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NXB i hc quc gia H Ni 2007.

Tkho:Giitchtonhc,giitch,tphp,sthc,nhx,Hmlintc,im gin on, lin tc, lin tc u, hm s cp, Gii hn, gii hn dy s, gii hn hm s, dy s, hm s, nguyn l Cantor, nguyn l Cauchy, gii hn trn, gii hn di, v cng b, v cng ln,hm s hp, hm s ngc, Php tch vi phn, o hm, vi phn,CngthcTaylor,KhaitrinMaclaurin,QuytcLhospital,tchphnkhng xcnh,tchphn,nguynhm,PhpthEuler,iukinkhtch,Hmkhtch, Dintch,thtch,Tchphnsuyrng,NguynlCanto,Tpcompact,Hmnhiu bin,Lintc,giihn,lintcu,ohm,cctrhmnhiubin,Phptchvi phn, S hi t. Ti liu trong Th vin in t H Khoa hc T nhinc thc s dng cho mc ch hc tp v nghin cu c nhn. Nghim cm mi hnh thc sao chp, in n phc v cc mc ch khc nu khng c s chp thun ca nh xut bn v tc gi. Mc lc Chng 1 Tp hp v s thc ............................................................................................... 7 1.1Khi nim v tp hp ................................................................................................. 7 1.2S thc........................................................................................................................ 9 1.3nh x ...................................................................................................................... 14 1.4Bi tp chng 1 ...................................................................................................... 16 Gii tch ton hc Tp 1 L Vn Trc 2Chng 2 Gii hn ca dy s v hm s.......................................................................... 19 2.1Gii hn ca dy s.................................................................................................. 19 2.1.1nh ngha dy s................................................................................................. 19 2.1.2Cc tnh cht ca dy hi t ................................................................................. 21 2.1.3Gii hn v hn.................................................................................................... 24 2.2Tiu chun hi t...................................................................................................... 25 2.2.1Cc nh l ........................................................................................................... 25 2.2.2S e....................................................................................................................... 26 2.2.3Nguyn l Cantor v dy cc on thng lng nhau v tht li ........................... 27 2.2.4S hi t ca dy b chn ..................................................................................... 28 2.2.5Nguyn l Cauchy v s hi t ca mt dy s................................................... 29 2.2.6Gii hn trn v gii hn di ............................................................................. 30 2.3Khi nim v hm s mt bin s ............................................................................ 32 2.3.1nh ngha............................................................................................................ 32 2.3.2 th ca hm s................................................................................................. 32 2.3.3Hm s hp .......................................................................................................... 34 2.3.4Hm s ngc ...................................................................................................... 34 2.3.5Cc hm lng gic ngc................................................................................... 36 2.3.6Cc hm s hypebol ............................................................................................. 38 2.3.7Cc hm hypebol ngc....................................................................................... 39 2.4Gii hn ca hm s................................................................................................. 41 2.4.1Ln cn ca mt im.......................................................................................... 41 2.4.2Cc nh ngha gii hn........................................................................................ 42 2.4.3Gii hn mt pha................................................................................................. 45 2.4.4Gii hn v cng.................................................................................................. 46 2.4.5Cc tnh cht ca gii hn.................................................................................... 47 2.4.6Tiu chun tn ti gii hn ca hm s................................................................ 47 2.4.7V cng b. V cng ln...................................................................................... 48 2.4.8Cc gii hn ng nh.......................................................................................... 51 2.5Bi tp chng 2 ...................................................................................................... 54 Chng 3Hm lin tc mt bin s .................................................................................... 61 3.1nh ngha s lin tc ca hm s ti mt im.......................................................... 61 3.1.1Cc nh ngha...................................................................................................... 61 3.1.2Hm lin tc mt pha, lin tc trn mt khong, mt on kn.......................... 62 3.1.3Cc nh l v nhng php tnh trn cc hm lin tc ......................................... 63 3.1.4im gin on ca hm s ................................................................................. 65 3.2Cc tnh cht ca hm lin tc ................................................................................. 68 3.2.1Tnh cht bo ton du ln cn mt im......................................................... 68 3.2.2Tnh cht ca mt hm s lin tc trn mt on ................................................ 68 3.3iu kin lin tc ca hm n iu v ca hm s ngc ..................................... 72 3.3.1iu kin lin tc ca hm n iu.................................................................... 72 3.3.2Tnh lin tc ca hm ngc................................................................................ 73 3.4Khi nim lin tc u ............................................................................................. 74 3.4.1M u ................................................................................................................. 74 3.4.2nh ngha............................................................................................................ 74 3.4.3Lin tc ca cc hm s s cp ............................................................................ 76 3.5Bi tp chng 3 ...................................................................................................... 77 3Chng 4 Php tnh vi phn ca hm mt bin................................................................ 81 4.1 o hm v cch tnh..................................................................................................... 81 4.1.1 nh ngha o hm................................................................................................ 81 4.1.2 Cng thc i vi s gia ca hm s ...................................................................... 81 4.2 Cc qui tc tnh o hm................................................................................................ 82 4.2.1 Cc qui tc tnh o hm......................................................................................... 82 4.2.2 o hm ca hm s hp ........................................................................................ 82 4.2.3 o hm ca hm s ngc .................................................................................... 84 4.2.4 o hm theo tham s............................................................................................. 85 4.2.5 o hm mt pha ................................................................................................... 85 4.2.6 o hm v cng..................................................................................................... 87 4.2.7 o hm cc hm s s cp .................................................................................... 87 4.3 Vi phn ca hm s........................................................................................................ 88 4.3.1 nh ngha............................................................................................................... 88 4.3.2 Cc qui tc tnh vi phn........................................................................................... 89 4.3.3 Vi phn ca hm s hp.......................................................................................... 89 4.3.4 ngdng ca vi phn ............................................................................................ 90 4.4 Cc nh l c bn ca hm kh vi ................................................................................. 90 4.4.1 Cc tr a phng................................................................................................... 90 4.5 o hm v vi phn cp cao........................................................................................... 96 4.5.1 nh ngha o hm cp cao ................................................................................... 96 4.5.2 Cc cng thc tng qut i vi o hm cp n ..................................................... 97 4.5.3 Vi phn cp cao....................................................................................................... 97 4.6 Cng thc Taylor ........................................................................................................... 98 4.6.1 Cng thc Taylor .................................................................................................... 99 4.6.2 Khai trin Maclaurin ............................................................................................. 101 4.7 Qui tc Lhospital kh dng v nh....................................................................... 103 4.7.1 Dng v nh......................................................................................................... 103 4.7.2 Dng v dnh......................................................................................................... 105 4.8 Kho st hm s ........................................................................................................... 108 4.8.1 Kho st ng cong cho di dng phng trnh hin....................................... 108 4.8.2 ng cong cho di dng tham s ..................................................................... 110 4.8.3 Kho st ng cong trong ta cc.................................................................. 114 4.9 Bi tp chng 4 .......................................................................................................... 117 Chng 5 Tch phn khng xc nh ............................................................................... 