin tc trn tp compact K th ffng l lin tc ngbc trn K.
3.1.18. Chng ta ni rng dy hm ffng xc nh trn A hi t lin tc trnA v hm f nu vi mi x 2 A v vi mi dy fxng nm trong A hi t v xth dy ffn(xn)g hi t v f . Chng minh rng nu dy ffng hi t lin tctrn A v hm f th
limk!1
fnk(xk) = f(x);
vi mi dy fxng nm trong A hi t v x 2 A v vi mi dy con ffnkg.
3.1.19. Chng minh rng nu ffng hi t lin tc trn A v f th f lin tctrn A (ngay c khi fn khng lin tc).
3.1.20. Chng minh rng nu ffng hi t u trn A v hm lin tc f thffng hi t lin tc trn A. iu ngc li c ng khng?
3.1.21. Cho ffng l dy hm xc nh trn tp compact K. Chng minh ccmnh sau l tng ng.
(i) Dy hm ffng hi t u trn K v hm f 2 C(K).
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3.1. Dy hm v s hi t u 83
(ii) Dy hm ffng hi t lin tc trn K v hm f .
3.1.22. Gi s ffng l dy hm tng hoc gim trn on [a; b] hi t imv mt hm lin tc trn [a; b]. Chng minh rng ffng hi t u trn [a; b].
3.1.23. Cho ffng l dy hm tng hoc gim trn R v b chn u trn R.Chng minh ffng cha mt dy con hi t im trn R.
3.1.24. Di nhng gi thit ca bi ton trn (3.1.23) hy ch ra rng: Nu
hm gii hn f ca mt dy hm con ffnkg hi tu im l lin tc th ffnghi t v f u trn mi tp con compact ca R. Dy hm ffng phi hi tu trn R khng?
3.1.25. Chng minh rng hm gii hn ca dy a thc hi t u trn Rl mt a thc.
3.1.26. Gi s rng fPng l mt dy a thc c dng
Pn(x) = an;pxp + an;p1xp1 + + an;1x+ an;0:
Chng minh ba mnh sau l tng ng:
(i) fPng hi t u trn mi tp con compact ca R,
(ii) C p+1 s phn bit c0; c1; : : : ; cp sao cho fPng hi t trn fc0; c1; : : : ; cpg;
(iii) Dy cc h s fan;ig hi t vi i = 0; 1; : : : ; p.
3.1.27. Chng minh rng nu ffng hi t im v lin tc ng bc trn tpcompact K th ffng hi t u trn K.
3.1.28. Cho ffng l dy hm lin tc trn on [a; b] v kh vi trn khong(a; b). Gi s ff 0ng b chn u trn (a; b), tc l c sM > 0 sao cho jf 0n(x)j M vi mi n 2 N v x 2 (a; b). Chng minh rng nu ffng hi t im trn[a; b] th ffng hi t u trn on .
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84 Chng 3. Dy v chui hm
3.1.29. Nghin cu s hi t v s hi t u ca ffng v ff 0ng trn A, vi
fn(x) =sinnxpn; A = R;(a)
fn(x) =x
1 + n2x2; A = [1; 1]:(b)
3.1.30. Gi s ffng hi t u trn A v hm f . Hn na gi s rng x0 lim t ca A v bt u t ch s n no , lim
x!x0fn(x) tn ti. Chng minh
limn!1
limx!x0
fn(x) = limx!x0
f(x):
V n