数学分析讲义第四版刘玉琏傅沛仁著高等教育出版社课后答案第四单元.pdf

8 111?1 ?SSS1.1(1)?SSS1.2(5) ?SSS1.3(9) 111?444? 13 ?SSS2.1(13)?SSS2.2(21) ?SSS2.3(36)?SSS2.4(40) ?SSS2.5(51) 111nnn?YYY 57 ?SSS3.1(57)?SSS3.2(64) 111ooo?YYY555 75 ?SSS4.1(75)?SSS4.2(83) 111? 95 ?SSS5.1(95)?SSS5.2(100) ?SSS5.3(109)?SSS5.4(113) ?SSS5.5(114) 111888?nnn444?AAA.125 ?SSS6.1(125)?SSS6.2(136) ?SSS6.3(142)?SSS6.4(148) 111? . 170 ?SSS7.1(170)?SSS7.2(172) ?SSS7.3(179)?SSS7.4(187) 111lll? . 194 ?SSS8.2(194)?SSS8.3(204) ?SSS8.4(210)?SSS8.5(232) 111? . 244 ?SSS9.1(?)(244)?SSS9.1(?)(252) ?SSS9.1(nnn)(266)?SSS9.2(?)(273) ?SSS9.2(?)(290)?SSS9.3(298) ?SSS9.4(309) 1 111?.323 ?SSS10.1(323)?SSS10.2(132) ?SSS10.3(334)?SSS10.4(352) 111? . 366 ?SSS11.1(366)?SSS11.2(375) ?SSS11.3(378)?SSS11.4(389) 111?222?CCC? 393 ?SSS12.1(393)?SSS12.2(400) ?SSS12.3(406) 111?nnn? . 425 ?SSS13.1(?)(425)?SSS13.1(?)(430) ?SSS13.2(493) 111?ooo?.465 2 1o?Y5 ? S K4.1 (56?1 136 ?) 1. ?e?8?(.?e(.(XJ?3),?y?: (4) n (1)n+1 ? 1 1 3n ? ? ?n = 1,2, o . ) l*?Jw?: sup n (1)n+1 ? 1 1 3n ? ? ?n = 1,2, o = 1 inf n (1)n+1 ? 1 1 3n ? ? ?n = 1,2, o = 1. fl ?, 1 ?n Nk(1)n+1 ? 1 1 3n ? 1; 2 ?,n0= 2k 1 N,?1 1 2 ? 1 + log3 1 ? ? kn0 log3 1 ? =?. u sup n (1)n+1 ? 1 1 3n ? ? ?n = 1,2, o = 1. 1 ?n Nk1 (1)n+1 ? 1 1 3n ? ; 2 ? 0,n0= 2k N,?(1)2k+1 ? 1 1 32k ? = 1 32k 1 1 2 ? 1 + log3 1 ? ? kn0 log3 1 ? =?. u inf n (1)n+1 ? 1 1 3n ? ? ?n = 1,2, o = 1. (5) x|xkn,x2 0,qkn?5,?3knx0,k2 0( 0,x0: 1 2,kx2 0 0,x0 ? 0, 2 ? (0,2),k1 0x0 A,ka inf B,=b 0,x : b 0,Kr 0,x a,b,kf(x) r. y? A3.2n5(?5). y fi ?f(x)3a,bY,.?n5f(x)3a,b?m, =x0 a,b,x a,b,kf(x) f(x0) = m. fi ?x a,b,kf(x) 0,Kf(x0) = m 0.u,r = m 2 0,x a,b,kf(x) r. y? Ak?CXn. y ?Y?5,y a,b,? f(y) 2 , y,x (y y,y + y) a,b,kf(x) = f(y) 2 . w,mm(y y,y + y)|y a,bCX4ma,b.?k?CXn,?3n?(k? ?)mm: (yi yi,y + yi)|yi a,b,i = 1,2, ,n ?CX4ma,b.x (yi yi,y + yi) a,b,kf(x) = f(yi) 2 ,i = 1,2, ,n.? m = minf(yi)|i = 1,2, ,n 0. l?,y a,b,yk a,b,x (yk yk,y + yk),kf(x) m 2 . u,?r = m 2 0,Kx a,b,kf(x) r. 10. y:E?: 0,x E,kx U(,). y =w,. =?1= 1,x1 Ekx1U(,1).w,x16= , ?2= min n1 2,|x1 | o 0x2 Ekx2U(,2).w,x2?u?x1 ?3= min n1 3,|x2 | o 0x3 Ekx3U(,2).w,x3?u?x1,x2 ?n= min n1 n,|xn1| o 0xn EkxnU(,n).w,xn?u?x1,x2, ,xn1, u,?8E?:8xn|n N,i,j N?i 6= jkxi6= xj. w,xn? ?,?lim xn = .u, 0,K:? +?U(,)?kE?:xn,=E?:. 11. y:eE?k?(.8,?supE = a,?a 6 EK?3?xn E,xn 0,x2,Ekx1 0,x3 Ekx2 0,xn,Ekxn1 0,x0 A,ka 2 0,y0 B,kb 2 0,x0+ y0 A + B,ka + b 0. u,x (0,1,Iak Iai|i = 1,2, ,n,x Iak,k|f(x)| = 1 x3(0,1vkk?IaC X(0,1 15. ef(x)3a,bY,?