? ?cKDf ?￥ ??? x2 ? ￥K 1.¨?l￡ ü/  ? ￥K1 , (5) limn!1(pn+1?pn) ￡ ü n5 jpn+1?pnj = (n+1)?npn+1+pn ? 1pn ?[8" > 0, |N = [ 1"2],5?n > N Hμ jpn+1?pnj? 1pn ? " ?K?l?V U limn!1(pn+1?pn) = 0 (6) limn!1 10nn! 4 U 10n n! = 10 1 ￠ 10 2 ￠￠￠ 10 1 0￠￠￠ 10 n ? 10 9 9! ￠ 10 n 2¨?l￡ ü (4) limn!1xn = 3 ?xn = 8> >>< >>>: 3; n = 3k 3n+1 n ; n = 3k +1(k = 1;2;￠￠￠) 2+ 1+n3?pn+n; n = 3k +2: 1 ￡ ü?n = 3k;k = 1;2;￠￠￠ H jxn ?3j = 0 ?n = 3k +1;k = 1;2;￠￠￠ H jxn ?3j = j3n+1n ?3j ? 1n ?n = 3k +2;k = 1;2;￠￠￠ H jxn ?3j = j2+ 1+n3?pn+n ?3j ? ?2+ pn 3?pn+n ? pn ?n2 +n; (?n ? 9 H) ? 2pn 8 ?n ? 9 H jxn ?3j? 1n + 2pn ? 4pn #8" > 0, |N = max9; 16"2,5?n > N Hμ jxn ?3j? 4pn < " ?[ limn!1xn = 3 8 p/ K (1) limn!1( 11￠2 + 12￠3 +￠￠￠+ 1n(n+1)) 4 U1n(n+1) = 1n ? 1n+1. 2 (2) limn!1( 1n2 + 1(n+1)2 +￠￠￠+ 1(2n)2) 4 U 1 n2 + 1 (n+1)2 +￠￠￠+ 1 (2n)2 ? 1n2 + 1n2 +￠￠￠+ 1n2 1 n (5) limn!1(1? 1np2)cosn 4 UK?ib (8) limn!1[(n+1)fi ?nfi]0 < a < 1 4 Uy10 < a < 1, ?[ (n+1)fi ?nfi = nfi[(1+ 1n)fi ?1] ? nfi[(1+ 1n)?1] = 1n1?fi (9) limn!1 12 ￠ 34 ￠￠￠￠￠ 2n?12n 3E1?? 2n = (2n?1)+(2n+1)2 ? p (2n?1)(2n+1) ?[ 1 2 ￠ 3 4 ￠￠￠￠￠ 2n?1 2n ? 1p1￠3 ￠ 3p3￠4 ￠￠￠￠￠ 2n?1p(2n?1)(2n+1) ? 1p2n+1 ?[? limn!1 1p2n+1 = 0 3 P? limn!1 12 ￠ 34 ￠￠￠￠￠ 2n?12n = 0 3E2: xn = 12 ￠ 34 ￠￠￠￠￠ 2n?12n yn = 23 ￠ 45 ￠￠￠￠￠ 2n2n+1 5^n xnyn = 12n+1 O xn ? yn ?[ xn ?pxnyn ? r 1 2n+1 #? limn!1 1p2n+1 = 0 P? limn!1 12 ￠ 34 ￠￠￠￠￠ 2n?12n = 0 16 !limn!1an = a￡ ü (1) limn!1 a1+a2+￠￠￠+ann = a  ?ù ?￥ I 5? ??$ ￡ ü8" > 0;9N1 2 N; P¤?n > N1 Hμ janj < "2 ? |?￥N1a1 +a2 +￠￠￠+aN1 ^B?%?￥ ?yN V[ |N > N1, P ¤?n > N Hμ ja1 +a2 +￠￠￠+aN1n j < "2 4 ? ^ ?¨ ???? T¤ ja1 +a2 +￠￠￠+ann j = ja1 +a2 +￠￠￠+aN1n + aN1+1 +aN1+2 +￠￠￠+ann j ? ja1 +a2 +￠￠￠+aN1n j+jaN1+1 +aN1+2 +￠￠￠+ann j < "2 + (n?N1) " 2 n < ": I 5?? ? è ?an = (?1)n. (2) ?an > 05limn!1 npa1a2￠￠￠an = a. 4 U ?¨ 5￥2 ?b 18¨?l￡ ü/  ? 1í kv  (4) 1+ 12 + 13 +￠￠￠+ 1n. 3 ?i?? ?ni?? ?k P¤2k ? n < 2k+1, ?[ 1+ 12 + 13 +￠￠￠+ 1n = 1+ 12 +(13 + 14)+(15 + 16 + 17 + 18)+￠￠￠ +( 12k?1 +1 + 12k?1 +2 +￠￠￠+ 12k?1 +2k?1)+ 12k +1 +￠￠￠+ 1n ? 1+ 12 +(14 + 14)+(18 + 18 + 18 + 18)+￠￠￠ +( 12k + 12k +￠￠￠+ 12k) | {z } 2k?1? = k +12 ? lnn2ln2 A ?18M > 0,?? T lnn 2ln2 > M?N?n > 2 2M 5 yN |N = [22M],5?n > M Hμ 1+ 12 + 13 +￠￠￠+ 1n > M 2 ?￡8b x3f ?￥K 3 !f(x) > 0￡ ü ?limx!x 0 f(x) = A5limx!x 0 n pf(x) = npA ? ?? ?n ? 2. 4 U ?A = 0,52 ?^￡b/ !A > 0,N H j n p f(x)? npAj = jf(x)?Aj( npf(x))n?1 +( npf(x))n?2 npA+￠￠￠+( npA)n?1 ? jf(x)?Aj( npA)n?1 5 p/ f ?3 ? U?￥P·K (5) f(x) = 8 >>>< >>> : 2x; x > 0; 0; x = 0; 1+x2; x < 0 x = 0. 4 U5¨?l￡ ü limx!02x = 1 18 !f ?f(x)(0;+1)  ?@Z?f(2x) = f(x) Olimx!+1f(x) = A￡ ü f(x) · A;x 2 (0;+1): ￡ ü ?i￥x 2 (0;+1)μ f(x) = f(2x) = f(22x) = ￠￠￠ = f(2nx);8?? ?n 6  T H 7n !1 |K'¤ f(x) = limn!1f(2nx) = limx!+1f(x) = A x4f ?￥ ??? 9 ?f(x)?g(x)?[a;b] ?? k￡ ümax(f(x);g(x)) ?min(f(x);g(x))?[a;b] ??. 4 U max(f(x);g(x)) = (f(x)+g(x))+jf(x)?g(x)j2 min(f(x);g(x)) = (f(x)+g(x))?jf(x)?g(x)j2 x5í kl Dí kv ￥1? 3?x ! 0 H/ ? T? ? ?$ (1) o(x2) = o(x) 3? ?.y1 limx!0 o(x 2) x = limx!0 o(x2) x2 ￠x = 0 (2) O(x2) = o(x) 3? ?.y1 limx!0 O(x 2) x = limx!0 O(x2) x2 ￠x = 0 (3) x￠o(x2) = o(x3) 3? ?.y1 limx!0 x￠o(x 2) x3 = limx!0 o(x2) x2 = 0 7 (6) o(x) = O(x2). 3?? ?. è ?f(x) = x32,5f(x) = o(x),?f(x) 6= o(x2) 8