Thiết kế bài giảng đại số và giải tích 11 nâng cao (tập 2) phần 2

B GlCfl HAN CUA HAM SO HAM SO LIEN TUC §4 D i n h n g h i a v a m o t s o d i n h l i v e g i d i h a n c u a h a m so' ( t i e t 7, 8, ) I MUC TIEU Kie'n thurc HS ndm dugc : • Dinh nghTa gidi ban ciia ham so Djnh If ve gidi han hiiu ban KT nang Sau hgc xong bai HS cdn giai thao cac dang toan ve gidi ban ciia ham so Van dung td't cac quy tac tim gidi ban cua ham sdi Thai • • Tu giac, tfch cue hgc tap Bie't phan biet rd cac khai niem ca ban va van dung tirng trudng hgp cu the - Tu cac van dd ciia toan hgc mdt each Idgic va he thd'ng II CHUAN BI CUA GV VA HS Chuan bi ciia GV • Chuan bi cac cdu hdi ggi md • Chudn bj phdn mau va mdt sd dung khac Chuan bj cua HS • Cdn dn lai mdt sd kie'n thiic da hgc ve gidi ban day sd III PHAN PHOI T H I L U O N G Bai chia lam tiet : Tii't : Tic ddu din hit phdn 138 Tii't : Tii'p theo di'n hi't dinh li Tii't : Phdn lgi vd bdi tap IV TIEN TRINH DAY HOC A OAT VAN DE Cau hdi Tim gidi han ciia cac day so sau ddy n-3" 1-3" a) Iim b) Iim 2" +'3" n + 3" Cau hoi Tfnh cac tdng sau a) Sn = l + - + - + " b ) S n = l - - + - + " B BAI Mdl HOATDONCl Gidi han cua ham so tai mgt didm • GV neu bai toan: GV treo bang X jr, = 10 fix) Axi) =6 H =9 A:^ M) M) ãô4=3 ^=1,9 /(JC4) M) ^2 > Sau dd GV dua cac cau hdi sau Hoat dgng ciia GV Cau hoi Hoat ddng ciia HS Ggi y tra ldi cau hdi Xac dinh f(Xn) /(^«) = ^ T ^ = 2(^n + 2) vdi mgi n 13< Cau hoi Ggi y tra ldi cau hdi Tirn limf(Xn) Iim/(Xn) = lim2(j:n + 2) = 2(IimjCn + 2) = 2(2 + 2) = GV neu djnh nghTa : Gid sic (a ; b) la mpt khodng chiia diem XQ vdf Id mpt hdm so xdc dinh tren tap hgp (a ; b)\{xQ} Ta ndi rdng hdm so fed gidi hgn Id sdtlncc L X ddn din XQ (hodc tgi diem XQ) ni'u vdi mgi ddy sd(x^) tap ligp (a ; b) \{XQ} (ticc la Xj^ e(a ; b) vd x^ ^XQ vdi mgi n) md limx^ = XQ, ta diu cd limf(x^) = L Khi dd ta vie't lim fix) = L hoac/(AT) -> L JC -> XQ Thuc hien vf du phiit Hoat ddng cua HS Hoat ddng cua GV Ggi y tra Idi cau hdi Cau hdi Vdi mgi day so (JC„) ma /(x„) = x„cos—,• JC,, ^ vdi mgi n hay xac dinh f(Xn) Cau hdi Vdi lim Xji = hay xac dinh Ggi y tra Idi cau hoi |/(^H)| = k l c o s — X ^w va lim|x„| = Iimf(Xn)' ndn Iim/(j:„) = Do dd Iim fix) = lim jccos— = ;c-*0 GV dua mdt so cau hdi ciing cd : 140 x-^0\ XJ HI Neu mdt VI du khac ve vf du ham sd H2 Ham sd khdng xac djnh tai a nhung cd gidi ban tai a Diing hay sai? • GV dua nhan xet: lim X = XQ ; lim c = c, vdi c Id hang so H3 Tim gidi ban ham sd sau bdng djnh nghTa: 2x + l f(x) = — X ddn de'n X +X + • Thuc hien [HIJ 5' Hoat ddng cua HS Hoat ddng ciia GV Ggi y tra Idi cau hdi Cau hdi Hay rut ggn f(x).' -'^ X^+3A: + /W=- ix + l)ix + 2) ; x+l x+l Ggi y tra ldi cau hdi Cau hdi Hay xac djnh Iimf(Xn) ; = = x+2 lim/(;cn) = lim (Xn + 2) = - + = i , J;^ + 3X + Vay hm : = Jf^-l x +l GV ndu nhdn xet: a) Ni'u fix) = c vdi mgi x e R, dd c Id mdt hdng sd, thi vdi mgi XQ e R, lim fix) - lim c = c b) Ni'u g(x) = X vdi mgi x e R thi vdi mgi XQ e R, lim gix) = Iim A: = XQ h) Gi&i hgn vo cixc IA: H4 Gidi ban vo cue ciia ham sd tai mdt diem dugc djnh nghTa tuong tu nhu gidi ban hiiu ban cua ham sd tai mgt di^in hay phat bieu djnh nghTa dd • GV ndu va hudng ddn HS thuc hien vf du HOATDONC 2 Gidi han cua ham sd tai vd cue • GV neu djnh nghTa 2: • Gid sic ham sdfxdc dinh tren khodng (a ; +ao) Ta ndi rdng hdm sdf cd gidi hgn Id sdthuc L x ddn din +oo ne'u vdi mgi ddy sd(x^) khodng (a ; +cc) (ticc ldXn> a vdi mgi n) md limx^y = +oo, ta diu cd limfix^) = L Khi dd ta vii't lim fix) = L liodc fix) -> L x -^ +oo, • Cdc giai hgn lim fix) = + 00, lim fix) = - co, lim fix) = L, lim fix) = + 00 vd lim fix) = - oo dicgc dinh ngliia tuang tie • GV neu vd hudng ddn HS thuc hien vf du • GV neu nhan xet: Ap dung dinh ngliia gidi hgn ciia hdm sd,cd the chieng minh dicgc rdng : Vdi mgi sd nguyin dicaiig k, ta cd ; ' a) hm Jf = + 00 ,• v^+oo , b) hm v->-+QO_f • GV neu dinh If 1: t X Gid sic Iim fix) = L vd lim ^(Jf) = M (L, M e R) Khi dd a) lim [fix) + gix)] =L+ M:, b) lim [fix)-gix)] =L-M; v->jr„ c) lim [fix)gix)] ^LM; Ddc biet, ni'u c Id mdt hdng sdthi ,' lim [C/(JC)] = cL ; x^.x^^, d) Ne'u M^O thi Iim 4 = ^ • A-*.r„ gix) M H5 Hay phat bieu bang ldi djnh If trdn • GV neu nhan xet: Ni'u k la mdt sd nguyen duang vd a Id mpt hang sdthi vai mgi XQ e R, ta cd k k k lim ax = Um a lim x Um x lim x =ai lim x) = OA^ • Ar-> v,) •f->-*'i) 'f^-^'o -A'() k thita so •>•—>- vdi mgi x e A (XQ) dd J Id mpt khodng ndo dd chica XQ, thiL>Ovd lim V / U ) = 4L x->x^•| GV neu va hudng ddn HS thuc hien vf du • Thuc hien |H4| 5' H o a t ddng cua H S H o a t ddng cua GV Ggi y t r a Idi cau hdi Cau hdi lim(A'^+7x) = - Tim lim (x + Ix) Ggi y t r a ldi cau hdi Cau hdi lim x^+lx Tfnh lim x" +lx va lim 4x^ + Ix x->-\ =8 x-^-\ va lim ylx^+7x = 4^ = -2 x^-l HOATDONC TOM TfiT Bfil HOC Gia sii (a ; b) la mdt