123 5.1Tch phn khng xc nh...................................................................................... 123 5.1.1nh ngha nguyn hm..................................................................................... 123 5.1.2Cc tnh cht....................................................................................................... 123 5.1.3nh ngha tch phn khng xc nh................................................................. 123 5.1.4Cc tnh cht ca tch phn khng xc nh....................................................... 123 5.1.5Bng cc tch phn c bn.................................................................................. 124 5.2Cch tnh tch phn khng xc nh....................................................................... 125 5.2.1Da vo bng cc tch phn c bn.................................................................... 125 5.2.2Tnh tch phn nh php i bin....................................................................... 126 5.2.3Phng php tnh tch phn tng phn............................................................... 127 5.2.4Cng thc truy hi.............................................................................................. 129 5.3Tch phn cc phn thc hu t .............................................................................. 130 45.3.1Tch phn cc phn thc hu t n gin nht ................................................... 130 5.3.2Tch phn ca cc phn thc hu t.................................................................... 132 5.4Tch phn cc biu thc cha hm lng gic v cc hm hypebol ...................... 134 5.4.1Tch phn cc biu thc cha hm lng gic................................................... 134 5.4.2Tch phn cc biu thc cha hm hypebol ....................................................... 136 5.5Tch phn mt vi lp hm v t ............................................................................ 137 5.5.1Tch phn dng ( , )m ax bI R x dxcx d+=+............................................................................ 137 5.5.2Tch phn dng ,( ) ,( ) ,p rq sax b ax bI Rx dxcx d cx d + + =+ + .................................................................... 138 5.5.3Tch phn cc nh thc vi phn .......................................................................... 138 5.6Tch phn cc biu thc dng 2R( , ) x ax bx c + + vi 0 a ...................................... 139 5.6.1Php th Eulerth nht ..................................................................................... 140 5.6.2Php th Euler th hai ........................................................................................ 140 5.6.3Php th Euler th ba ......................................................................................... 141 5.6.4Tch phn eliptic................................................................................................. 142 5.7 Bi tp chng 5 .................................................................................................... 143 Chng 6 Tch phn xc nh........................................................................................... 145 6.1nh ngha tch phn xc nh................................................................................ 145 6.1.1Bi ton din tch hnh thang cong..................................................................... 145 6.1.2Bi ton tnh khi lng..................................................................................... 146 6.1.3nh ngha tch phn xc nh............................................................................ 146 6.1.4 ngha hnh hc ca tch phn xc nh ........................................................... 148 6.2iu kin kh tch .................................................................................................. 148 6.2.1iu kin cn hm kh tch ........................................................................... 148 6.2.2Cc tng Darboux............................................................................................... 149 6.2.3Cc tnh cht ca tng tch phn Darboux ......................................................... 150 6.2.4Du hiu tn ti ca tch phn xc nh ............................................................. 151 6.3Cc lp hm kh tch.............................................................................................. 152 6.4Cc tnh cht c bn ca tch phn......................................................................... 154 6.4.1Cc tnh cht ca tch phn xc nh.................................................................. 154 6.4.2Cc nh l gi tr trung bnh .............................................................................. 158 6.5Nguyn hm v tch phn xc nh........................................................................ 159 6.5.1Cc nh ngha.................................................................................................... 160 6.5.2Tch phn xc nh nh hm ca cn trn.......................................................... 160 6.6Tnh tch phn xc nh.......................................................................................... 162 6.6.1Php i bin trong tch phn xc nh.............................................................. 162 6.6.2Php ly tch phn tng phn ............................................................................. 164 6.6.3Tnh gn ng tch phn xc nh...................................................................... 168 6.7Mt s ng dng hnh hc, vt l ca tch phn xc nh...................................... 172 6.7.1Tnh din tch hnh phng................................................................................... 172 6.7.2Tnh di ng cong phng........................................................................... 177 6.7.3Tnh th tch vt th............................................................................................ 180 6.7.4Din tch mt trn xoay...................................................................................... 183 6.8Tch phn suy rng................................................................................................. 186 6.8.1Tch phn suy rng loi 1................................................................................... 186 6.8.2Tch phn suy rng loi 2................................................................................... 195 6.8.3Thay bin s trong tch phn suy rng ............................................................... 199 56.9Bi tp chng 6 ..................................................................................................... 200 Chng 7 Hm s lin tc trong nR............................................................................... 206 7.1Tp hp trong nR.................................................................................................. 206 7.1.1Khong cch trong nR...................................................................................... 206 7.1.2Ln cn ca mt im........................................................................................ 207 7.1.3im trong, im bin, im t ca tp hp...................................................... 208 7.1.4Tp m, tp ng................................................................................................ 210 7.1.5Tp lin thng..................................................................................................... 210 7.2S hi t trong nR, cc khi nim c bn ca hm s nhiu bin s................... 211 7.2.1S hi t trong nR............................................................................................. 211 7.2.2Dy c bn.......................................................................................................... 212 7.2.3Nguyn l Canto ................................................................................................. 213 7.2.4Ch .................................................................................................................. 213 7.2.5Tp hp compact ................................................................................................ 214 7.2.6nh ngha hm nhiu bin s............................................................................ 214 7.2.7Tp xc nh ca hm nhiu bin s .................................................................. 214 7.2.8ng mc v mt mc..................................................................................... 215 7.3Gii hn ca hm s trong nR............................................................................... 216 7.3.1Gii hn ca hm s ti mt im...................................................................... 216 7.3.2Gii hn lp........................................................................................................ 217 7.3.3Quan h gia gii hn theo tp hp cc bin v cc gii hn lp...................... 218 7.3.1Ch .................................................................................................................. 219 7.4Hm s nhiu bin s lin tc ................................................................................ 221 7.4.1Hm s lin tc ti mt im............................................................................. 221 7.4.2Hm s lin tc u............................................................................................ 222 7.4.3Lin tc theo tng bin....................................................................................... 223 7.5Php tnh vi phn ca hm s nhiu bin s .......................................................... 224 7.5.1o hm ring v vi phn cp mt..................................................................... 224 7.5.2o hm v vi phn cp cao............................................................................... 230 7.6o hm ca hm s n.......................................................................................... 233 7.6.1Khi nim v hm s n mt bin s ................................................................. 233 7.6.2Khi nim hm s n ca hai bin s................................................................. 235 7.7o hm theo hng .............................................................................................. 237 7.7.1o hm theo hng .......................................................................................... 237 7.7.2Gradien............................................................................................................... 238 7.8Cng thc Taylor. Cc tr ca hm s nhiu bin s............................................. 239 7.8.1Cng thc Taylor ............................................................................................... 239 7.8.2Cc tr ca hm nhiu bin s............................................................................ 241 7.8.3Gi ln nht v nh nht ca hm s nhiu bin s trn compac ...................... 244 7.9Cc tr c iu kin ................................................................................................ 245 7.9.1nh ngha: ......................................................................................................... 245 7.9.2Phng php tm cc tr ..................................................................................... 245 7.10ng dng ca php tnh vi phn trong hnh hc.................................................... 250 7.10.1Tip tuyn ca ng cong............................................................................ 250 7.10.2Mt phng tip xc ca mt cong................................................................... 251 67.10.3 cong .......................................................................................................... 253 7.10.4Bao hnh ca mt h ngcong ................................................................... 255 7.11Bi tp chng 7 .................................................................................................... 258 7.12Hng dn gii bi tp v p s........................................................................... 262 Ti liu tham kho............................................................................................................ 302