“:8G = x|f(x) = 0 6= ,KsupG G,inf G G. y “:8G a,b,=GQk?.qke.,? supG = p?inf G = q. yp G?q G,= yf(p) = 0?f(q) = 0. XJGk?,?p = supG,Kf(p) = 0 XJGvk?,KG7?8.fi ?supG = p.k?(.?, = 1,x1 G,kp 1 0,k N,k 0,x1,x2 (1,1),?“ |f(x1) f(x2)| = |x2 1 x 2 2| = |x1+ x2|x1 x2| (|x1| + |x2|)|x1 x2| 0, = 2 0,x1,x2 (1,1) : |x1 x2| 0, 0,n, n + 1 R : |n + 1 n| =1 n + 1 +n 1 42 ? ,k |f(n + 1) f(n)| = n + 1 n = 1 1 2 = 0, =f(x) = x23R?Y. (3) f(x) = x31,+)?Y y 0,x1,x2 1,+),?“ |f(x1) f(x2)| = |x1 x 2| = |x1 x2| x 1+ x 2 1 2|x1 x2| 0, = 2? 0,x1,x2 1,+) : |x1 x2| 0, 1 0,x1,x2 I : |x1 x2| 0,x1,x2 I : |x1 x2| 0, = min1,2 0,x1,x2 I : |x1 x2| 0, 0,x1,x2 (a,b) : |x1x2| 0, 0,x1,x2 (a,b) : ? a 0, 0,x,y (a,b) : |x y| 0,N N,n N,k|xn yn| 0, 0,x,y I : |x y| 0. ? = 1,x1,y1 I : |x1 y1| 0, = 1 2,x2,y2 I : |x2 y2| 0, = 1 n,xn,yn I : |xn yn| 0, l?3mI?E,?xn?yn.w, lim n(xn yn) = 0? lim nf(xn) f(yn) 6= 0,?fi ?g. u,f(x)3mI?Y. ?Y?7?,kf(x)3mI?Y mI?3, ?xn?yn,? lim n(xn yn) = 0?,k lim nf(xn) f(yn) 6= 0. eyf(x) = ex3R?Y. y n N,?xn= ln(n + 1),yn= lnn.l?3N?E?xn?yn,k lim n(xn yn) = lim nln(n + 1) lnn = limn ln ? 1 + 1 n ? = 0, ?, lim nf(xn) f(yn) = limn(n + 1 n) = 1 6= 0 =f(x) = e x3R?Y. 8. yef(x)3a,+)Y,? lim n+ f(x) = b,Kf(x)3a,+)?Y. y fi ? lim n+ f(x) = b,?OK,k 0,A a,x1,x2 A,k|f(x1) f(x2)| 0, 0( 0, 1 0,x1,x2 a,b : |x1 x2| 0,x1,x2 b,c : |x1 x2| 0, = min1,2 0,x1,x2 a,c : |x1 x2| 0,?a,bk?m: x0,x1,x1,x2, ,xn1,xn,x0= a,xn= b, x0i,x00 i xi1,xi,i = 1,2, ,n,kf(x0i) f(x00 i)|. y fi ?f(x)3a,bY,?n4,f(x)3a,b?Y,= 0, 0,x,y a,b : |x y| 0,x1,x2 a,b,?“ |Pn(x1) Pn(x2)| = |a0(xn 1 xn 2) + a1(x n1 1 xn1 2 ) + + an1(x1 x2)| |a0|xn 1 xn 2| + |a1|x n1 1 xn1 2 | + + |an1|x1 x2| (nMn1|a0| + (n 1)Mn2|a1| + + |an1|)|x1 x2| =A|x1 x2| 0, = ? A 0,x1,x2 I : |x1 x2| 0,A a,x1,x2 A,k |bx1 f(x1) bx2 f(x2)| 0,2 0,x1,x2 a,A + 1 : |x1 x2| 0, = min1,2,1 0,x1,x2 a,+) : |x1 x2| 0,x0 a,bk|f(x0)| M. ?M1= 1,x1 a,b,k|f(x1)| M1, M2= max2,|f(x1)| 0,x2 a,b,k|f(x2)| M2, M3= max3,|f(x2)| 0,x3 a,b,k|f(x3)| M3, Mn= maxn,|f(xn1)| 0,xn a,b,k|f(xn)| Mn, w,xi6= xj,i,j N.l?,?E?k.?:8A = xn|n N a,b.?: n,:8A?3?:,? a,b(y). ?,f(x)3:Y,=0= 1 0, 0,x (a,b) : |x | nk.d unk N,?nk +,l?,f(x)3U(,)q.,?cyg. uf(x)3a,bk 15. A?5ny:ef(x)34ma,bY,Kf(x)3a,b?m? ?M. 11 y ?n1f(x)3ma,bk.,=?8f(x)|x a,bQk?.qke? (.n,?8f(x)|x a,bQk?(.qke(.,? supf(x)|x a,b = ? inff(x)|x a,b = eyf(x)3a,b?,=x0 a,b,f(x0) = . d?(., 0,x0 a,b, 0,x1 a,b, 1 0,x2 a,b, 1 2 0,xn a,b, 1 n 0?f(b) 0,x a,b.A?(a A)k? .(b?.)?8.?(.n,8A?(.,? supA = supx|f(x) 0,x a,b = c (a,b). eyf(c) = 0.?y.b?f(c) 6= 0.?Y?5, ef(c)