khoang chiia diem XQ v a / l a mdt ham sd xac djnh tren tap hgp (a ; ^)\(xo} Ta ndi rang ham s d / c d gidi ban la so thUe L x dan den XQ (hoac tai diem XQ) neu vdi mgi day sd (Xn) tap hgp ia;b)\ {XQ} (tiic la Xn e(a ; 6) ya Xn ^ XQ vdi mgi n) ma limXn = Xg, ta deu cd lirri/(Xn) = L Khi dd ta viet lim fix) = L hoac/(x) -^ L x -> XQ a) Ne'u/(x) = c vdi mgi x G M, c la mdt hdng sd, thi vdi mgi XQ G R , 10-TKBGDSVGTIINCT2 145 lim fix) = lim r = c •V >.v„ V ->.v„ b) Neu g{.\) F A N'di moi v M ihi vdi mgi xg e R, lim ^v(-v) = lim x = XQ ,V >.V„ \ ->.V|| • Gia su' ham so/'xac djnh tren khoang (c/ ; '+00) Ta ndi vdng ham ,sd/c6 gidi han la sd ihuc / x dan den +co neu vdi mgi day sd (x^) Irong khoang (c/ ; +00) (tiic la Xn > a vdi moi 11) ma lim.Vn = +co la deu cd !im/(Xn) = L Khi dd la vicl iim /(x) =Lhoac / ( x ) -^ L x —> +QO .\ • > * ' ' •' • • Cac gioi ban lim /'(x) = + a;, lim /'(x) = - c o , • > A„ =L-M: • c) lim [/(.vX!>(x)] =LiV/; V >.V|, Dac bicl, neu r la mdl hdng sd ihi lim [ r / ( x ) ] = cL ; X >.1„ d) Neu M ^ thi 146 lim ^ ^ = — • V > r„ gi.x) M' .Neu k la mol sd nguyen duong \'a a la mol hdng so thi vdi mgi-xg J R, la cd lim ax^ = lim c/ lim v lim A; lim x =ô( lim x)^' = C/.VQ ãi->.f|| v ->.r|, v >.V|| ,-> >.vji ' v->-Vn •>'->-»(i /: thfra s o Gia sir Um /(x) = L.Khidd v->.v„ a) lim |/(x)| = |/.| ; b) lim 4M = 4L ; c) Nc'u/(x) > vdi mgi X e A {xol, dd / la mol khoang nao chUa XQ Ihi L > va lim J/Cv) = 41 HOATDONC MQT SO Cfia HOI TR^C NQHIEM ON TfiP Bfil X x+ ,bang: > x (b>- (a)l; 1 2x + l bang: x2-2 (a) (b)2; (c) I ; (d) Trd ldi (c) 147 Ggi y tra ldi cau hdi Cau hdi Giai phuang trinh da cho X = k2TC b) Boat ddng ciia GV Boat ddng cua BS Ggi y tra Idi cau hdi Cau hoi Chflng minh: HS tu chflng minh 1-cos^x tan x= r— 1-sin X Ggi y tra Idi cau hdi Cau hoi Giai phuong trinh da cho X = 71 + ^271, X = — + kn Bai Hudng ddn Dua vao cac quy tdc de'm, to hgp, chinh hgp va hoan vi cua mdt tap hgp a) Boat dgng cua GV CSu hoi Mdi each xep ngudi vao tga mdi toa mdt ngudi la mdt hoan vi, td hgp hay chinh hgp cfla tap hgp hanh khach ? Cau hoi Boat ddng ciia BS Ggi y tra Idi cau hdi Mdi Cach xd'p ngudi vao toa, mdi toa mdt ngudi la mdt hoan vi cfla tdp hgp hanh khach Ggi y tra ldi cau hdi 3! = trUdng hgp Cd bao nhieu trudng hgp a) Boat dgng cua GV Om hoi Boat ddng cua BS Ggi y tra ldi can hoi Cd bao nhidu each chgn hai hanh Cd C3 = each chgn khach di chung toa ? 