7 Chng 1 Tp hp v s thc 1.1Khi nim v tp hp 1.1.1Tp hp Cho tp hp M, ch x l phn t ca tp M ta vitx M (c l x thuc M), ch x khng phi l phn t ca tp M ta vit x M (c l x khng thuc M). Tp hp M ch c mt phn t a, k hiu l { } a . Tp hp M khng c phn t no gi l tp rng, k hiu l . Cho hai tp A v B. Nu mi phn t ca A u l phn t ca B ta ni rngA l tp con ca B v ta vit A B . Nu A l tp con ca B v A Bta ni rng A l tp hp con thc s ca tp hp B v vitl A B .TrongtrnghpnytntitnhtmtphnttrongBmkhngphil phn t ca A. V d nh tp hp cc s nguynZ l tp con ca tp hp cc s hu t. Cho A, B, C l ba tp hp. Khi c tnh cht sau: a) A (1.1.1)

)) bcv = (1.1.2)v (1.1.3)A B B A A B A BB C A C. 1.1.2 Mt s tp hp thng gp Trong cc gio trnh i s trng ph thng trung hc ta lm quen vi tp hp cc s t nhin` `={ 0,1,2,, n,}(1.1.4) `*={1,2, n,}.(1.1.5) xt nghim ca phng trnh x+n = 0 trong ` nta a thm tp cc s nguyn ]: { } 0, 1, 2,..., ,... = ] n .(1.1.6) xt nghim ca phng trnh mx + n = 0 trong , m n ]ta a thm tp cc s hu t 8| , 0, = = mx x n m,nn_ ] .(1.1.7) Tabitbnphptoncs(cng,tr,nhn,chia)cashutvcchspxp chngtheoln(nua,blhaishut,thmttrongchngbhnsthhai).Tng a+b, hiu a - b, tch a.b, thng( 0)abbca hai s hu t a,b li l s hu t, nhng vi cc php ton khc nu ch xt trn tp cc s hu t, ta thy nhng iu nu trn khng cn ng na. V d php ly cn l php ton nh vy. Ta hy tm cn bc hai ca s 2, tc l tm mt s x m bnh phng ca n bng 2. Ta khng nh rng khng c s hu t no m bnh phng ca n bng 2. Gi srng s hu t x nh vy tn ti, ta c th vit di dng phn s ti gin pq, trong p v q ch c c s chung l1 . Khi 22 222; 2 = =pp qq cho nnp2lschnvdopcnglschn,p=2m,trongmlsnguyn,do 4m2=2q2,2m2=q2chonnq2lschnvvthqlschn.Nhvyp,qlccschn, iu ny mu thun vi gi thit l p,q ch c c chung l1 . Mu thun nhn c chng minh khng nh trn. T nguyn nhn ny, trong ton hc ta a thm vo nhng s mi, l cc s v t. V d v s v t l2, 3, lg3, , sin20o Tp cc s hu t v cc s v t c gi l tp cc s thc v k hiu l\. Nh vy ta c bao hm thc: . N Z R (1.1.8) 1.1.3 Cc php ton trn tp hp a)HpA B ca tp hp A v tp hp B, c l A hp B l tp hp c nh ngha bi: { | } = A B x x A B hoc x . (1.1.9) b)Giao A B ca hai tp hp A v B, c l A giao B l tp hp nh ngha bi:{ |} A B x x A = v x B . (1.1.10) c)Hiu= | { |v} A B x x A x B .(1.1.11) Ta ni rng cc tp A v B l ri nhau nuA B = . d)B sung CAB ca B trong A ( B A) l tp hp nh ngha bi= { |v}ACB x x A x B(1.1.12) Php giao, hp v b sung c cc tnh cht sau: i)( ) ( ) = A B C A B C(1.1.13) ii)( ) ( ) = A B C A B C(1.1.14) iii)( ) ( ) ( ) = A B C A C B C(1.1.15) iv)( ) ( ) ( ) = A B C A C B C(1.1.16) 9v)\ , \ A= = A A (1.1.17) vi) 1 2 1 2( ) = A A AC B B CB CB (1.1.18) vii) 1 2 1 2( ) = A A AC B B CB CB . (1.1.19) 1.1.4 Tch cc Cho hai tp hp A,B khng rng. Tch cc ca hai tp hp A v B, k hiu l AB l tp hp cc cp (x,y) trong , x A y B , ng thi (x,y)= (a,b) khi v ch khi x = a, y = b. Nh vyAB ={(x,y)|, x A y B }(1.1.20) Thay cho AA ta vit l A2 V d: {1,2}{2,3,4} = {(1,2); (1,3); (1,4); (2,2); (2,3); (2,4)} Ngoi ra {1,2}2 ={(1,1); (1,2); (2,1); (2,2)}. 1.1.5 Cc k hiu lgic By gi gi s M l mt tp hp v t l mt tnh cht no ca cc phn t ca tp M. Nu phn tx M c tnh cht t ta vit t(x). Gi c(t) l tp hp ca tt c cc phn t ca tp M c tnh cht t: c(t) ={x M |x c tnh cht t} (1.1.21) hay c(t) ={x M |t(x)} (1.1.22) khi nu c(t) = M th mi phn t ca M u c tnh cht t, ta ni rng vi mix M , x c tnh cht t v ta vitx M : t(x) hay( )x Mt x . K hiu gi l k hiu ph bin. Nu( ) c t , th c t nht mt phn tx M , x c tnh cht t v vit : ( ) hay( )x Mx M t x t x K hiu gi l k hiu tn ti. 1.2S thc 1.2.1Php cng v nhn cc s thc Xt tp hp cc s thcR. Ta c th xc nh php cng v nhn hai s thc bt k a v b.Phptoncngchotngnghaisthcavbvisthcckhiula+b,php 10nhn cho tng ng hai s thc a v b vi s thc c k hiu l a.b sao cho tho mn cc tnh cht sau: Vi mi s thc a,b v c. a)a+b = b+a (tnh cht giao hon), b)a+(b+c) = (a+b)+c (tnh cht kt hp), c)a.b = b.a (tnh cht giao hon ), d)a(b.c) = (a.b).c (tnh cht kt hp), e)(a+b).c = a.c+b.c (tnh cht phn phi), f)Tn ti duy nht s 0 sao cho a+0 = a R a , g)Vi mi a, tn ti s a sao cho a + ( a) = 0, h)Tn ti duy nht s 1 0 sao cho a.1 = a R a , i)Vi mi s a 0, tn ti s a-1 sao cho a.a-1= 1, s a-1 cn c k hiu l 1a. Ch : S ( a) v s a-1 ni trong tnh cht g) v i) l duy nht. Tht vy, v d nh nu tn ti s b a tho mn iu kin a+b =0, th a+b+ ( a)= a, t y a+ ( a)+b= a hay 0+b = a v b= a, mu thun. 1.2.2So snh hai s thc a v b Cho hai s thc bt k a v b. Khi ch c th xy ra mt trong ba trng hp sau: a = b (a bng b), a > b (a ln hn b) hay b > a (b ln hn a). Mnh = c tnh cht: nu a=b v b=c th a=c. Mnh > c tnh cht sau: Vi mi s thc a,b v c. a)Nu a > b v b > c th a > c b) Nu a > b tha+c > b+c. c)Nu a > 0, b > 0 th ab > 0. Mnh ab ngha l hoc a=b, hoc a>b. Cc mnh a < b, a b, a > b, a b c gi l cc bt ng thc. Cc bt ng thc a < b, a > b c gi l cc bt ng thc thc s. S thc a tho mn bt ng thc a>0 c gi l s dng.S thc a tho mn bt ng thc a 0,sao cho> sup x Mx M - . Tht vy, nu s x nh vy khng tn ti th ssupM cng l cn trn v khi s supM khng phi l cn trn ng ca tp M. Ni mt cch khc, tnh cht ny ni ln supM l s nh nht trong s cc cn trn ca M. V d 1: Tm cn trn ng ca tp 1 1 1{1, , ,..., ,...}.2 3Mn=Gii:Tathy< N*10 1 nn,vthtphpMbchntrn,dthys1lcn trn. Ta hy chng minh s 1 l cn trn ng ca M. Tht vy0 > , ta phi tm c s t nhin n sao cho 11n > . S n ny, v d l n = 1. V d 2: sup(0,1) = sup[0,1] = 1. By gi ta c th nh ngha cn trn ng ca tp M mt cch khc nh sau: S supM c gi l cn trn ng ca tp M b chn trn nu a) supx M x M (1.2.4) b)> 0, sao cho sup x M x > M -(1.2.5) 12Tp hp s M c gi l b chn di, nu tn ti s g sao cho x g x M .(1.2.6) Mi s g c tnh cht ny gi l cn di ca tp hp M. Do tp M b chn di, nu n c t nht mt cn di. S ln nht trong cc cn di ca tp M gi l cn di ng ca M v c k hiu l inf M. V d 3: Xt tp M=(a,b) HinnhinsavsbtkbhnalcndicaM.Hinnhinsalcndi ng ca tp M, tc l a= inf M. Tng t nh i vi cn trn ng, cn di ng c tnh cht sau: , x Msao cho x < inf M + . (1.2.7) V d 4: Xt tp 1 1 1{1, , ,..., ,...}2 3Mn=Ta chng minh rng s 0 l cn di ng ca tp M.Tht vy,0 > , ta phi tm c s t nhin n sao cho< + 10 ,n hay< > 1 1nn. iu ny ngha l s 0 l cn di ng ca tp M, tc l inf M = 0. V d 5: inf(0,1) = inf[0,1] = 0. Trong cc v d trn, ta thy sup M, inf M c th thuc M, cng c th khng thuc M. nh l 1.2.2 Tp hp s khng rng bt k b chn trn (di) c cn trn (di) ng. Chng minh: Gi s X l tp hp s khng rng b chn trn. Khi tp hp Y cc s l cn trn ca tp X khng rng. Theo nh ngha ca cn trn suy ra rng i vi bt k x Xv bt k y Yta c bt ng thc. x y . Da vo tnh cht lin tc ca tp hp cc s thc, tn ti mt s c sao cho x c y x X, y Y .(1.2.8) T bt ng thc th nht trong (1.2.8) suy ra s c chn trn tp hp X, t bt ng thc th hai trong (1.2.8) suy ra c l s b nht trong cc cn trn ca X, tc l cn trn ng ca tp X. V d 6: Chng minh rng tp hp cc s nguyn X= {,3, 2, 1,0,1,2,3,} khng b chn trn, cng khng b chn di, tc l supX= + v inf X=. Tht vy, gi s ngc li, tp hp X b chn trn. Khi theo nh l trn, n c cn trn ng 13c = sup X. Theo tnh cht ca cn trn ng, i vi1 = , ta tm c mt s nguynx X sao cho x > c 1 nhng khi x+1> c. Bi v1 x X + , iu ny c ngha l c khng phi l cn trn ng ca tp hp X, mu thun vi iu ni trn. V d 7: Gi s X v Y l hai tp hp s. Hy chng minh rng nuY X th supX supY. Gii: Gi s supX = A, supY = B. Ta phi chng minh BA. Gi s ngc li B > A. Khi da vo tnh cht cn trn ng, > 0, sao cho >y Y y B . Bi v: B A >0, nn ta c th lyB A = . Ta nhn c y >B = B B +A, tc l y > A. Nhngy Y vY X nny X , theo nh ngha sup suy ray A . Mu thun nhn c chng t rngB A .Ta c th chng minh khng nh trn bng cch khc nh sau: Bi vY X nnx X vy Y ta c sup ,sup x X y Xv sup y Y . Nhng supY l s thc nh nht trong cc cn trn ca Y v supX l mt trong s cn trn ca Y nn sup Ysup X. Nu tp M ng thi b chn di v b chn trn, ta gi l tp b chn. Cui cng nu tp M khng b chn trn, th ta ni rng cn trn ng ca tp l +, sup M = +. Tng t nu tp M khng b chn di, ta ni rng cn di ng ca tp l , inf M=. V d nh sup(0,+) = +, inf(,0)= . Gi s M l tp hp cc s thc, nu tn ti mt phn t ln nht trong cc phn t ca tp M, th ta k hiu phn t l maxM. Tng t ta k hiu phn t nh nht ca tp M l minM. V d nh max{2, 3, 5,0} = 2, min{2,3, 5,0}=5,|x|=max {(x,x)}x . 1.2.5Trc s thc By gi ta tm cch biu din hnh hc tp cc s thc. Ta ly mt ng thng nm ngang v trn ta ly mt im 0 no lm gc. Ta chon mt di thch hp lm n v v t di lin tip nhau t im 0 sang tri v sang phi sao cho tri khp ng thng. V d nh s 2 c biu din bng im 2, tc l im bn phi im 0 vi khong cch 2 n v. Tagingthngnitrnlngthngthchaytrcs.Btkmtsthcno cngcngvimtimtrnngthngthcvngcli,btkmtimnotrn 14ng thng thc cng c ng vi mt s thc. S thc a ng vi im M trn trc s c gi l to ca im M. Thng thng ngi ta khng phn bit im a nm trn ng thng thc v s thc a (l to ca im ). Tp hpR khng c phn t cc i v phn t cc tiu, bi v i vi mt s thc x bt k lun lun tn ti hai s y v z sao cho y< x< z (v d y = x 1, z = x+1). V th ta hy b sung vo tpR hai phn t mi m ta k hiu l +, v ta gi chung l cc im v tn ca trc thc. Ta k hiu tp mi xut hin nh vt lR*. Nh vy l R*=R {, +}.(1.4.2) Tp hp R* ta s gi l trc thc m rng. Cui cng ta ch thm l < a < +, R a .(1.4.3) 1.3nh x Trong phn ny chng ta s trnh by mt vi khi nim v nh x m n rt c ch cho vic nghin cu l thuyt hm s sau ny. 1.3.1nh ngha Cho hai tp hp A v B. nh x t tp hp A ti tp hp B l mt quy lut f cho tng ng mi phn tx A vi mt v ch mt phn ty B . V d 1: Cho A = B =R. Qui lut y = x3 cho tng ng miR xvi mt v ch mtR y , nn qui lut trn l mt nh x tR tiR. V d 2: Cho A = B = {x | R x ,0 x }. Qui luty x =cho tng ng mix A vi mt v ch mt y B, nn l mt nh x t A ti B din t f l nh x t tp hp A ti tp hp B ta vitf: AB hayfA B v gi A l tp xc nh ca nh x f. Phnty B tngngvix A biquilutfgilnhcaxvxcgil nghch nh ca y v ta vit: ( )hay( ) y f x x y f x = = . Ta gi tp( ) { | ( ), } = = f A y y f x x A (1.3.1) hay( ) { | , ( )} = = f A y x Ay f x (1.3.2) l nh x ca tp A qua nh x f. Ch rng ta lun c( ) f A B. Nu f(A)=B, ta ni rng f l nh x t tp hp A ln tp hp Bhay nh x: f A B l mt ton nh. 15V d 3: nh x cho bi qui lut= R ( ) si n , f x x xl nh x tpR ti tpR v ng thi nh x tpR ln tp hp tt c cc s thc y sao cho1 1 y . Nu nh N M Ath( ) ( ) N f M f . (1.3.3) 1.3.2n nh, song nh nh x: f A B gi l nh x n nh nu1 2 1 2( ) ( ) f x f x x x = =(1.3.4) v d nh nh x c cho bi sinx l n nh t tp hp{ | 0 }2 < < R x x ln tp hp{ | 0 1} < < R y y . V d 4: Xt nh x cho bi qui lut 2y x = . V phng trnh= R2, y x yc hai nghim khc nhau x1 v x2 nu y > 0, c ngha l f(x1) = f(x2) nhng 1 2x x , vy nh x ny khng phi l n nh. nh x: f A B gi l mt song nh nu n va l n nh va l ton nh. V d 5: nh x: R R fcho bi qui lut 3y x =l mt song nh V d 6: nh x:+ R R fcho bi qui lut 2y x =khng phi l song nh, nhng nh x :+ + R R fcho bi qui lut 2y x =l mt song nh. V d 7: ChoR, x[x] k hiu phn nguyn ca x (ngha l [x] l s nguyn ln nht khng ln hn x chng hn [4,5] = 4; [2] = 2; [2,5] = 2; [2,7] = 2). Ta c [x] x[x]+1. nh x R Z : fcho bi qui lut y=[x] khng phi l song nh. 1.3.3nh x ngc Gi s f l mt nh x tp hp A ln tp hp B. Khi ng vi mi phn t c mt v ch mtx A sao cho y f x ( ) = . nh x cho tng ng phn ty B vi phn tx A sao choy f x ( ) =gi l nh x ngc ca nh x f k hiu lf1 . Nh vyf B A1:f y x f x y1( ) ( )= =(vi x A , y B ).