Đại số và Giải tích 11 nâng cao và hướng dẫn thiết kế bài giảng (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 ... lim X^->1 2x + l bang: x2 -2 (a) (b )2; (c) I ; (d) Trd ldi (c) 147 ,., , Cau r X + V2 Iim -^; bang: - / X^ x-» (a)l (b )2; (c) (d) 42 2V2 ' Trd ldi (c) X-1 Cdu lim-^; x^ix^-1 bang: (a) (b )2; (c) Trd... 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... ->0 X • Thuc hien H2 Irone Hoat ddng ciia GV Cau hdi Ggi y tra Idi cau hdi ChUng minh I-V - >2 yJ2- Hoat ddng cua HS X Dal fix) — — V2-X (x„ ) Irong khoang (-oo, 2) ma lim x„ = 2, ta cd 158 Vdi