335 Cau hdi Ggi y tra Idi cau hdi Vdi mdi each dy Iai cd bao nhieu Vdi mdi each ay lai cd each chgn each chgn toa tau cho hg ? toa tau chohg cau hdi Ggi y tra Idi cau hdi C d b a o nhieu each chgn hai Cd 3 = each chgn hai hanh khach hanh khdch va toa tau cho hg di va toa tau cho hg di chung chung.? Cau hdi Cd bao nhieu each chgn? Ggi y tra Idi cau hdi Mdi each d'y, hanh khach thfl ba cd the chgn mdt hai toa tau cdn Iai Ap dung quy tdc nhan, ta cd 9.2 = 18 trudng hgp cd the xay Bai Hudng ddn Dua vao cac quy tac ddm, td hc^, chinh hgp va hoan vi cua mdt tap hgp Boat ddng ciia GV Boat ddng cua BS Cau hdi Ggi y tra Idi c^u hdi Vdi hai phdn tfl x va y cua A cho X > y; ta lap dugc bao nhidu cap (x, y) thoa man de bai? Cau hoi Vdi hai phdn tfl x va y cua A cho Cd bao nhieu each chgn? X > y, ta chi ldp dugc mdt cap duy, nhdt (x, y) thoa man de bai Ggi y tra Idi cau hdi Do dd mdi cap nhu vdy cd the xem la mdt td hgp chdp cua n phdn tfl Bai Hudng ddn Dua vao cac quy tdc de'm, td hc^, chinh hgp va hoan vi cua mdt tap hgp Cac quy taC tinh xdc, sudt a) GV nen ggi mdi HS thuc hien mdt y Boat ddng cua GV Cau hdi 356 Boat ddng cua BS Ggi y tra Idi cau hdi Hay tfnh sd trudng hgp cd the Sd trudng hgp cd the laC^g cau hdi Goi y tra Idi cdu hdi Sd trudng hgp rflt dugc ca hai Sd trudng hgp rflt dugc ca hai vien bi vidn bi den labao nhieu ? den la C^ cau hdi ' Tfnh xac sudt de lay dugc vieh Ggi y tra ldi cau hdi bi den Do dd xac sudt de rut dugc vien bi den la = r^ A-16 Boat ddng cua GV cau hdi Hay tfnh sd trudng hgp cd the ° Boat ddng cua BS Ggi y tra ldi cau hdi So trudng hgp cd the la Cjg cau hoi Ggi y tra Idi cau hdi Sd trudng hgp rflt dugc vien bi Sd trudng hgp rflt dugc ca hai yidn bi trang, vien bi den Id bag nhieu? Sd trudng hgp rflt dugc vien bi Cau hoi trdng, vien bi den la CJ.Cg = 42 Tfnh xac sudt de Idy dugc vien Ggi y tra ldi cau hdi , bi trang, vien bi den xac sudt rflt dugc vien bi trdng, 42 vien bi den la -^r- = — C2, 20 b) Boat ddng cua GV cau hoi Hay tinh sd trudng hgp cd the Boat ddng ciia BS Ggi y tra ldi cau hdi So trudng hgp cd the la Cjg 357 Cau hdi Ggi y tra Idi cau hdi Sd trudng hgp rut dugc vien bi Sd trudng hgp rflt dugc vidn bi la dd la bao nhieu ? C^=l Ggi y tra ldi cau hoi Cau hdi vay xac sudt rflt dugc vien bi dd Tfnh xac sudt rflt dugc vien bi 1 la dd 560 '16 Ggi y tra ldi cau hdi can hdi Xac sua't rflt dugc vidn bi cd mdu Tfnh xac sudt rflt dugc vidn