(1.3.5) 16 Hnh 1.3.1 Vd8:NuAltphpccvngtrnngtmnmtrncngmtphngvf(x)lbn knh ca vng trn x, khi f l nh x n tr tp A ln tp cc s thc dng. Khi nh x ngcf1 tng ng mt s thc dng x vi vng trn nm trong tp A m bn knh ca n l x.1.3.4Hp (tch) ca hai nh x Cho hai nh x: : v : g M A f A B . Xt nh x t tp M ti tp hp B c xc nh nh sau: ( ( )) x M z f g x B = .(1.3.6) nh x ny gi l hp ca nh x g v nh x f (hay tch ca g v f ), k hiu lf g Nh vy: f g M B ( ) ( ( )), f g x f g x x M = .(1.3.7) Vd9:nhxchobiquilutsinx2,R x lhpcanhxtrongchobiquilutx2, R xv nh x ngoi c cho bi qui lut siny,R ynh x sin2x,R xl hp ca nh x trong cho bi sinx,R xv nh x ngoi cho bi y2,R y . 1.4Bi tp chng 1 1.1 Cho a l sv t, r l s hu t 1)Hy chng minh rng a+r v a r l cc s v t 2)Gi s0 rhy chng minh rng cc s, ,a rarr a l cc s v t. 1.2 Cho a,bR, gi s 17( , ) | | = d a b a bl khong cch gia hai im a v b ca trc s Hy chng minh rng 1)d(a,a) = 0 2)d(a,b)>0 khia b 3)d(a,b) =d(b,a) 4)d(a,b) + d(b,c) ( , ) d a c . 1.3 Hy chng minh mnh tp hp R Ml b chn khi v ch khi tn ti s thc r>0 sao cho: |x| rx M . 1.4 Cho R X . nh ngha: (X) = {x|xX} Hy chng minh: 1)inf(X) = sup X 2)sup(X)= inf X 1.5 Cho R , XY . nh ngha X+Y = { = + R| , , a x X y Ya x y} { R| , , } XY a x X y Ya xy = =Ngha l X+Y l tp hp cc s thc c dng x+y vi, x Xy Y , cn XY l tp hp cc s thc c dng xy vi, x Xy Y . 1)Gi s X,Y b chn trn, chng minh: sup(X+Y) = supX+ supY 2)Gi s X, Y b chn di, chng minh: inf(X+Y) = inf X + inf Y 3)Gi s X, Y b chn trn, + + R R , X Y .Chng minh:sup(XY)= (sup X)(sup Y) 4)Gi s X, Y b chn di,,+ + R R X Y . Chng minh: inf(XY) = (inf X)(inf Y). 1.6 Gi s N R *M . Chng minh rng: i nf i nf sup sup N N M M . 1.7 Cho R Av{ | : } F f f A A = . Chng minh rng nu f,g,hF v i l nh x ng nht trn tp A, tc l i(x) =x,xA th: 1)( ) ( ) f g h f g h = , 2)f i f =. 1.8 Cho F l tp hp ni trn v 18*{ | : F f f A A = v f l n nh} Chng minh rng nu f,gF* th1)f g F* 2) 1f f i= 19 Chng 2 Gii hn ca dy s v hm s 2.1Gii hn ca dy s 2.1.1nh ngha dy s Cho *N ={1,2,3,}ltphpccstnhin.Mtnhxf: *N Rcgil mt dy s thc. Nu t xn= f(n) th ta c th biu din dy s di dng: 1 2 3, , ,..., ,...nx x x x (2.1.1) Phn t xn c gi l s hng th n ca dy s. chogntaskhiudysbng{xn}.Chsntrongshngxnchvtrcas hng ny trong dy (2.1.1). Trc ht ta hy nu ra mt vi v d v dy: 1 2 3 41 1 1 1 1: 1, , , ,..., ,...2 3 4 = = = = = nx x x x xn n(2.1.3) 1 21 2: , ,..., ,...1 2 3 1 = = = + + nn nx x xn n (2.1.4) 1 2 3 2 1 21 1 1 1 1 1, : , 1, ,..., , ,...2 2 1 2 4 2 2 1 = = = = = n nx x x x xn n n n (2.1.5) Tathyrngccshngcady(2.1.3)vdy(2.1.5)gn0tukhintng,ccs hng ca dy (2.1.4) gn 1 tu khi n tng. Ta ni rng dy (2.1.3) v dy (2.1.5) c gii hn 0, cn dy (2.1.4) c gii hn 1. By gi ta a ra nh ngha chnh xc v gii hn ca dy. nh ngha 1: Ta ni rng s a l gii hn ca dy {xn} nu i vi mi s dngb tu u tm c mt sN*psao cho> N*, n p nta u c: |xn a| > n n . Nu ta ly 11 1 = + p (phn nguyn ca (11 )) thn p>ta c: |xn 1|p ta c |xn a|p, cho nn|xn+1 a|p |xn xn+1|= |(xn a)+ ( a xn+1)| | xn a|+| xn+1 a|n< 1+1 = 2 iu ny mu thun vi tnh cht ca cc s hng ca dy (2.1.8) l: | xn xn+1|= 2 N*n . 212.1.2 Cc tnh cht ca dy hi t a) Tnh duy nht nh l 2.1.1Mi dy hi t u c gii hn duy nht. Chng minh:Gi s dy: (2.1.1) c hai gii hn khc nhau a v b vi a< b Ta ly s = >1( ) 02b a . Bi v, a l gii hn ca dy (2.1.1), ta tm c s p1 sao cho vi n> p1 ta c: |xn a|p2 ta c: |xn b| max(p1, p2) th b < xn < a+ b < a+ ba < 2 , iu ny mu thun vi gi thit b a = 2 . nh l 2.1.2 Mi dy hi t u b chn. Chng minh: Gi sl i mnnx a= . Theo nh ngha vi =1, ta tm c mt s t nhin p sao cho vi mi s t nhin np, ta c: n x a < 1. Do n x a n x a nn nx 1 v do k22, bi v k2 l s t nhin. Mt cch tng qut gi s ta chng minh c nk n, ta nhn c 1 nk n+ >v do + +11nk n . Cc v d v dy con l: = = =2 4 6 8 1 2, , , ,...,( 2, 4,..., 2 )nx x x x k k k n(2.1.14) = = = 1 3 5 7 1 2, , , ,...( 1, 3,..., 2 1)nx x x x k k k n(2.1.15) = = =21 4 9 16 1 2, , , ,...( 1, 4,..., )nx x x x k k k n (2.1.16) =1 3 5 7 11 13 17, , , , , , ,...( ,n n nx x x x x x x x p p l s nguyn t) (2.1.17) nh l 2.1.3 Mi dy con ca mt dy hi t l mt dy hi t v c cng gii hn.Chng minh: Gisdy(2.1.1)ccnggiihnav{nkx }lmtdyconcady(2.1.1).Tahy chng minh dy {nkx } cng c gii hn l a. t yn=nkx . Gi s cho trc >0 vl i m ,=nnx ann tn ti s p sao cho vi n>p ta c | | p th kn>p (v kn n) v do | | | | = 0tatmcp1vp2saocho:khin>p1th|xn a|p2 th|yn b|< 2. Gi p = max(p1 , p2) th khi n>p ta c:| ( ) ( )|| | | |n n n nx y a b x a y b + + + < . T y suy ra iu phi chng minh. (ii) Chng minh tng t nh trn (iii)Ta c ng thc:( )( ) ( ) ( ) = + + n n n n n nxy ab x a y b ay b b x a . V, n nx ay b, nn vi > 0 cho trc, tm c p1, p2 sao cho: khi n>p1 th |xn a|< , khi n>p2 th |ynb|< .Gi p =max(p1,p2) th khi n>p ta c: | | | | . | | . , < + +n nxy ab a b t l i m( ) 0. =n nnxy ab(iv) Do y bn, nn ta c th chn m sao cho khi n>m th | ny b |0 cho trc tm c p >m sao cho khi n > p ta c| ny b |p ta c: 21 1 2| || | = < n pth < < +nb y b . Nu gi p =max(1 2, p p ) th bt ng thc: , < + = < >n nx a b y n p c tho mn. Do :, < >n nx y n p . Ch : Trng hp c bit khi *, = Nny b nta c khng nh sau: Nu nhl i m= n pta cnny b a, thpsao cho> n pta c. >ny anh l 2.1.6 Cho hai dy s {xn} v {yn}. Khi : (i)Nu = = vl i m , l i mt hn n n nn nx y x a y b a b (ii)Nu {zn} l mt dy tho mn = = =vl i m l i mt h l i m .n n n n n nn n nx y z n x z a y aChng minh: (i) Hy chng minh khng nh ny bng phn chng. Gi s a < b, khi tn ti s r tho mn a< r p1 th xn < r. Tng t ta tm c p2 sao cho khi n > p2 th yn> r. Nugip=max(p1,p2)thkhin>ptacxn r,nghalxncho trc tm c p1 sao cho khi n >p1 th:| |hay n nx a a x a < < < + . Tng t, v nz a , ta tm c p2 sao cho khi n>p2 ta c < < +na z a . T y, t p = max(p1,p2), th khi n > p ta c < < +n n na x y z a . Suy ra < < +na y a , tc l ny a , iu phi chng minh. 2.1.3Gii hn v hn nh l 2.1.7 Cho dy s {xn}. Nu vi mi M > 0 ln tu , bao gi cng tn ti mt s p sao cho, n p >ta c xn>M, th ta ni rng dy {xn} c gii hn cng v cng v k hiu l 25l i mnnx= +. Nu vi mi M >0 ln tu , bao gi cng tn ti mt s p sao cho, n p >ta c xn< M, th ta ni rng dy {xn} c gii hn tr v cng v k hiu l l i mnnx= . Cui cng ta hy ch rng mt dy hi t khi v ch khi n c gii hn hu hn. Dy c gii hn l khng c xem l dy hi t. 2.2Tiu chun hi t 2.2.1Cc nh l nh ngha 1 Dy 1 2 3, , ,... x x xgi l tng nu(2.2.1) + N*1 n nx x n (2.2.2) Nu+< N*1 n nx x n (2.2.3)ta ni rng dy (2.2.1) l dy thc s tng. Tng t, nu nh *1 + Nn nx x n (2.2.4)dy (2.2.1) l gim. Nu nh *1 +> Nn nx x n (2.2.5) th ta ni rng dy (2.2.1) thc s gim. Cc dy ni trn gi chung l cc dy n iu. Tt c cc dy n iu to nn mt lp dy rt quan trng. By gi i vi nhng dy ny ta c hai nh l quan trng sau. nh l 2.2.1 Gi s dy (2.2.1) l khng gim. Nu nh dy khng b chn trn th l i m= +nnx .(2.2.6) Nu nh dy b chn trn th c gii hn hu hn 1,2,...l i m sup==n nnnx x .(2.2.7) Chng minh: (i)Gisdy(2.2.1)khngbchntrn.Khi0 > M lntu,tautmc mtstnhinpsaochoxp>M(tcltnhtmtshngcadylnhnM).Biv, dy l khng gim, nn khi n>p, ta c:n px xv do xn>M, cho nn l i m= +nnx . (ii) Gi s dy(2.2.1) b chn trn, ta t261,2,...supnna x== . (2.2.8) Theo nh ngha * Nnx a nMt khc,0, > ta tm c chng hn phn t xp ca dy sao cho> px a . Vi n>p ta cn px x , nn: < < +na x a a tc l |xn a| + + n nn n.(2.2.13) Tht vy, bt ng thc (2.2.13) tng ng vi bt ng thc: 1 21 21+ ++ + > + n nn nn n(2.2.14) hay2 21 2 1.1+ ++ + + > + n nn n nn n n(2.2.15) hay 22( 1) 11( 2)+ +> + + nnn n n(2.2.16) hay 21 11 1( 2)+ + > + + nn n n (2.2.17) bi v (n+1)2 =n (n+2)+1. Mt khc, theo bt ng thc Bernoulli vi h >0, k >1, k nguyn ta c: (1 ) 1 + > +kh kh(2.2.18) Bng cch thay 1,2( 2)= = ++h k nn n vo (2.2.18) ta thy rng bt ng thc (2.2.17) hin nhin c chng minh v do dy {yn} l dy gim. Hn na dy {yn} b chn di (bi v yn>0 * N n ). Do tn ti gii hnl i m .nnyMt khc 1 11 11 l i m 11l i m l i m 1 l i m l i m1 11 l i m 1+ + + + = + = = = + + n nnnn nn n nnnn nx ynn n T y suy ra gii hnl i mnnx tn ti. Theo Euler gii hn ny c k hiu bi ch e: e =1l i m 1 + nnn.(2.2.19) S e l mt s v t, 15 s u trong khai trin thp phn ca n l e =2,718281828459045 2.2.3Nguyn l Cantor v dy cc on thng lng nhau v tht li nh l 2.2.4 Gi s 1 1 2 2[ , ] [ , ] ... [ , ] ... n na b a b a bl dy v hn cc on thng lng nhau v tht li: l i m( ) 0. =n nnb a (2.2.20) 28Khi tn ti duy nht mt phn t 1[ , ]= n nna b .(2.2.21) Chng minh: Xtccdy{an}v{bn},tathyrngbn*1 N b n v < n na b n nn 1 cho trc bao gi cng * N p sao cho, > m n p ta c | | cho trc tn ti s p sao cho> n p ta c:| |2 n m p ta c | | | | | | | |2 2 = + + < + =m n m n m nx x x a a x x a x a . Vy {xn} l dy c bn. ii) iu kin Gis{xn}ldycbn.Trchttahychngminhdy{xn}bchn.Thtvy 1, = p sao cho, > m n pta c:m nx x n px x n p. T y suy ra: *| | , nx M n N . Nh vy {xn} l mt dy b chn, theo nguyn l Bolzano Weierstrass tn ti mt dy con {knx } hi t, gi sl i m=kknnx a . Khi 0 > cho trc, 1 >kn p ta c | |2 m n pta c | m nx x |p th> n pta c:.2 2 = + + < + =k n n nk k kn n n nx a x x x a x x x a Vyl i m=nnx a. V d 1: Dng tiu chun Cauchy xt s hi t ca dy 1 11 ...2= + + +nxn Ta thy, vi n bt k, m = 2n th: 3021 1 1 1 1 1| | ... ...1 2 2 2 2 2 = + + + > + + =+ +n nx xn n n n n. Do 2112 > n nx x n , vy dy {xn} phn k. V d 2: Dng tiu chun Cauchy xt s hi t ca dy 2 2 21 1 11 ...2 3= + + + +nxn Gi s m > n, ta c 2 2 21 1 1...( 1) ( 2)1 1 1 1 1 1 1 1 1...1 1 2 1 = + + + btu,nuchns 11 p> +(khiu 1 lphn nguyn ca 1), khi :, > m n pth | | > nu a n . Theo tnh cht ca supremun ta c, n k + < n ka x a . Tytathyrngdy{xn}cvsshngnmtrong( , ] a a .Vyalgiihn ring ca dy {xn}. Sau y ta chng minh a l gii hn ring ca ln nht. By gi gi s ngc li c mt dy con{ }knxca dy {xn} ml i m=knkx b v b > a. Khi 2+< nhxkhong( , ) + lnkhong ( 0, + ). Ngc li, i vi mi y > 0 tn ti x sao cho xa =y, ta tl og =ax y. Hm ngc 1f c xc nh nh sau:1( ) l oghayl og= = =xa af y y x y a y . 36Bygihyvthcahmsl og =ay x.Hmsnynhncthms l og =ax y bng cch i x thnh y nn th ca hm sl og =ay x i xng vi th hm s y=xaqua ng phn gic th nht. Ch ta c h thc sau 1( ) = f f x x x Y(2.3.8) ngc li 1( ) = f f x x x X .(2.3.9) V d nh vi0, 1 a ata cl ogv, 0 = = >al ogx xa a x x a x x . 2.3.5Cc hm lng gic ngc a) Hm s y =arcsinx Hm s y=sinx c xc nh trong khong X=( + , ) v gi tr ca n lp y on Y=[1,1]. ng thng song song vi trc Ox ct ng sin, tc th ca hm s y = sinx ti mt tp v hn cc im, ni mt cch khc mi gi tr y[1,1] s ng vi mt tp v hn cc gi tr ca xX. V vy, hm ngc m ta k hiu l x = Arcsiny, s l hm a tr.= ar csi n y xyx22ar ccos y x =2 Hnh 2.3.5 Thng thng ta ch xt mt nhnh ca hm s ng vi x bin thin giav 2 2 . Mi gi tr y[1,1] s ng vi mt gi tr,2 2x , n c k hiu bng x = arcsiny v gi l nhnh chnh ca hm Arcsiny. Bng cch ly i xng ng sin qua ng phn gic th nht ta c th ca hm atry=Arcsinx.Bngcchthuhpthtrnvi,2 2y tacthhms y=arcsinx c xc nh trn on [1,1] vi tp gi tr,2 2y v l mt hm s tng (xem Hnh 2.3.5).37Ta c cng thc cho tt c cc gi tr ca hm ngc ar csi n 2Ar si nv i 0, 1, 2,...