数学分析试题集锦（一）：limit

1 B5C0BP 2002 CIARA1B7AGB5C0BOBNAQAZATDBA7ARA9 A6B6AMCJC2B4CEBRA0B6AMCJD8BYC2B4BOBDAA 2002AA11CKA8 A2A2D7DBDFDIDDDJ APAYBDA6C6A8 17(2) and (3).ProofAC Set e n =1+1+ 1 2! + ... + 1 n! ,n∈ N + . For any given n ∈ N + , because e n+p = e n + 1 (n +1)! + 1 (n +2)! +···+ 1 (n + p)! , (i) where p ∈ N + ,p≥ 3, and n +2 (n + j)! < 1 (n + j ?1)! ,j=3,···,p,wemayget e n+p ?e n = 1 (n +1)! + 1 n +2 braceleftBig 1 (n +1)! + n +2 (n +3)! ···+ n +2 (n + p)! bracerightBig < 1 (n +1)! + 1 n +2 braceleftBig 1 (n +1)! + 1 (n +2)! ···+ 1 (n + p?1)! + 1 (n + p)! bracerightBig = 1 (n +1)! + 1 n +2 (e n+p ?e n ). Thus, there holds the following inequality e n+p ?e n < 1 (n +1)! + 1 n +2 (e n+p ?e n ). (ii) Furthermore, (ii) implies that e n+p ?e n < n +2 (n +1)!(n +1) . (iii) Notice that e n+p → e(as p →∞). Letting p →∞in (iii). Then, there holds the inequality e?e n ≤ n +2 (n +1)!(n +1) < 1 n!n , (v) which is the second inequality in (2). The first inequality is easy. Set θ n = n!n(e?e n ). From (2), it is easy to see that (3) is true. BOBD1.3BBBD3BD(CPA4COA3B4BZBNBMA5A8 Ideas of Proof: (1). First, to give the formula for S n : S n =2 n ·3·sin 360 ? 2 n ·3 ; (2). By using continuously the formula sin(2x)=2sinxcos x, to give the equality: sin 60 ? = 2 n?1 cos 60 ? 2 ·cos 60 ? 2 2 ···cos 60 ? 2 n?1 sin 60 ? 2 n?1 ; (3). Notice that the facts sin 360 ? 2 n ·3 =sin 60 ? 2 n?1 ,and1> cos 60 ? 2 j ≥ cos 60 ? 2 = √ 3 2 ,j=1,2, ..., n?1. It may be obtained: sin 360 ? 2 n ·3 < √ 3 2 · 1 2 n?1 · parenleftbig 2 √ 3 parenrightbig n?1 = √ 3 2 · parenleftbig 1 √ 3 parenrightbig n?1 . 2 (4). By (3), it is easy to see that 0 <S n+1 ?S n =2 n+1 ·3·sin 360 ? 2 n+1 ·3 ?2 n ·3·sin 360 ? 2 n ·3 =2 n+1 ·3·sin 360 ? 2 n+1 ·3 parenleftBig 1?cos 360 ? 2 n+1 ·3 parenrightBig =2 n+1 ·3·sin 360 ? 2 n+1 ·3 ·2·sin 2 parenleftBig 360 ? 2 n+2 ·3 parenrightBig < 9 4 · parenleftbig 2 3 √ 3 parenrightbig n+1 (5). To prove that {S n } is a Cauchy sequence: 0 <S n+p ?S n < n+p summationdisplay k=n |S k+1 ?S k | < 9 4 · n+p summationdisplay k=n parenleftbig 2 3 √ 3 parenrightbig k+1 . (6). The conclusion is obvious from (5). DAA2D5D8A0DI 1A0CUA3ACAMC5BECFBBA7BEAKBEAZCXCFA3 2A0AU lim x→x 0 f(x)=a,?f(x)CMx 0 CCBGC9A3BJCMx → x 0 BBC1AXD1AAx CZBPAJB9x 0 ?AZBPALCC a = f(x 0 )? 