bi khac la —r- = — cd mau khac 40 -16 Bai 10 Hudng dan Dua vao quy tac tim xac sudt cua bidn ngau nhien a) E(X) = 5,96 b) Diem trung binh ban 48 ldn la 48 5,96 = 286,08 Bai 11 Huang ddn Sfl dung phuong phap quy nap HS tu giai Bai 12 Hudng ddn Sfl dung phuong phap quy nap HStugiai Bai 13 De y rang («„) la mdt cdp sd cdng cd sd hang ddu M, = va cdng sai d = -2, ta dugc a)M„ =1 -2n; b)5,oo =-9400 14 Sfl dung tfnh cha't cua cdp sd cdngva capsoos nhan De y rang (M„) la mdt cdp sd nhan cd sd hang ddu MJ = va cdng bdi ^ = 3, ta dUOC a)//„=2.3"~'; 10 b)5,o = ' " - l Bai 15 Hudng ddn Sfl dung tfnh chdt cua cdp sd cdngva cS'p sd nhan 358 (x;y) = (3; 1), (x;y) = I 13 13, Bai 16 Hudng ddn Six dung tfnh cha't gidi ban cua day so a)l; b)0; c) +00 Ggi y Chia ca tfl va mdu cho n g d) — Gai y.-Chia ca tfl va mdu cho 7" ' • Bai 17 Hudng ddn Sfl dung tfnh chdt gidi ban cua day sd a) +00 ; c) b ) -00 ; Boat ddng ciia BS Boat ddng cua GV cau hdi Xac djnh bilu thflc lien hgp Cau hoi Tfnh gidi ban dd Ggi y tra Idi cau hdi Vn"^ +n^ +l+n^ Ggi y tra Idi cau hdi 2" d) Boat ddng ciia GV cau hoi Xac dinh bieu thflc lidn hgp Cau hdi Boat ddng ciia BS Ggi y tra Idi cau hdi 4n^ +2n.+ n Ggi y tra Idi cau hdi Tfnh gidi ban dd Iim =^ yjn^.+ 2n-n , 4ri^+2n+n hm 2« 35 P-+1 -lim^ " =1 Bai 18 Hicdng ddn Sfl dung tfnh chdt cua cdp sd nhan lui vd ban Boat ddng cua BS Boat ddng cua GV cau hdi Ggi y tra ldi cau hdi Tim he phuong trinh ve md'i quan he gifla cac yeu to qUi 12 UI =_ 15 [l-q Cau hdi Ggi y tra Idi cau hdi I M, = 12 ;q = — hoac M, = ; q=— Xac dinhq, u, Bai 19 Hicdng ddn Sfl dung tfnh chd't cua gidi ban ham sd a) ; b) 10 c)l; d)0; e) Boat dgng cua GV cau hoi Boat dgng cua BS Ggi y tra ldi cau hdi Chflng minh — yj2x^+4x^ +3- HS tu chflng minh X cau hdi Tfnhd gidi ban dd 360 Ggi y tra Idi cau hoi , 42x'^+Ax'^+3 hm ' ;c->- vdi mgi x va chi J3m > 0, |A' = l-3m — ' 363 Bai 23 Hudng ddn Sfl dung cdng thflc tfnh dao ham va cdng thflc nghiem a) Boat ddng cua BS Boat ddng cua GV cau hdi Tfnh dao ham cua ham so c a u hoi Ggi y tra ldi cau hdi y ' = cos 2x + cos x = 2cos x + c o s x - Ggi y tra ldi cau hdi Giai phuang trinh y' = x= TC+ k2TC ; X = ± — - k2tc b) Boat ddng cua BS Boat ddng cua GV c a u hgi Tfnh dao ham cua ham sd c a u hoi Giai phuang trinh y' = Ggi y tra ldi cau hdi y ' = 3cos 3x + 6sin 3x - Ggi y tra Idi cau hoi '2a ,2TC x=k—;x= ,2TC ,.