(2 1) ar csi nx kc x kk x+ = = + (2.3.10) b) Hm s y = arccosx Ta thy y = cosx, = 0 ar ccos x x y. Hm s ngc ca hm s y = cosx l hm s y = arccosx. Hm s y = arccosx c min xc nh l tp [1,1] v min gi tr l [0, ] v l hm s gim. (xem hnh v Hnh 2.3.5) Dosi n cos( )2= x x nn d dng suy ra cng thc: ar csi n ar ccos2+ = x x. (2.3.11) c) Hm s y = arctgx Hm s y=tgx l mt n nh tp,2 2 ln tpR, hm s ngc ca n l x=arctgy: R y , x ,2 2 . Do hm s y=arctgx c tp xc nh l tpR v tp gi tr l khong m,2 2 v l hm s tng. d) Hm s y=arccotgx Hm s y = arccotgx l mt n nh tp X=(0, ) ln tp Y=( , ) +nn c hm ngc l x = arccotgy, y( , ) + ,x X=(0, ). Do hm s y = arccotgx c tp xc nh l X=R, v tp gi tr l Y(0, ) v l mt hm s gim. Ta c th chng minh cng thc ar ct g ar ccot g2x x+ =(2.3.12) 22y ar ct g = x y ar ccot g = x2 38Hnh 2.3.6Hnh 2.3.7 e) Khi nim cc hm s cp Trongtonhccchmssauycgilcchmsscpcbn:hmslu tha: R , x x ; hm s m:, 0, 1xx a a a > ; hm s logarit:l ogax x ; cc hm s lng gic: xsinx, xcosx, x tgx, xcotgx v cc hm s lng gic ngc. Ngi ta gi hm s cp l nhng hm s c to thnh bi mt s hu hn cc php ton s hc (cng,tr,nhn,chia),ccphplyhmshpivicchmsscpcbn.Trong phn ny, ngoi cc hm s s cp nu trn, ta cn nghin cu lp cc hm s hypebol. 2.3.6Cc hm s hypebol Xt t hp tuyn tnh ca hm s m 1 1v i>12 2 2+= + =x xx xa ay a a a (2.3.13) Ta thu c th ca hm s y bng cch cng cc th ca cc hm 12xav 12xa . 12xa2x xa ay+=12xa12 2=x xa ay12 Hnh 2.3.8Hnh 2.3.9 l ng cong i qua im (0,1) v i xng qua trc Oy. Hm y l hm s chn. Trong gc to th nht khi x + th ca hm s (2.3.1) dn n th ca ng 12xav trong gc to th hai, khi x th ca hm s (2.3.1) dn n th hm s 12xa. Hm s = =1 1v i 12 2 2x xx xa ay a a a > (2.3.14) l mt hm s l. th ca hm s ny i qua gc to v i xng qua gc 0. t: 1 2( ) ( )2 2x x x xa a a ah x h x + = =39ta c th thit lp nhng h thc n gin nh sau: 2 21 2( ) ( ) 1 = h x h x2 21 2 1( ) ( ) (2 ) + = R h x h x h x xNi chung 1( ) h xc nhng tnh cht tng t vi cosx, 2( ) h xc nhng tnh cht tng tvisinx.Khia=etagihmthnhtlcosinhypebol,khiulchx,hmthhail sinhypebol, k hiu l shx, nh vy ch ,sh2 2x x x xe e e ex x + = = .(2.3.15) Ta cn c cc hm hypebol khc xc nh tng t nh cc hm lng gic tng ng, chng hn sh cht h ,ct hch shx xx xx x= = (2.3.16) Trn hnh v Hnh 2.3.10 biu din th ca cc hm thx v cthx. t h y x =ct h y x =ct h y x = Hnh 2.3.10 T nh ngha ta suy ra cc cng thc nu di y, mang tn l nhng nh l cng: 2 22 2ch sh 1ch sh ch2ch( ) ch .ch sh .shsh( ) sh .ch ch .sha aa a aa b a b a ba b a b a b =+ =+ = ++ = + (2.3.18) V d, ta c th chng minh hai cng thc cui ca (2.3.18) nu vit: ch( ) ,sh( )2 2a b a b a b a be e e e e e e ea b a b + + = + =v s dng h thc: ch sh , ch shch sh , ch sh .a ab be a a e a ae b b e b b= + = = + = 2.3.7Cc hm hypebol ngc 40a) Hm y= Argshx HmsshxnhxtpRlnRnnnchmngc,takhiuly=Argshx.Vy y=Argshx x=shy vi R R , x y . Ar gsh y x = Hnh 2.3.11 By gi ta biu din hm y=Argshx di dng lga. Ta thy y=Argshx tng ng vish2y ye ex y= = . t ye t = hayy=lntv 2112 2= =tttxt.Vynuxchotrcthtlnghimca phng trnh bc hai t2 2tx1=0. VimiR x ,phngtrnhtrnchainghim,trongcmtnghimdng 2( 0) : 1 = > = + +yt e t x x . Vy vi miR xta c 2Ar gsh l n( 1) = + + x x x.(2.3.19) b) Hm y =Argchx Hm y=chx nh x khong[0, ) +ln khong[1, ) + , vy ta c th xc nh mt hm ngc, k hiu l Argchx. Vyy=Argchx,[1, ), [0, ) + + x y ch , = x y [0, ), + y [1, ) + x .thca hmsy= Argchxsuytthcay=chx,0 x bngphplyixngquangphn gic ca gc phn t th nht. 41Ar gch y x =x Hnh 2.3.12 By gi ta hy biu din ngc di dng lga. Ta thyAr gch chv i 02+= = = y ye ey x x y yt ye =t hay y=lnt, ta c 2112 2++= =tttxt Vy vi1 x cho trc th t l nghim ca phng trnh bc hai t22tx+1=0. Vi1 x phng trnh trn c hai nghim dng m tch ca chng bng 1. Vy trong hai nghim th nghim ln hn c lga dng v ta c ye = t =21 + x xy=Argchx =ln(21 + x x ). Vi1 x th ca hm sy=Argchx trng vi th hm s y= ln(21 + x x ). Mt cch tng t, do hm thx nh x khong( , ) +ln khong (1,1) ta c th xt hm ngc ca n y=Argthx. Trong khong (1,1) hm y=Argthx tng ng vi hm. 1 1l n2 1+= xyx cn trong khong x 1, hm y=Argcthx tng ng vi hmy= Argcthx= 1 1l n2 1+ xx. 2.4Gii hn ca hm s 2.4.1Ln cn ca mt im a) Ln cn ca mt im 42Cho mt imR0x . Mt tp hp conR Uc gi l ln cn ca im 0xnu c mt s0 > sao cho 0 0( , ) + x x U . T nh ngha, ta suy ra: i)Nu U l ln cn ca 0xth mi tp V Ucng l ln cn ca im 0x . ii) Nu U1, U2 l ln cn ca im 0xth 1 2 U Ul ln cn ca im 0x . iii) Khong l ln cn ca mi im ca n. Ta thng gi khong 0 0( , ) + x x lln cn ca 0xv k hiu l 00 ( ) x. b) im t Ta xt tp hp R A . im a c gi l im t (im gii hn) ca tp A nu ln cn bt k( , ) + a a ca im a u cha t nht mt im ca A m im khc a. imtcatpAcththucAhoccthkhngthucA,vdnuA=[a,b]hoc A=(a,b],ima,trongchaitrnghp,limtcaA,nhngtrongtrnghpun thuc A, cn trong trng hp th hai th khng. NuA=(0,1),thmiim(0,1) a ulimtcaA,ngoiraccim 0 10, 1 a a = =tuy khng thuc A nhng vn l im t ca A.Mt dy im {xn} c gi l dy phn bit nu nh mi cp phn t bt k ca dy u khc nhau, tc l m nx x num n . Ta thy rng nu a l im t ca A th c th trch ra t A, theo v s cch, mt dy im phn bit 1 2 3 4, , , ,.... x x x xcc phn t ca A khc a hi t n a. Tht vy, bng cch chn mtdyccsdng0n ,ritrongmilncn( , ) +n na a caima(vin= 1,2,3,)tachnmtimx=xnthucAkhca.Khiv0 n v|xna|cho trc b tu bao gi cng tn ti mt s0 >sao cho 0,0| | x A x x < < , ta c | ( ) | f x L < , th ta ni rng L l gii hn ca hm s f khi x tin n a. nh l 2.4.1 Hai nh ngha trn l tng ng nhau. Chng minh: Gisl i m ( )x aL f x= theonhngha2vgis nx a tahychngminh ( )nf x L . Tht vy, theo gi thit0, 0 > >sao cho , x A < | | x ath | ( ) | f x L < . V nx a nnp sao cho> , n p < | |nx a . T suy ra,| ( ) |nn p f x L > < , vy l i m ( )x af x L= . By gi gi s vi mi dy nx a , ta u c dy cc gi tr tng ng ca hm s( )nf x L . Taschngminhl i m ( )x af x L= theonhngha2.Gisngcli,khi 0, 0,j jx A > sao cho0| |v0j khij , khi ta chn c dy {jx } c tnh cht jx a khij nhng f(jx ) khng tin n L, tri vi gi thit. V d 1: Cho f(x)= x. Ta chng minh rng00l i m ( ) .x xf x x=Tht vy, do 0 0| ( ) | | | f x x x x = nn cho trc0 > , ta cn chn =th khi0| | x x |0x |0| || |2xx > v 4420 01 1 2| || | x x x < .Bygitahychn = 20 0| | | |mi n ,2 2x x,th 0,| | x x x < tac01 1| | sao cho, x A x K > , ta c ( ) f x L Mln tu , tn ti mt s K > 0 sao cho, > x Ax Kta c f(x) > M th ta ni rng f c gii hn l + khix + v k hiu ll i m ( )+= +xf x . Nu0 M >ln tu , tn ti mt s K > 0 sao cho, > x Ax Kta c f(x) < M th ta ni rng f c gii hn l khix + v k hiu ll i m ( )+= xf x . nh ngha 4 Cho tp A c im t l, : R f A . Ta ni rng f(x) c gii hn hu hn l L khix nu0 >cho trc,0 K >sao cho, x A x K < , ta c | ( ) | < f x L v k hiu l = l i m ( )xf x L.Nu0 M >ln tu , tn ti mt s K > 0 sao cho, < x Ax Kta c f(x) > M th = + l i m ( )xf x . Nu0 > M lntuchotrc,baogicngtntimtsK>0saocho , < x Ax Kta c f(x) < M thl i m ( )= xf x . 47V d 7: Chng minh rng sin x khng c gii hn khix +. Tht vy, chn dy s(2 1)2= +nx n. R rng rng khin th nx + Khi dy {sin xn} nhn cc gi tr 1,1, 1,(1)n,... Dy s ny phn k, t suy ral i msi n xxkhng tn ti. Tng tl i m cos xxkhng tn ti. 2.4.5Cc tnh cht ca gii hn Ta c th d dng chuyn cc kt qu i vi gii hn ca dy sang trng hp gii hn ca hm, c th, ta c cc nh l sau. nh l 2.4.3 Nu hm f(x) c gii hn ti a th gii hn l duy nht. nh l 2.4.4 Gi s trn tp A cho hai hm f(x), g(x) v a l im t ca A (a c th hu hn hoc v hn). Ngoi ra gi s c hai hm c gii hn hu hn 1 2l i m ( ) ,l i m ( ) = =x a x af x L g x LKhicchmcf(x)viclhngs( ) ( ), f x g x ( ). ( ), f x g x( )( )f xg xcngcgiihn hu hn (trong trng hp thng c thm gi thit 20 L ), c th l: 1a) l i mc ( )=c ; xi f x L[ ][ ]1 21 2122)l i m ( )+ ( ) ;)l i m ( ). ( ) . ;( ))l i m ( 0).( )= == x ax ax ai i f x g x L Li i i f x g x L LL f xi v Lg x L Ch rng nh l trn cha c kt lun g trong cc trng hp sau: Trong trng hp ii), khi L1=+ v L2=v mt hnh thc ta c dng v nh. Trongtrnghpiii),khiL1=0( )vL2=(0)vmthnhthctacdngvnh 0. . Cui cng trng hp iv), khi L1=0( ) v L2=0( ), v mt hnh thc ta c dng v nh 00 hoc . 2.4.6Tiu chun tn ti gii hn ca hm s Sau y ta pht biu cc nh l ni v s tn ti gii hn ca hm s. nh l 2.4.5 ChoA R v x0 l im t ca tp A. Gi s ba hm s tho mn ng thc:( ) ( ) ( ) f x g x hx x A48Khi nu 0 0l i m ( ) l i m ( ) = =x x x xf x hx Lth 0l i m ( )=x xg x L . nh l 2.4.6 (Tiu chun Cauchy) Hmsfcgiihntix0khivchkhi0, 0 > > saochox v 0 0,0 | | ,0 | | < < < < x A x x x x ta c| ( ) ( )| < f x f x . nh l 2.4.7 (Tiu chun gii hn ca hm n iu) i) Gi s hm f(x) l hm tng trn [ )0, xx , khi nu f b chn trn trn [ )0, xx th hm f c gii hn tri ti x0. ii) Nu f(x) l hm gim trn [ )0, xx , khi nu f b chn di trn [ )0, xxth f c gii hn tri ti x0. Chng minh: Gi s f(x) l hm tng trn [ )0, xx . Gi ) 0,sup .=x xL fKhi [ )> 1 00, , x xxsao cho 1( ) L f x L < . Do hm tng nu 1 0( , ) x x x ta c 1( ) ( ) L f x f x L < . Chnsao cho 1 0x x = . Khi 0, 0 > >sao cho 0 0( , ) x x x th( )| ( ) | L f x L f x L < < + 0)) l cc v cng b cng bc. Chng hn VCB 1- cosx l VCB cp hai i vi VCB x khi0 x v VCB tgx sinx l VCB cp ba i vi VCB x khi0 x bi v 512 30 01 cos 1 t g si n 1l i m ,l i m2 2 = =x xx x xx x. c) So snh cc v cng ln Ta hy quay li trng hp xt cc v cng ln. V d 1: Hm s 2 si n( )+=xf xx l VCL khi0 xVd2:Xthms 1 1( ) cos = f xx xkhi0 x .Ticcim 12=+kyk,k=1,2,3, f(yk)=01,2,3... = kKhi, k 0, ( ) 0 k ky f y , cho nn f(x) khng phi l VCL khi 0 x . Cho R A vx0limtcaA.NgoirachocchmR , : f g A lVCLkhi 0x x . i) Nu fg l VCL khi 0 x xc ngha l nu 0l i m= +x xfg, th ta ni f l VCL bc cao hn g khi 0 x xii) Nu 0( )l i m 0( )= x xf xlg x, th ta ni rng f v g l hai VCL cng bc khi 0 x xcbitkhil=1,tanirngfvglVCLtngngkhi0x x .Trongkhixt nhiu i lng VCL ngi ta chn mt trong cc i lng lm c s v so snh cc VCL khc vi lu tha ca i lng c s . Chng hn, nu tt c cc i lng u l hm ca x v tr thnh VCL khi 0x x th dng lm VCL c s ngi ta ly|x|nu x0= v ly 01 x x khi x0 l hu hn. 2.4.8Cc gii hn ng nh 0si n 1) l i m 1 ) l i m 1 + = + = xx xxa b ex x 101) l i m 1) l i m(1 ) . + = + = xxx xc e d x ex Chng minh: Gii hn a) c trnh by bc ph thng. By gi ta hy chng minh gii hn b). Gii hn b): Tht vy, ta chng minh c 1l i m 1 . + = nnen 52By gi ta rng vi s dng bt k x no cng tn ti s t nhin n ( 0 n ) sao cho 1 + n x nngha l 1 1 11 + n x n t y 11 1 11 1 11+ + + + + n x nn x n Chuyn qua gii hn bt ng thc kp trn ta suy ra 1l i m 1+ + = xnex Gii hn c): S dng php bin i x = y ta c: 11 1 1l i m 1 l i m 1 l i m l i m11 1 1l i m 1 l i m 1 11 1 1 + + ++ + + = = = = + = + + = y x x xy x x xx xx xx xy x x xex x x (theo gii hn b)). T y suy ra gii hn c) c chng minh. Gii hn d): chng minh gii hn d) ta t 1yx= v p dng gii hn b) v gii hn c) Ni chung 1( ) h xc nhng tnh cht tng t vi cosx, 2( ) h xc nhng tnh cht tng t visinx.Khia=etagihmthnhtlcosinhypebol,khiulchx,hmthhail sinhypebol, k hiu l shx, nh vy ch ,sh2 2x x x xe e e ex x + = = (2.3.15) Ta cn c cc hm hypebol khc xc nh tng t nh cc hm lng gic tng ng, chng hn sh cht h ,ct hch shx xx xx x= =(2.3.16) Trn hnh v Hnh 2.3.10 biu din th ca cc hm thx v cthx. t h y x =ct h y x =ct h y x = 53Hnh 2.3.10 T nh ngha ta suy ra cc cng thc nu di y, mang tn l nhng nh l cng: 2 22 2ch sh 1ch sh ch2ch( ) ch .ch sh .shsh( ) sh .ch ch .sha aa a aa b a b a ba b a b a b =+ =+ = ++ = + (2.3.18)