3A0B0CBε ?δ CHC2D1BWB3B8 x → x 0 AV f(x)ARB1D9CF a. 4A0CBBNBMBGC9CUA3AC (1) lim x→1 x?3 x 2 ?9 = 1 2 ; (2) lim x→3 x?3 x 2 ?9 = 1 6 ; (3) lim x→1 x?1 √ x?1 = 2; (4) lim x→1 (x?2)(x?1) x?3 =0; (5) lim t→1 t(t ?1) t 2 ?1 = 1 2 ; (6) lim x→∞ x?1 x +2 =1; (7) lim x→3 x x 2 ?9 = ∞; (8) lim x→∞ x 2 + x x +1 = ∞. 5A0BQDCA4BDAZBPCTAL?APCTALAAAFBXB0CUA3ADAPARCTALAAAFCWB0BKD6A3 (1) lim x→x 0 f(x)=aAYBNALC4BFCQAZ lim x→x 0 |f(x)| = |a|; (2) AP lim x→x 0 f(x)=a,CN lim x→x 0 [f(x)] 2 = a 2 ; (3) AP lim n→∞ f( 1 n )=a,CN lim x→0 + f(x)=a; (4) AP lim x→x 0 f(x)CG lim x→x 0 (f(x)+g(x)) BHB3CMAACN lim x→x 0 g(x)B3CMAD (5) AP lim x→x 0 f(x)CG lim x→x 0 (f(x)g(x)) BHB3CMAACN lim x→x 0 g(x) B3CMAD (6) APCMx 0 BBA5DDCIA7f(x) > 0AAAO? lim x→x ? 0 f(x)=a AAA6A0ALCC a>0. 3 6A0CUA3AC (1) lim x→∞ f(x)=a ?? lim x→+∞ f(x) = lim x→?∞ f(x)=a; (2) lim x→x 0 f(x)=a ?? lim x→x + 0 f(x) = lim x→x ? 0 f(x)=a. 7A0CUA3ACC2B4CEBRBBBHC6BVA0CVASAJC3BVCGCVASAJBXBVA3 8A0BQDCCLB8CCBKB4BL?B4CMC5B1? (1) lim x→0 sin x x = lim x→0 sin x lim x→0 x = 0 0 = 1; (2) lim x→∞ sin x x = lim x→∞ sin x lim x→∞ x =0; (3) lim x→0 xsin 1 x = lim x→0 x lim x→0 sin 1 x =0. 9A0AU m, n ∈ N + AAAHBQDCCEBRAC (1) lim x→0 (1 + mx) n ?(1 + nx) m x 2 ; (2) lim x→1 parenleftBig m 1?x m ? n 1?x n parenrightBig ; (3) lim t→0 (x + t) n ?x n t (x ∈ R); (4) lim x→0 n √ 1+x?1 x ; (5) lim x→0 sin 2x?sin 3x x ; (6) lim h→0 cos(x + h)?cos x h ; (7) lim x→0 ( √ 1+x 2 + x) n ?( √ 1+x 2 ?x) n x ; (8) lim x→0 + x bracketleftBig 1 x bracketrightBig ; (9) lim x→1 (1?x)tan πx 2 ; (10) lim x→0 x 2 1?cos x ; (11) lim x→0 sin 5x?sin 3x sin 2x ; (12) lim x→0 cos x?cos 3x x . (13) lim x→0 x √ 1?2x; (14) lim x→∞ parenleftBig 1+ 2 x parenrightBig ?x ; (15) lim x→∞ parenleftBig x 2 ?1 x 2 +1 parenrightBig x 2 ; (16) lim x→+∞ parenleftBig cos a x parenrightBig x 2 (a negationslash=0); (17) lim x→0 cos(narccos x) x (nBIACB4); (18) lim n→+∞ parenleftBig cos √ n +1?cos √ n parenrightBig ; (19) lim n→+∞ cos x 2 ·cos x 2 2 ···cos x 2 n ; (20) lim n→+∞ sin(π radicalbig n 2 +1); (21) lim x→ π 2 (sin x) tan x ; (22) lim x→∞ parenleftBig sin 1 x +cos 1 x parenrightBig x ; (23) lim n→+∞ parenleftBig n + x n?1 parenrightBig n ; (24) lim n→+∞ parenleftBig n +lnn n?ln n parenrightBig n ln n . 