^ _ — + k— vaitana=2 3 Bai 24 Huang dan Six dung cdng thflc tfnh dao ham va y righTa hinh hgc cfla nd a) Boat ddng cua GV c a u hdi Tfnh dao hiimcua ham so c a u hdi Vie't phuOn g trinh tie'p tuye'n (T) tai diem A f a ; 1'^ - 364 Boat ddng cua BS Ggi y tra ldi cau hoi /'w=-4x^ Ggi y tra Idi cau hoi y = — z1 - ( x - a )x+i- hu a y y = — j1x + 2- a\ Cl a^ a b) Boat dgng ciia BS Boat ddng cua GV cau hdi Tim cac giao diem I va J cfla (T) vdi cac true toa dd cau hdi Hay ve (T) Ggi y tra Idi cau hoi '\ / /(2a ; 0), / ; - a) Ggi y tra Idi cau hdi HS tu ve c) Boat ddng cua GV Cau hdi Tfnh dien tfch tam giac GU Boat ddng ciia BS Ggi y tra Idi cau hdi = 1|0/|.|0/| = 12a.2' cau hoi Giai cau c) " ' Ggi y tra ldi cau hdi HS tu giai Bai 25 Hudng dan Sfl dung cdng thflc tfnh dao ham va y nghia hirih hgc cfla nd Boat ddng ciia GV cau hdi Boat ddng cua BS Ggi y tra Idi cau hdi Vie't phuong trinb dudng thang di Phuang trinh dudng thdng id) di qua qua P P(2 ; 0) vdi he sd gdc bang k la y = ^(x-2) cau hdi Ggi y tra Idi cau hdi Tim die'u kien di (d) la tie'p | - x ^ + 17x-66 = /:(x-2), tuye'n \-2x + ll =k cau hdi Ggi y tra Idi cau hoi Hay ke't Iuan 365 Khfl k, ta dugc x^-4x-32 = ' » 'x:j=-4, _X2=8 Vay ngudi quan sat cd the nhin dugc cac di^m M thudc parabol da cho, neu hoanh dd diem M thudc doan [-4; 8] 366 MUC LUC Trang Chuang HI DAY SO CAP SO CONG VA CAP SO NHAN §1 Phuang phap quy nap toan hgc kk'r-f §2 Day sd 18 §3 Ca'p so cdng 41 §4 Ca'p so nhan 56 cau hoi va bai tap on tap chuong III 75 Chuang N 89 Gl(5l HAN A-Gidl HAN CliA DAY SO § Day so cd gidi ban 91 §2 Day sd cd gidi han hiiu ban 102 §3 Day sd cd gidi ban vd cue 120 - Gidl HAN cCiA HAM SO HAM S LIEN TUG §4 Dinh nghia va mdt sd dinh If ve gidi ban cfla ham so 138 §5 Gidi ban mdt ben 154 §6 Mdt vai quy tdc tim gidi han vd cue 172 367 §7 Cac dang vd dinh 187 §8 Ham sd lidn tuc 206 cau hoi va bai tap on tap chuong IV 225 OAO HAM 237 § Khai niem dao ham 239 §2 Cac quy tdc tfnh dao ham 258 §3 Dao ham cfla cac ham so Iugng giac 280 §4 Vi phan 300 §5 Dao ham cap cao 309 Cau hoi va bai tap on tap chuong V 326 cau hdi va bai tap on tap cud'i nam 338 Chuang V Utl^^ \ ^B»^ s6*C.^ m aW :\XN ^\ ^c^-HWô>;>9 ^0ĐS5 Gid:39.000d ... cau hdi x^ +24 i x^ -2 ix + 42) ix^-x 42 +2) '' ix + 42) ix- 42) x^-x 42+ 2 x- 42 Ggi y tra Idi cau hdi Cau hdi ^, ^ unh Hoat ddng ciia HS ^ x^ + 2^ /2 hm — x^S x^ - x-^^ x^ +2^ /2_ x^ -2 '' -24 2 _ 3V2 b) Hoat... X0 Ggi y tra Idi cau hdi ^jx^ + X +2x jr—>^