5252V d, ta c th chng minh hai cng thc cui ca (2.3.18) nu vit: ch( ) ,sh( )2 2a b a b a b a be e e e e e e ea b a b + + = + =v s dng h thc: ch sh , ch shch sh , ch sh .a ab be a a e a ae b b e b b= + = = + = 2.4.9Cc hm hypebol ngc a) Hm y= Argshx HmsshxnhxtpRlnRnnnchmngc,takhiuly=Argshx.Vy y=Argshx x=shy vi R R , x y . Ar gsh y x = Hnh 2.3.11 Bygitabiudinhmy=Argshxdidnglga.Tathyy=Argshxtngngvi sh2y ye ex y= = . t ye t =hay y =lnt v 2112 2= =tttxt. Vy nu x cho trc th t l nghim ca phng trnh bc hai t2 2tx1=0. VimiR x ,phngtrnhtrnchainghim,trongcmtnghimdng 2( 0) : 1 = > = + +yt e t x x . Vy vi miR xta c 2Argsh l n( 1) = + + x x x . (2.3.19) b) Hm y =Argchx Hm y=chx nh x khong[0, ) +ln khong[1, ) + , vy ta c th xc nh mt hm ngc, k hiu l Argchx. 5353Vyy=Argchx,[1, ), [0, ) + + x y ch , = x y [0, ), + y [1, ) + x .thca hm s y = Argchx suy t th ca y=chx,0 x bng php ly i xng qua ng phn gic ca gc phn t th nht. Ar gch y x = Hnh 2.3.12 By gi ta hy biu din ngc di dng lga. Ta thyAr gch chv i 02+= = = y ye ey x x y yt ye =t hay y=lnt, ta c 2112 2++= =tttxt Vy vi1 x cho trc th t l nghim ca phng trnh bc hai t22tx+1=0. Vi1 x phng trnh trn c hai nghim dng m tch ca chng bng 1. Vy trong hai nghim th nghim ln hn c lga dng v ta c ye = t =21 + x xy=Argchx =ln(21 + x x ). Vi1 x th ca hm sy=Argchx trng vi th hm s y= ln(21 + x x ). Mtcchtngt,dohmthxnhxkhong( , ) + lnkhong(1,1)tacthxt hm ngc ca n y=Argthx. Trong khong (1,1) hm y=Argthx tng ng vi hm. 1 1l n2 1+= xyx cn trong khong x 1, hm y=Argcthx tng ng vi hmy= Argcthx= 1 1l n2 1+ xx. 54542.5Bi tp chng 2 2.1Chngminhrngdy ( 1,2,3,...)nx n = cgiihnbngkhngbngcchsdng ngn ng" " . 1) 1=+nnxn 2) 1( 1) +=nnxn 3) 321=+nnxn 4) 1!=nxn 5)( 1) .0,999 = n nnx . 2.2S dng nh ngha hy chng minh rng cc dy sau c gii hn v cc khin + 1) xn=(1)n n(n=1,2,3,) 2)2( 1,2,3,...)nnx n = =3)l n(l n )( 1,2,3,...)nx n n = = . 2.3Tml i mnnx+ vi 1) ( 1) .1nnnxn=+2) 8cos .24nnxn=+ 3) 2 ( 1)nnnxn+ = 4)2 . . cosnnx a n= , a=const. 2.4Tm cc gii hn sau 1) ( )1l i m 12nn n n++ +2) ( )3 3l i m 1nn n++ 3) 1 1 1l i m ...1 2nn n n+ + + + + 4) 2 21 1 1l i m ...( 1) (2 )nnn n n+ + + + + 5) 2 2 21 1 1l i m ...1 1nn n n n+ + + + + + + . 55552.5Chng minh rng +=+ =0l i m 0kinia n inu ==00kiia2.6Tm cc gii hn sau 1) 1 1 1l i m ...1.2 2.3 ( 1)nn n+ + + + + 2) ( )2 4 8l i m 2 2 2... 2nn+ 3)l i m2nnn+ 4) 2 2 23 3 31 3 (2 1)l i m ...nnn n n+ + + + 5) + + + + 21 3 2 1l i m ...2 2 2nnn 6) 11 2 3 ( 1) .l i m ...nnnn n n n+ + + + . 2.7Chng minh rng 1) 1 3 2 1l i m . ... 02 4 2+ = nnn, 2)l i m 0nu | | = + . Hy tml i mnnx. 2.12 Cho dy 11 ; 1,2,3,...nx nn= =Hy tml i mvl i mn nnnx x . 2.13 Cho dy 11( 1) 2nnxn = + Hy tml i m , l i mn nn nx x . 2.14 Cc dy s: 0 10 1, ,..., ,..., ,..., ,...nnx x xy y y c xc nh bng cng thc: 0 0, x a y b = =vi0, 0; a b > >1,n n nx xy+ =12n nnx yy ++= . Chng minh rng: 1) {xn} l dy tng, {yn} l dy s gim. 2) Cc dy s trn hi t v c gii hn bng nhau. 2.15 Tm tp xc nh v tp gi tr ca cc hm sau 57571) 22 y x x = + 2)l g(1 2cos ) y x = 3) 22ar ccos1=+xyx4)ar csi n l g10 = xy . 2.16 Cho f xc nh trn tp A=(0,1). Tm tp xc nh ca hm s 1)(si n ) f x 2)(l n ) f x 3) | | xfx 2.17 Tm cn di ng v cn trn ng ca hm s ( )1xf xx=+ trong min0 x < + 2.18 Chohmf(x)xcnhtrnkhong( , ) + vthamnngthc ( ) ( ) f x T kf x + = ,trongkvTlnhngsdng.Chngminhrngkhi ( ) ( )xf x a x = , trong a l hng s dng v( ) x l hm tun hon vi chu k T. 2.19 Xt tnh tun hon v chu k ca cc hm s sau 1) ( ) 2t g 3t g2 3x xf x = 32) ( ) cos cos2 2x xf x =3) ( ) si n si n v il sf x x x = + v t 24) ( ) | si n | | cos |5) ( ) cos6) ( ) si n .f x x xf x xf x x x= +== + 2.20 Chng minh rng hm 1nu hu t ( )0nu v t = xxxl hm tun hon. 2.21 Cho 1( )1f xx= Hy tnh[ ( )], { [ ( )]}. f f x f f f x2.22 Gi s( ) ((... ( )))nf x f f f x = . Hy tnh( )nf xnu 2( )1xf xx=+. 2.23 Tm hm s ngc ca cc hm s sau = + < < +21) 3v i)0,) 0y xa x b x 2 312) , 113) 1 v i)1 0,) 0 1.= += < 2.25 1) Cho hm s R R : fc xc nh bi ( )1 | |xf xx=+ Chng minh hm s f(x) c hm ngc v tm hm ngc n. 2) Tm hm f(x) bit rng1 1( 1) ( ) ( )1x f x fx x + = khi0, 1 x x 2.26 Tm hm s f(x) bit rng n c xc nh vi mi gi tr x v tho mn h thc( ) ( ) 2 ( )cos f x y f x y f x y + + =vi mi x v y. 2.27 Tm cc gii hn sau 1) 11l i m1mnxxx (m,n l nhng s nguyn dng) 2) ++ 1 cosl i msi nxxx 3) 4si n 2 cos2 1l i mcos si nxx xx x. 2.28 Tm cc gii hn sau 22121( 2)222220ar csi n(1 2 )1) l i m4 1si n(2 )2) l i m 24ar si n( 2)3) l i m2si n4) l i m .1 si n cos + +++ xxxxxxxxxc xx xxx x x 2.29 Tm cc gii hn sau 2021 11) l i msi n4si n2 x xx 5959222si n2) l i m t gcos xxxx 2.30 Tm cc gii hn sau + 201 si n cos21) l i msi nxx x xx 402) l i m1 2 1+ xxx 30( 1)3) l i mcos1++xxx 201si n4) l i m xxxx 015) l i m ( 1 ).2+ xx x x2.31 Tm cc gii hn sau 011) l i m xxx 2) l i m v i0.>x ax ax aax a 2.32 Tm cc gii hn sau 21) l i m si n( 1)++nn 2 22) l i m si n ( )++nn n 213) l i m si n .+=nnkkan 2.33 Chng minh 2l i m(cos )mmx tn ti vi mi gi tr x v tm gii hn. 2.34 1) Tm gii hnl i msi n si n...si nnx. 2) Tm gii hn 2lim[ lim(cos( !) ) ] mn mn x 6060 6161Chng 3 Hm lin tc mt bin s Khi nim lin tc ca hm s l khi nim rt c bn, ng mt vai tr rt quan trng trong vic nghin cu hm s c v l thuyt v ng dng. Trc ht, ta hy tm hiu v tnh lin tc ca hm s. 3.1nh ngha s lin tc ca hm s ti mt im 3.1.1Cc nh ngha nh ngha 1 ChoR : f Av 0x A . Ta ni rng hm f lin tc ti im 0xnu vi bt k s 0 >cho trc c th tm c s0 >sao chox A m 0| | x x , chn = , th < R0,| | x x x ta c 0| ( ) ( )| f x f x < . Vy hm( ) | | f x x = lin tc ti miR. xV d 2: Dng ngn ng( ) hy chng minh 22l i m(3 2) 10.xx =Ta thy: 2| ( ) 10| | 3 2 10| 3| 2| | 2|3| 2| | 2 4| 3| 2| (| 2| 4) 3 ( 4)f x x x xx x x x = = + == + < + < + khi | 2| x < 6262Trc ht, nu chn1 0 ta c + +++=++22l i m l i m111nxnxnxn nnxxxx xeeee 6565Vi x >0, khi ch l nxe khin + ta c++=+22l i m1nxnxnx xexe Vi x Hin nhin l khi0 x hm s lin tc. Mt khc ta thy 0 0l i m ( ) l i m ( ) 0 (0)x xf x f x f + = = = , vy hm s cng lin tc ti x=0. Do f(x) lin tc R x . V d 10: Cho 12 khi 21( ) khi 2.1xa xf xxe = = + Tm a hm lin tc R x . D thy khi2 x hm s lin tc. hm s lin tcx R th n phi lin tc ti x=2. Ta thy 12 221l i m ( ) l i m 01x xxf xe+ + = =+ (bi v 12 xe + khi2 x+ ) 12 221l i m ( ) l i m 11x xxf xe = =+ (bi v 120xe khi2 x ). Vya hm s khng th lin tc ti x = 2. 3.1.4 im gin on ca hm s Hm s f(x) c gi l gin on ti x0 nu ti x = x0 hm f(x) khng lin tc. Vy x0 l im gin on ca hm s f(x) nu: hoc x0 khng thuc tp xc nh ca f(x), hoc x0 thuc tp xc nh ca f(x) nhng 00l i m ( ) ( )x xf x f x , hoc khng c 0l i m ( )x xf x. im gin on x0 ca hm s f(x) c gi l im gin on loi 1 nu cc gii hn mt pha 0l i m ( )x xf x+, 0l i m ( )x xf x tn ti hu hn v t nht mt trong hai gii hn ny khc f(x0). im gin on khng phi loi 1 s c gi l im gin on loi 2. Nh vy im x0 l im gin on loi 2 nu hm s khng c gii hn mt pha hay mt trong hai gii hn l v hn. 6666Gi s x0 l im gin on ca hm s y = f(x). Nu tha mn ng thc 0 0lim ( ) lim ( )x x x xf x f x+ =th im gin on x0 gi l kh c. Nu t nht mt trong cc gii hn mt pha ni trn bng th x0 gi l im gin on v cng. V d 11: Cho hm s: khi 0( )1khi 0.x xf xx x = + < im x = 0 l im gin on loi 1 ca hm s. V d 12: Cho hm s 1( ) f xx=im x=0 l im gin on loi 2, bi v0 01 1l i m , l i mx x x x x x+ = + = . V d 13: Cho += l i m 0xxxe (trong chng 4 ta c th tnh gii hn ny mt cch d dng nh qui tc Lhospital). Xt hm s 2121 khi 0( )khi 0xe xf xxa x = = Vi gi tr no ca a hm s gin on ti x = 0. Khi hy xt xem x = 0 l im gin on loi g? t 21,khi 0 t xx= tht + . Ta thy: 0l i m ( ) l i m 0,tt xf x te++ = =0l i m ( ) l i m 0tt xtf xe+ = = . Vy vi0 a hm s gin on tai x = 0. Mt khc do 0 0l i m ( ) l i m ( ) 0,x xf x f x+ = =nn x = 0 l im gin on kh c. V d 14: Cho hm s: si n khi 1( ) ar csi n khi 1 1,coskhi 1.x xf x x xa x x = < th0 > sao cho f(x) >00 ( ) x x A ,trong 0 00 ( ) { :| | } x x x x = < . (3.2.1) ii) Nu 0 sao cho 0( ) 0 ( ) . f x x x A < (3.2.2) Chng minh: i) Do hm s f lin tc ti 0x A nn0, 0 > >sao chox A 0| | x x 00 ( ) x x A . By gi hy chn = >0( ) 0 f xkhi 0 0( ) ( ) ( ) f x f x f x > + =00 ( ) x x A . ii) Tng t hy chn 0( ) f x = . ngha: Hm lin tc f(x) bo ton du ca n trong mt ln cn ca im x0. 3.2.2 Tnh cht ca mt hm s lin tc trn mt on nh l 3.2.2 (nh l Weierstrass th nht) Nu hm f xc nh v lin tc trn on [a,b] th n b chn, tc l tn ti cc hng s m v M sao cho ( ) [ , ] m f x M x a b . (3.2.3) Chng minh:Ta hy chng minh nh l bng phn chng. Tht vy, ta gi s rng hm s khng b chn. Khi vi mi s t nhin n ta tm c trn [a,b] gi tr x = xn sao cho: ( )nf x n >(3.2.4) Theo b Bolzan - Weierstrass t dy {xn} c th trch ra mt dy con{ }knx hi t n mt gii hn hu hn: 0knx x khik +, trong hin nhin 0. a x b 6969V hm lin tc ti x0 nn 0( ) ( )knf x f x . Nhng khi t (3.2.4) ta suy ra ( )knf x +, khc f(x0) l mt hm s hu hn. Mu thun ny suy ra nh l c chng minh.Ta ch rng nh l khng cn ng i vi nhng khong khng ng. V d nh hm 1x lin tc trn khong (0,1) nhng trong khong ny hm s khng b chn. nh l 3.2.3 (nh l Weierstrass th hai) Nu hmR : [ , ] f a blin tc th n t cn trn ngv cn di ng trong [a,b], tc l tn ti 1 2, [ , ] c c a b sao cho 1[ , ]sup ( ) ( )x a bf x f c=v 2[ , ]i nf ( ) ( )x a bf x f c= . (3.2.5) Chng minh: Theo nh l trn, do hm lin tc nn n b chn. Ta c M=[ , ]sup ( )x a bf x< +. Theo nh l supremum ta c mt dy {xn} [a,b] sao chol i m ( ).nnM f x=Dy {xn} b chn nn n cha mt dy con {knx } hi t, c th 1knx c khik . Mt khc t cc bt ng thc kna x b suy ral i mknka x b tc l 1[ , ] c a b . theo gi thit hm f lin tc ti c1, nn 1l i m ( ) ( )knkM f x f c= = . Hon ton tng t 2c sao cho: 2[ , ]i nf ( ) ( )x a bf x f c= . c) Ch : i) Cui cng ta ch rng gi tr c1, c2 ni trn khng phi l duy nht. V d trn hnh 3.2.1 Hnh 3.2.1 V d trn hnh 3.2.1 hm f(x) trong on [a,b] nhn gi tr ln nht ti hai im c1, c2 v nhn gi tr b nht ti nhiu v hn im ca on [d1, d2]. 7070nh l khng cn ng i vi nhng khong khng ng, v d hm f(x) = 2x2 nh x khong [0,1) ln khong [0,2), do sup ( ) 2 f x =nhng hm f(x) khng nhn gi tr 2 trong khong [0,1). ii) Nu hm f(x) khi bin thin trn mt khong X no d l b chn th ta gi dao ng ca n trong khong l hiuM m = gia cn trn ng v cn di ng ca n. Ni cch khc ,sup {| ( ) ( )| }.x x Xf x f x = Nu xt hm f(x) lin tc trn on hu hn X=[a,b], th theo nh l va chng minh trn, dao ng ca hm f(x) s ch n gin l hiu gia gi tr ln nht v b nht ca hm trn on . nh l 3.2.4 (nh l Bolzano - Cauchy th nht) Gi s hm f(x) xc nh v lin tc trn [a,b] vf(a).f(b) < 0. Khi tn ti t nht mt im c(a,b) sao cho f(c)=0. Chng minh: xc nh ta gi s f(a)< 0 cn f(b)>0. Ta chia i on [a,b] bi im 2a b +. C th xy ra l f(x) trit tiu ti im , khi nh l c chng minh, ta c th t c =2a b +. By gi gi s f(2a b +) 0, khi ti cc u mt ca mt trong cc on [a, 2a b +], [2a b +,b] hm s ly cc gi tr khc du nhau (c th gi tr m ti mt tri v gi tr dng ti mt phi). Gi on l [a1,b1], ta c (xem hnh 3.2.2) f(a1)< 0, f(b1)> 0 Hnh 3.2.2 By gi ta li chia i on [a1,b1]. C th xy ra hai kh nng: Hoc l f(x) trit tiu ti trung im 1 12a b + ca on , khi ta c th chn im c = 1 12a b +, nh l c chng minh. Hoc l ta thu c on [a2,b2] l mt trong hai na ca on [a1,b1] sao cho f(a2) < 0, f(b2) > 0. (Xem hnh 3.2.2) Ta tip tc qu trnh lp cc on . Khi hoc sau mt s hu hn bc ta s gp trng hp im chia l im ti hm trit tiu v khi nh l c chng minh. 7171Hoc c mt dy v hn cc on cha nhau. Khi i vi on th n, [an,bn] (n=1,2,3) ta s c f(an) 0 v di ca onr rng bng bn an=2nb a . Dy cc on ta lp c tho mn cc iu kin ca b v dy cc on lng nhau, bi v theo trnl i m( ) 0n nnb a = . V vy, c hai dy {an}, {bn} dn ti gii hn chung = = l i m l i mn nn na b c, m r rngc[a,b]. Ta hy chng minh im c ny tho mn yu cu ca nh l. Tht vy, do tnh lin tc ca hm s ti x = c, ta c ( ) l i m ( ) 0nnf c f a= v = ( ) l i m ( ) 0nnf c f b . Vy f(c)=0, nh l c chng minh. nh l 3.2.5 (nh l Bolzano - Cauchy th hai) Gi s hm f(x) xc nh v lin tc trn on [a,b] v ti cc u mt ca on hm f(x) nhn cc gi tr khng bng nhau f(a) = A, f(b) = B. Khi vi s C bt k nm trung gian gia A v B, ta c th tm c im( , ) c a b sao cho f(c)=C. Chng minh: Khng mt tnh tng qut ta c th gi thit rng Ach ph thuc vosao cho, x x A m | | x x < , th | ( ) ( )| f x f x < .(3.4.2) V d 1:i) Chng minh rng hm f(x) = x lin tc u trn ton trc s. Tht vy0 >ly =ta thy, x x Rm | | x x cho trc bt k, ta ch cn chn =th khi , x x R , | | x x nh ty , ta ch cn chn 2 = , khi x x , ( 1,1) mx x | | sao cho n nx x | | 0 th n nf x f x | ( ) ( )| . (3.4.3) nh l 3.4.1 (nh l Cantor): Nu: [ , ] f a b R lin tc th n lin tc u trn [ , ] a b . Chng minh: Ta hy chng minh nh l bng phn chng. Gi s f(x) khng lin tc u trn [ , ] a b , tc tn ti mt s dngsao cho0, >tn ti, [ , ] x x a b mx x | | nn l n( ) ( )x x aa e = , t u = xlna, l n( )' . l n l nu x a xe e a a a = =Do ta c cng thc sau ln ( )x xa a = avi0 a > .(4.2.6) ii) Ta c0l nx e x = >x v RDo :1 1l n l n( ) ( ) . . . .x xx e e xx x = = = . V ta c cng thc sau: 1( ) . x x = .(4.2.7) V d 4: Tnh 11cosxxdI edx+=vi1 x t11cos xux=+ 1111cos. . cosxu uxxd xI e eu edx x+ = = = + Li t 11xvx=+ ta c 21 1 1 21 1 1 1(cos ) si n . si n si n .( )x x xv v vx x x x = = = + + + + Cui cng 84841121 121 1cos. .si n( )xxxI ex x+ = + + . V d 5: Cho, : f gU R trong f(x)>0,x U v tn ti( ), ( ) f x g x vix U . Khi ( ) ( )l n ( ) ( )l n ( )( )( ( )) ( ( ). l n ( ))( ) =( ( )) . ( ) l n ( ) ( ). .( )gx gx f x gx f xgxd d df x e e g x f xdx dx dxf xf x g x f x g xf x = = + 4.2.3 o hm ca hm s ngc nh l 4.2.4 Gi s hm f(x) kh vi lin tc trn (a,b) vi0 ( ) f x ( , ) x a b . Khi hm f(x) n iu thc s nn c hm ngc x = g(y),: ( ( ), ( )) ( , ). g f a f b a b Khi g(y) cng kh vi ti y = f(x) v c o hm g(y) tho mn h thc:1( )( )g yf x =(4.2.8) hay gn hn:

1yxxy =.(4.2.9) Chng minh: Do (g.f)(x) = x( , ) x a b Hay( ( )) gf x x = ( , ) x a b . Ly o hm hai v ng thc trn theo x ta c 1 hay 1 ( ( )). ( ) ( ). ( ) g f x f x g y f x = =suy ra 1( )( )g yf x =( , ) x a b . V d 6: i) Xt hm s y = arcsinx vi 1< x nn 21yx x = , suy ra 211xyx =. Tng t, tao c cc cng thc sau: ii) y = arccosx vi 1< x sao cho ( )( )f tA Agt < < + khit c < . V d 6: Tnh 2 201 1l i m( )si nxIx x= 2 22 2 2 20 0si n 2si n cos 2l i m l i msi n 2 si n 2 si n cosx xx x x x xx x x x x x x = =+ 2 20si n 2 2l i m2 si n si n 2xx xx x x x=+ 2 2022 22 os2 2l i m2si n 4 si n 2 2 cos22si nl i msi n 2 si n 2 cos2xc xIx x x x xxx x x x x=+ += + + 108108 202 22 1l i m3 si n 2 cos21 2si n si nxx x xxx x= = + +. V d 7:l n 00 0) l i m l i m 1,x x xx xa x e e+ + = = =1 l n1 11 1) l i m l i m .xx xx xb x e =Ta thy 1 11l nl i m l i m 11 1x xxxx = =, nn 111l i mxxx e= . Cui cng ta hy xt tng ca cc hm, vl ogx maa x xV d 8: Cho a>1 v nu m l s tu , th l i mxmxax+= +(4.7.12) Do2 12 1l n l n l n l n1 ...1! 2! ! ( 1)!n x nx n na a a a aa x x x xn n ++= + + + + ++. By gi ta hy chn s t nhin n sao cho n>m. Vi x>0, t (4.7.13) ta thyl n,!nx naa xn>hay l n!x nn mma axn x> . T y suy ral i mxmxax+= + . V d 9: cho a>1 v m>0, l ogl i m 0amxxx+= (4.7.14) Thtvy,tl og . l og ,m y ma ax a y x m x = = = khix +thy +.Do l og 1,am yx ym x a=theo v d trn l og 1l i m l i m 0am yx yx ym x a+ += = , iu phi chng minh. T cc v d trn ta thy khix +, hm ax vi a>1 tng nhanh hn bt c hm lu tha nocax.Khix +,hml og , 1a x a > tngchmhnbtkhmluthaxmvism dng. 4.8Kho st hm s 4.8.1 Kho st ng cong cho di dng phng trnh hin Xt hm s( ), ( , ) y f x x a b = . (4.8.1) 109109 trng ph thng, kho st s bin thin ca hm s ta thng tm cc i, cc tiu ca hm s theo qui tc I v quy tc II v tm im un ca th. V d 1: Ta hy xt hm s 2( )1xf xx=+. Min xc nh ca hm s l( , ) Df = + . Ta c 2 22 2 2 31 2 ( 3)( ) , ( )( 1) ( 1)x x xf x f xx x = =+ +. Theoducaohmf(x)tathyrnghmstngtrongkhong(1,1),gimtrong cc khong ( ,1 ) v (1, +). Do ti im 1 hm s t cc tiu, ti im 1 hm s t cc i: 1 1( 1) , (1)2 2f f = = . ohmcphaimtrongkhong( , 3 )vtrongkhong(0,3 ),dngtrong khong ( 3 ,0) v ( 3 ,+). Do nhng im (33,4 ); (0,0); ( 3 ,34) l nhng im un ca th. Ngoi ra ta ch rng 2 2l i m 0 vl i m 01 1x xx xx x+ = =+ +. Sauychngtasgiithiuhainhlkhif(x)cohmcpcaomphnchng minh ca n c gi c th c trong cun [1]. nh l 4.8.1 Gi s trong ln cn ca im x = c hm f(x) c o hm cp n v o hm cp n ti im x = c lin tc. Ngoi ra gi s ( ) ( )( ) 0 v i1v( ) 0.k nf c k n f c = < Khi i) Nu n chn, ( )( ) 0nf c> , th hm f(x) t cc tiu a phng ti x = c ii) Nu n chn, ( )( ) 0nf c< , th hm f(x) t cc i a phng ti x = c iii) Nu n l th hm f(x) khng t cc tr ti x = c. nh l 4.8.2 Gi s trong ln cn im x = d hm f(x) c o hm cp m v o hm cp m ti im x = d lin tc. Ngoi ra gi s ( ) ( )( ) 0 v i1 ,nh- ng( ) 0k mf d k m f d = < . Khi nu m l th (d, f(d)) l im un ca th. Kt hp hai nh l trn ta c iu cn nh sau y: Nu 0 0( ) 0, ( ) 0 f x f x = thhmstcctrtiimx=x0(tcctiunu 0( ) f x>0, cc i nu 0( ) f x

1 0 1 0 ( ) ,( ) . f f = >Cho nn hm s t cc i ti x = 0, cc tiu ti x = 1, im un ti x = 1, 1 2, . Chivihmsnytacthtmcci,cctiutheoquitcIvxtduo hm cp hai tm im un. 4.8.2 ng cong cho di dng tham s Cho h hai phng trnh( ), ( , )( )x x tty y t = =. (4.8.2) Khivimigitr( , ) t hphngtrnh(4.8.2)chotamtimM(x,y)tng ng trong mt phng Oxy v khi t bin thin trong( , ) im M vch nn mt ng cong no trong mt phng, v th ta gi h phng trnh (4.8.2) l h phng trnh tham s ca ng cong, trong t l tham s. V d 3: Phng trnh tham s ca ng thng iqua hai im A(a,c) v B(b,d) l , x mt aty nt b= + = +R (4.8.3) trong m = b a, n = d c. V d 4: Phng trnh tham s ca ellip 2 22 21x ya b+ =l 0 2cosv i[ , ]si nx a tty b t= =.(4.8.4) kho st ng cong cho di dng tham s ta cn thc hin cc bc sau y: 111111a)Tm tp xc nh ca cc hm s x = x(t), y= y(t) b)Xt chiu bin thin ca x, y theo t c)Tm cc ng tim cn: i ) Nu 0( )l i mt tty= v 0( )l i mt ttx a=(hu hn) th ng cong c tim cn ng l x = a. ii) Nu 0( )l i mt ttx= v 0( )l i mt tty b=(hu hn) th ng cong c tim cn ngang l y = b. iii) Nu 0( )l i mt ttx= , 0( )l i mt tty= , v nu 0 0( ) ( )l i m , l i m ( )t t t tt tya b y axx = = th ng cong c tim xin l y = ax+b. V d 5: Kho st v v ng cong axtrit2 2 23 3 30 , x y a a + = > . D thy phng trnh tham s ca ng cong ni trn l 330cos,( , ),si nx a tt ay a t = + >= . (4.8.5) Trcht,tathyngcongykhngctimcn.Hnnax,ylcchmtunhon vi chu k2 nn ta ch cn kho st ng cong cho trong on[0, 2 ] . 2233 0 0 22 233 0 0 22 2( ) cos .si nkhi ; ; ; ;( ) si n . coskhi ; ; ; ; .x t a t t ty t a t t t = = = = = = Cuicngtanhnxtrngivingcongaxtrittac 223tg3si n coscos si ntty dy a t ttdx x a t t= = = . Do 0dydx =ti0 2 ; ; t =v ti cc im ny tip tuyn thng ng (xem hnh 4.8.1). 112112 Hnh 4.8.1 V d 6: Kho st v v ng cong cho bi phng trnh 3 33 0 0 , x y axy a + = > . (4.8.6) Ta thy rng khi thay x bi y v y bi x th phng trnh (4.8.6) khng thay i. Do , th ca n i xng nhau qua ng phn gic th nht. Ta t y = tx v thay vo phng trnh trn ta c 3 3 3 23 0 x x t ax t + = , t y suy ra 23 33 31 1, at atx yt t= =+ + vi1 t . (4.8.7) Ly vi phn ta thu c 33 231 2 1( ) 3 ; ( ) 0 khiv(1 ) 2tx t a x t tt = = =+ 113113333 22( ) 3 ; ( ) 0 khi 2(1 )ty t at y t tt = = =+. Khi1 t th x, y u dn tiv 1 1l i m l i m 1t tya tx = = = 33 31 13 3l i m( ) l i m[ ]1 1t tat atb y axt t = = + =+ + 23 21 11 ( 1)3 l i m 3 l i m1 ( 1)( 1)t tt t ta a at t t t + += = = + + +. Vytimcnxincangcongly=xa.Tccktqutrntacbngbin thin. Cui cng ta hy ch thm rng 33(2 )1 2dy t tdx t= Do 3310 khi 0, 2,khi ,2dy dyt t t tdx dx= = = = = = . Cc tip tuyn ca ng cong ng vi hai gi tr= 0, t =32 tsong song vi trc Ox. Cc tip tuyn ng vi hai gi tr= =31 v 2t tsong song vi trc Oy (xem hnh 4.8.2) 114114 Hnh 4.8.2 Hnh 4.8.3 4.8.3 Kho st ng cong trong ta cc a) H ta cc Trongmtphngchn mtimOcnhvtiaOxiquaimO.TagiimOl cc, tia Ox gi l trc cc. H ta xc nh bi cc v trc cc gi l h ta cc. Gi ,OPl vc t n v nm trn tia Ox. V tr ca im M trong mt phng c xc nh bi vc t ,OM , ngha l xc nh bi gc =,,( , ) OP OMv=,| | r OM ,c gi l gc cc, r c gi l bn knh cc. Gc gi l gc nh hng, ly gi tr dng nu chiu quay ,OPn trng vi ,OMngc chiu kim ng h v ly gi tr m nu ngc li. Cp s c th t ( , ) rgi l ta ca im M trong mt phng. b) Mi lin h gia ta Descartes vung gc v ta cc By gi ta ly trc honh trng vi trc cc v trc tung ng vi tia =2 ta c h ta Descartes vung gc. Gi (x,y) v ( , ) rln lt l ta ca im M ni trn trong h ta Descartes vung gc v h ta cc. Khi , ta c cosv i0 2 , 0si nx rry r = < = Chohms( ) r f = .Trckhikhosthmstacnhnxtsau.Gischoim M(x,y)nmtrnth.Gi lgcdnggiavctOM,vvctchphngcatip tuyn vi th ti im M. Gi l gc dng gia trc cc v tip tuyn, ta c. = =+t g t g,t g1 t g t g. (4.8.8) Mt khc theo ngha hnh hc ca o hm 115115' si n cost g' cos si ndy r rdx r r += =, trong 'drrd= (4.8.9) Thay vo (4.8.9) vo (4.8.8) ta c Hnh 4.8.4 ' si n cos si n' cos si n cost g' si n cos si n1' cos si n cosr rr rr rr r +=++ Sau cc php bin i n gint g'rr = .(4.8.10) V d 7: Hy v ng xon c lgarit c phng trnh: , 0, 0br ae a b= > > . (4.8.11) Tathyhmsrxcnhvimi .Khi tngrcng tng,khi0 = thr =a,khi +thr +, khi th0 r v khi ng cong qun v hn quanh cc O; O c gi l im tim cn ca ng cong. Theo cng thc (4.8.9) ta c 1t g'rr a = = , Do vc t ch phng ca tip tuyn vi ng cong lun lun to viOM, mt gc khng i (xem hnh 4.8.5). 116116 Hnh 4.8.5 V d 8: Hy v ng hoa hng ba cnh c phng trnhsi n 3 , 0 r a a = > . y r l mt hm tun hon vi chu k 23 v th ch cn kho st hm s trong mt on c di bng chu k, chng hn on [0, 23]. Ta c' 3 cos3 ; ' 0 khi ,,6 2r a r = = = =1t g t g3' 3rr = = . th ng vi khong [0, 23] gm hai cnh, sau o cho th quay cc gc quanh cc ta s c ton b th (xem hnh 4.8.6) Hnh 4.8.6 117117

4.9 Bi tp chng 4 4.1Chohm 1( )1xxf xe=+vi0 x vf(0)=0.Chngminhrnghmf(x)lintcti mi im, nhng(0) 0, f+ = (0) 1 f = . 4.2 Chngminhrng hm 1( ) si n f x xx=vi0 x , v f(0)=0 lin tc ti x = 0, nhng khng c o hm bn tri v bn phi ti x = 0. 4.3Chng minh rng hm 2nu l hu t ( )0 nu v t x xf xx= ch c o hm ti x=0. 4.4Da vo nh ngha hy tnh f(a) nu ( ) ( ) ( ), f x x a x = trong ( ) x lin tc ti x=a. 4.5Cho hm y=sgnx c nh ngha sau 1 nu0,sgn 0 nu0,1 nu0.xx xx < = => Chng minh rng |x|=xsgnx 4.6Tnh o hm ca cc hm s sau 1) y=|x| 2) y=x|x| 3) y=ln|x| vi0 x . 4.7Tnh o hm cc hm sau 1)y x x x = + +2) 2 3| ( 1) ( 1) | y x x = + . 4.8Tnh y nu 1) y=f(x2) 2) y=f(sin2x)+ f(cos2x) 3) y=f(f(f(x))), trong f(x) l hm kh vi. 1181184.9Chng minh rng hm 21si nnu0( )0nu0x xf x xx= = c o hm gin on. 4.10Tnh o hm cc hm sau 1) y=xx 2) y=xslnx, trong s l hng s. 4.11Tnh o hm yx nu hm s y c cho di dng 2 2 2 2cos , si nt tx e t y e t = = . 4.12Xc nh min tn ti hm ngc x = x(y) v tm o hm ca n nu1) y=x+lnx (x>0)2) y=x+ex 4.13avdngF(x,y)=0(hayy=f(x))phngtrnhccngcongchodidng tham s. 1) x = acost, y = bsint2) x = acos3t, y = asin3t 3),2 2t t t te e e ex y + = = 4) x = tgt, y = cos2t. 4.14Cho hm f(x)=x32x+1. Hy xc nh(1) f v(1) dfnu0,1 x = . 4.15Tm vi phn ca hm 1)( ar ct g )a xdx a+ , a l hng s 2)(1 cos ) d u 3)( )btd bt e , b l hng s. 4.16Hy tnh 1) 3 6 92( 2 )( )dx x xdx 2) 2si n( )( )d xx dx 3) (t g )(cot g )d xd x 4.17Hy tm hm( , ) x x = sao cho( ) ( ) f x x f x + == '( ), xf x x + (0 1) < . Chng minh rng1 21 2 1 2

1( ) ( )( ) ( )x xf ff x f x x x = , trong 1 2x x < < . 4.21Cho hm f(x) lin tc trn [a,b] v c o hm ti mi im( , ) x a b . Chng minh rng bng cch p dng nh l Rolle i vi hm ( ) 1( ) ( ) 1( ) 1x f xx b f ba f a =ta s thu c nh l Lagrange 4.22Cho: f R R chng minh rng nu ( 1)(0) '(0) (0) ... (0) 0nf f f f = = = = =th ( )( ) ( )!nnf x f xn x=trong 0 1 < < . 4.23Tm o hm cp 2 ca cc hm sau 1)cosxy e x =2) 3 xy ax =3) 2si n y x x = . 4.24Tm o hm cp 3 ca cc hm sau 1)si nxy e x=2) 2l n y x x =3)cos y x x = . 4.25Cho( )xaf x xe = . Tm (4) ( ) ( )( ), ( ), (0)n nf x f x f4.26Cho( )1xf xx=+. Chng minh rng vi2 n ( ) 111.3.5...(2 3)(0) ( 1)2n nnnf n= . 1201204.27Cho 21( )1f xx=. Chng minh rng ( )!khi2(0)0 khi 2 1.nn n mfn m= = = 4.28Cho hm s 2( ) , 0xaf x xe a= . Chng minh rng( )2( 1)( 1)(0)nnnn nfa = . 4.291) Cho *( ) ,nf x x n N = . Chng minh rng( )(1) (1) (1)(1) ... 21! 2! !nnf f ffn + + + + =2) Cho( si n ), (1 cos ) x a t t y a t = = . Tnh 22dydx 3) Cho= =2 3, 3t tx e y . Tnh 22dydx. 4.30Cho a thc Legendre: 2 ( )1( ) [( 1) ]( 0,1,2...)2 . !m mmmP x x mm= =Chng minh rng ( )mP xtha mn phng trnh 2(1 ) ( ) 2 ( ) ( 1) ( ) 0.m m mx P x xP x m m P x + + =4.31Cho a thc Lague: ( )( ) ( )x m x mmL x e x e=tha mn phng trnh( ) (1 ) ( ) ( ) 0m m mxL x x L x mL x + + =4.32Chng minh cng thc: 11( l n ) !(l n )n nnnkdx x n xk dx== + vi x>0. 4.33Khai trin hm 22( )x xf x e=n s hng cha x5. 4.34Khai trin hm( )1xxf xe= n s hng cha x4.4.35Tm 3 s hng u tin ca khai trin Taylor trong ln cn ca im x=0. 1) (sinx)2 2)cos x . 1211214.36Cho n s 1 2, ,...,na a a . Xc nh x sao cho hm s. 21( ) ( )nkkx a x == c gi tr b nht. 4.37 Tm3shngcakhaitrinhm( ) f x x = theocclythanguyndngca hiu x1. 4.38Khai trin hm f(x)=xx1 theo cc ly tha nguyn dng ca nh thc (x1) n s hng cha (x1)3. 4.39p dng qui tc Lhospital tm cc gii hn sau: 1) 11 1l i m( )l n 1xx x2) 210si nl i m( )xxxx 3) 202 l n(1 2 )l i m2xx xx +4) 2201 2l i m2xxe xx . 4.40Hy tm gii hn sau, xt xem c th p dng qui tc Lhospital hay khng 1) 201si nl i msi nxxxx2) si nl i msi nxx xx x+ 3) si n1 si n cosl i m( si n cos )xxx x xx x x e+ ++. 4.41Tm cc gii hn sau: 1) 01 1l i m( )1xxx e2) 0l n(si n )l i ml n(si n )xaxbx 3) 2240l i msi nxxe xx4) 0l i m(si n )tgxxx 5) 1210l i m(1 )xe xxx + 6) 2cos2l i m( )xxtgx. 4.42Chng minh rng khi0 x ta c 1) 3ar ct g3xx x 2)~ l nx xaa b xb3) 2 21 2 ~ 2xe x x 4) 22 l n(1 2 ) ~ 2 x x x + . 4.43Cho hm 61( ) si n g x xx=vi0; (0) 0 x g =v 6( ) 2 ( ). f x x g x = +Xt cc tr a phng ca cc hm g, f ti im x=0. 4.44Kho st v v th hm s 22( 1)1xyx=+. 1221224.45Kho st v v th hm s l n.xyx=4.46Kho st v v th hm s 2 xy x e= . 4.47Kho st v v th hm s 2 21 1 y x x = + + . 4.48Kho st v v th hm s 1 2cos r = + . 123123 Chng 5 Tch phn khng xc nh 5.1Tch phn khng xc nh Trong nhiu vn ca khoa hc v k thut, ta cn tm hm s khi bit o hm ca n. V d nh bit gia tc a=a(t) ca chuyn ng, ta cn tm vn tc v=v(t) ca chuyn ng bit rng dvadt= . Sau khi bit vn tc v, ta cn tm qung ng s=s(t) ca chuyn ng, bit rng dsvdt= . 5.1.1nh ngha nguyn hm Hm F(x) c gi l nguyn hm ca hm f(x) trn tp D, nu c hai hm cng c xc nh trn tp D v( ) ( ), F x f x x D D = R .(5.1.1) 5.1.2 Cc tnh cht Ta c cc tnh cht sau m c th d dng thu c t nh ngha. a)NuFvGlccnguynhmcahmfvhmgtngngtrntpD,thF G trong v l cc hng s, l nguyn hm ca hmf g trn tp D. b) Nu F l nguyn hm ca hm f trn tp D, th hm F+C, trong C l hng s tu cng l nguyn hm ca hm f trn tp D. Ta gi biu thc F(x) + C, trong C l hng s tu , l h nguyn hm ca hm f trn tp D. 5.1.3 nh ngha tch phn khng xc nh HttcccnguynhmcahmftrnmtkhongInocgiltchphn khng xc nh ca hm ny trn khong I v c k hiu l( ) : f x dx ( ) ( ) f x dx Fx C = +. (5.1.2) 5.1.4 Cc tnh cht ca tch phn khng xc nh a)( ) ( ) Af x dx A f x dx = trong A l hng s(5.1.3) 124124b) 1 2 1 2( ( ) ( )) ( ) ( ) f x f x dx f x dx f x dx = . (5.1.4) Vic tm mi nguyn hm ca mt hm s c gi l php ly tch phn ca hm v bi ton ny l bi ton ngc ca php tnh vi phn. 5.1.5 Bng cc tch phn c bn 11)0 , 2)( 1, )1du C dx x Cuudu C R += = += + + 3) l n| |4)u uduu Cue du e C= += + 5) cos si n udu u C = + 6) si n cos udu u C = + 217)cosdu tgu Cu= + 218) cot gsi ndu u Cu= + 9) ch sh udu u C = + 10) sh ch udu u C = + 2111)t hchdu u Cu= + 2112)ct hshdu u Cu= + 2ar csi n 113) ar ccos1u Cduu Cu+ = + 2ar ct g du14) ar ccot g . 1u Cu C u+ = + + thun tin cho vic p dng, ta b sung vo bng trn cc cng thc sau: 4 )( 0, 1)l nuuaa a du C a aa= + > 2 2ar csi n13 ) ar ccos+= +uCduaaua uCa 1251252 21ar ct gdu14 ) 1ar cot g+= + +uCa aau a uCa a 2 2du 115)l n2a uCa a u a u+= + 22du16)l n . u u a Cu a= + + ++ 5.2 Cch tnh tch phn khng xc nh 5.2.1 Da vo bng cc tch phn c bn V d 1: a)Tnh 21(1 ) I x dx = +. Bi v 2(1 ) 1 2 ,n n x x x + = + + 1213222(1 2 ) 24 12 .32 3 22I x x dx dx xdx xdxx xx C x x x x C= + + = + += + + + = + + + b) Tnh 2 2 2si n cosdxIx x=. Bi v 2 22 2 2 2 2 21 si n cos 1 1si n cos si n cos cos si nx xx x x x x x+= = +nn 2 2 2 2 21 1 1 1cos si n cos si nI dx dx dxx x x x = + = + t g cot g x x C = + . c) Tnh 236si n2xI dx = 33(1 cos ) 3( si n ) I x dx x x C = = + d) Tnh 4 21xxeI e dxx = + 24 211.x xxI e dx e dx x dxxe Cx = + = + = + 1261265.2.2 Tnh tch phn nh php i bin Giscntnhtchphn( ) f x dx.Tahyavobinmi( ) x u = hay( ) u x = , trong cc hm( ) u ,( ) x l cc hm s ngc ca nhau a) Php i bin th nht S dng( ) x u = , khi ( ) dx u du =v nhn c cng thc( ) [ ( )]. ( ) ( ) f x dx f u u du g u du = = (5.2.1)trong ( ) [ ( )]. ( ) g u f u u =b) Php i bin th hai Gi s hm f(x) c vit di dng ( ) [ ( )]. ( ) . f x g x x dx =Khi ( ) [ ( )]. ( ) ( ) f x dx g x x dx g u du = = (5.2.2) trong ( ) u x = . Nu( ) ( ) ,t h g u du G u C = + ( ) ( ) ( ) f x dx g u du G u C = = + (5.2.3) V d 2:1221ar csi na) Tnh11ar csi n (ar csi n ) (ar csi n ) .2xI dxxI xd x x C== = + 22) Tnh t gsi n (cos ) (cos )cos cos cosl n| cos | .b I xdxx x d xI dx dxx x xx C== = = = + 3 23 2) ( 2 2)ar ct g( 1)[( 1) 1]ar ct g( 1)dxc Ix x xdxIx x=+ + +=+ + + 3 2 t1, ,t a c(ar ct g )l n| ar ct g | .ar ct g (1 )ar ct gu x du dxdu d uI u Cu u u= + == = = ++ 22 422) 1xxxedxd Ie=+ 12712722222424 2 2Do,tt a c1ar ct g1 ( ) 1 ( )xxu uxu uedxI u xee du deI e Ce e= =+= = = ++ +

105)(1 ) eI x dx = + 2 t1+ ( -1) , 2( 1) . x = u,x =u dx u du = Khi ( )1011 1052 ( 1) 2 I u u du u du u du = =

12 11212 11u uI C = + 111(11 12)66u u C = + 111(1 ) (11 1) .66x x C = + +5.2.3 Phng php tnh tch phn tng phn Theo qui tc ly o hm mt tch: ( ) d uv udv vdu = + . Ly tch phn c hai v ta c . uv udv vdu = + T y ta c cng thc sau udv uv vdu = . (5.2.4) Cng thc (5.2.4) gi l cng thc tch phn tng phn. V d 3: a) Tnh tch phn 3 21l n I x xdx = t 2l n u x =12l n du x dxx=3dv x dx =44xv =4 4 42 21124 4 4l n l n l ndx x xI x x x dxx= = 42 314 2l n l nxx x xdx = 4 4214 2 4l n l nx dxx x = 1281284 4 4211 14 2 4 4l n l nx x xI x x dxx = 42 4 41 14 8 32l n l nxx x x x C = + + . b) Tnh 2ar csi n I xdx = t u = arcsinx, dv = dx, ta c 211du dxx=v x =221ar csi nxI x x dxx= 221 ar csi n I x x x C = + + . c) Tng t 231ar ct g ar ct g l n(1 )2I xdx x x x C = = + + d)24I x bdx = + t u = 2x b + , dv = dx v = x. Ta c 24222222xI x x b x dxx bx b bx x b dxx b= + ++ = + + 2242 22 2 2l n( ) .2x b dxI x x b dx bx b x bbx x b x bdx x x b C+= + ++ += + + + + + + Suy ra2 2 2l n( ) .2 2x bx bdx x b x x b C + = + + + + + V d 4: a) TnhcosaxI e bxdx = t u = cosbx, du = bsinx dv = eaxdx, 1axv ea=cos cos si nax axaxe e bI bxd bx e bxdxa a a= = + (5.2.5) Mt khc 129129si n si naxaxee bxdx bxda= si n cosaxaxe bbx e bxdxa a= (5.2.6) Thay (5.2.6) vo (2.2.5), sau mt vi php bin i n gin, ta c 2 2cos si ncosax axa bx b bxe bxdx e Ca b+= ++(5.2.7) b) Tng t 2 2si n cossi nax axa bx b bxe bxdx e Ca b= ++.(5.2.8) 5.2.4 Cng thc truy hi a) Xt tch phn cosnnI xdx = vi *n N . Ta c1 1cos cos cos cos si nn n nxdx x xdx xd x = =

1 2 21 211 1cos si n ( ) cos si ncos si n ( ) cos ( ) cos .n nn n nx x n x xdxx x n xdx n xdx = + = + T y chng ta nhn c cng thc truy hi 1 21 1 ( )cos cos si n cosn n nnxdx x x xdxn n = + . (5.2.9) Cng thc ny cho php gim s m ca lu tha ca hm di du tch phn mi ln 2 n v cho n khi ta nhn c tch phn tu theo n l l hay chn: hay cos si n . xdx x C dx x C = + = + Tng t ta nhn c cng thc truy hi 1 21 1si n si n cos si nn n nnxdx x x xdxn n = + . (5.2.10) b) Xt tch phn 2 2( )n ndxJx a=+ vi0*, n N a Ta bit 1 2 21arctgdx xJ Ca a x a= = ++. xy dng cng thc truy hi cho tch phn Jn, ta hy xt 2 2 11 2 2 12 22 2 11 2( )( )( )( )( )nn nnndxJ x a dxx axx n x a xdxx a + = = ++= + ++ hay21 2 2 1 2 22 1 ( )( ) ( )n n nx xJ n dxx a x a = + + + 1301302 2 21 2 2 1 2 221 1 2 2 12 12 1( )( ) ( )( )[ ]( )n n nn n n nx x a aJ n dxx a x axJ n J a Jx a + = + + += + + Suy ra 21 2 2 12 1 2 3 ( ) ( )( )n n nxn a J n Jx a = + + hay1 2 2 1 2 22 32 1 2 1( )( )( ) ( )n n nx nJ Jn x a a n a = + + (5.2.11) 5.3 Tch phn cc phn thc hu t 5.3.1Tch phn cc phn thc hu t n gin nht Cc phn thc hu t n gin nht l cc phn thc c dng 2 2I ) , I I ) , I I I ) , I V)( ) ( )k kA A Mx N Mx Nx a x a x px q x px q+ + + + + + trongA,M,N,p,qlccsthc,k=2,3,4,cntamthcbchaikhngcnghim thc, tc l p2 4q < 0 . By gi ta hy kho st tch phn cc phn thc hu t trn. a) Tch phn dng I l n| |Adx A x a Cx a= +. b) Tch phn dng II 1111( )( ) ( )k kA Adx C kk x a x a= + . c) Tch phn dng III 2 222 2( ) ( )M Mpx p NMx Ndx dxx px q x px q+ + +=+ + + + 2 222 2( )M x p Mp dxdx Nx px q x px q+= + + + + + . (5.3.3) Ta hy xt tch phn th hai v phi (5.3.3). t 222 4, , ,p px t q dx dt + = = =2 2 2 222 4( )dx dx dtx xp q p p tx q= =+ + ++ + 22 21 2 2arctg arctg4 4.dx t x pC Cx xp qq p q p += + = ++ + 131131Cui cng ta c 2++ +Mx Ndxx px q22 22 2arctg24 4l n( ) . += + + + + M N Mp x px px q Cq p q p (5.3.4) d) Tch phn dng IV 2 222 2( ) ( )( ) ( )k kM Mpx p NMx NI dx dxx px q x px q+ + += =+ + + + (5.3.5) Tchphnvphitrong(5.3.5)ctchthnhhaitchphn.tnhtchphnth nht ta t 2x px q t + + = 1 22,( )+= =+ +k kx pJ dxx px q

1 2 11 11 1 ( ) ( )( )k k kdtt k t k x px q = = = + +(5.3.6) Cn tch phn th hai, k hiu bng 2,kJ , ta c 2 2 2 2 ,( ) ( )k k kdx dtJx px q t = =+ + + trong 2,pt x = +224pq = (5.3.7) V d 1: Tnh 26 54 9xI dxx x+=+ + Gii: Ta c 2 24 9 2 5 ( ) . x x x + + = + +t x+2 = t, x = t2, dx = dt, ta nhn c

2 26 5 6 2 52 5 5( )( )x tI dx dtx t+ += = =+ + + 2 2 226 7 23 75 5 573 5 arctg5 5l n( ) .t t dtdt dtt t ttt C= = =+ + += + + Tr v bin x, ta nhn c 226 5 7 23 4 9 arctg2 55 5l n( )( )x xdx x x Cx+ += + + ++ + V d 2: Tnh 2 21( 1)xI dxx x+=+ + Gii: 2 2 2 2 2 21 1(2 1)1 (2 1) 1 12 22 2 ( 1) ( 1) ( 1)xxI dx dx dxx x x x x x+ ++= = ++ + + + + + 132132Ta c 1 1 2 2 22 2 22 2(2 1) 1( 1) 111 3( 1)[( ) ]2 4xJ dx Cx x x xdxJ dxx xx+= = ++ + + += =+ ++ + t 1,2t x dt dx = + =ta