10A0AU [a,b] AZC6BVCCBRAKAICNAAAOC0 ?x 0 ∈ [a,b], lim x→x 0 f(x) B3CM?CCBRAACUA3AC f(x) CM [a,b] ASCCCUA3 11A0AU f :(a,b) → R AZBKCUC2B4AACUA3ACB3CMB4DC{x n }?(a,b), AWBA lim n→∞ f(x n )=∞. 12A0AU f :[a,+∞) → R,CUA3AC lim x→+∞ f(x) B3CM?CCBR?? ?ε>0,?M>0 AWBA?x 1 ,x 2 >M, C7CC|f(x 1 )?f(x 2 )| <ε. 4 13A0AU f :[a,+∞) → R AZB6BFCOC2B4AACUA3BQA2AQBVA4BDAZBCCMBBAC (1) lim x→+∞ f(x n )B3CM?CCBRAD(2) {f(n)}AZB1D9B4DCAD(3)f CM[a,+∞)ASCCASCUA3 14A0D5CBBKAGBSBBBCCMB5CBAHBQDCCEBRAC (1) lim x→0 1?cos x sin 2 x ; (2) lim x→1 5x 2 ?2(1?cos 2 x) 3x 2 +4tan 2 x ; (3) lim x→0 radicalbig 1+sin 2 x?1 xtan x ; (4) lim x→0 tan(tan x) sin(2x) ; (5) lim x→0 ( 3 √ 1+tanx?1)( √ 1+x 2 ?1) tan x?sin x ; (6) lim x→0 ? (1? n √ cos x)tanx (1 ?cos x) 3/2 . 15A0AU f,g : A ? R → R AZDABVC2B4AA??x ∈ A, f(x) > 0, CNAVBUAOf(x) g(x) BBC2B4BIDEA3DC DHA3APlim f(x)=1,lim g(x)=∞,CNCEBRlim f(x) g(x) B2CF1 ∞ A1D6D9DGA3BJCFCRD2ARBGAXAAC6AHD5 CBBCAX f(x) g(x) = e g(x)lnf(x) D2CABIBBDG0·∞BTARBGAX lim g(x)lnf(x)BBCEBRBJBDAB A61A8AU g 1 (x) ～ g 2 (x). CUA3ACAP lim f(x) g 1 (x) B3CMAACN lim f(x) g 1 (x) = lim f(x) g 2 (x) ; A62A8CLBG lim x→x 0 ln x =lnx 0 AACHBJB4C2B4y =lnxAZD8BYBBA6CPCCBZCPCTA8AACUA3ACAP lim x→x 0 f(x)= a>0, lim x→x 0 g(x)=b,CNlim f(x) g 1 (x) = a b ; A63A8AHBQDCCEBRAC lim x→0 (1 + sin x) 1 x ; lim x→∞ (cos 1 x ) x 2 . 16A0CUA3ACC2B4 f CMx 0 BED8BY?? CMx 0 CJD4D8BYCECDD8BYA3 17A0DABVCM x 0 B1ARD8BYC2B4CVC4CM x 0 AZBPC6BGARD8BY?APABD1C6BVCM x 0 B1D8BYAAC6BVCM x 0 B1ARD8BYAACNBAA1BBC4CM x 0 B1AZBPC6BGARD8BY? 18A0CUA3ACAP f D8BYAACN|f|C5D8BYA3A9A4BDAWD7DH? 19A0CUA3ACAPD8BYC2B4CMCCD4BEASC2B4CWBIDEAACNB2C2B4C7BIDEA3 20A0APC2B4 f CM[a,b]D8BYAAC7CTAAAFCQBGC9CUA3 1 f CM[a,b]D8BYA3 21A0APf C4g BHCM[a,b]D8BYA3B0CUA3ACmax{f(x),g(x)}C7CGmin{f(x),g(x)}BHCM[a,b]D8BYA3 22A0AP f AZD8BYBBA3CUA3ACBJCFAMC5 c>0AAC2B4 F(x)= ? ? ? ? ? ? ? ? ? ? ? ?c, AP f(x) < ?c f(x), AP|f(x)|≤c c, AP f(x) >c AZD8BYBBA3 23A0AP f CM[a,b]D8BYAA a<x 1 <x 2 < ···<x n <b,CNCM[x 1 ,x n ]D1ALCC ξ AW f(ξ)= f(x 1 )+f(x 2 )+···+ f(x n ) n . 24 A0AUC2B4 f CM (?∞,+∞) DID3 Lipschitz BFCQAC ?M>0, AWBA ?x, y ∈ (?∞,+∞) C7CC |f(x)?f(y)|≤M|x?y|. CUA3AC f CM(?∞,+∞)ASC6CZD8BYA3 5 25A0CUA3ACC2B4f : I → RCMx 0 ∈ I B1D8BY?? ?x n ∈ I,x n → x 0 (n →∞),CC lim n→∞ f(x n )=f(x 0 ). 26A0AUC2B4 f : I → R CM x 0 ∈ I D8BYAA? f(x 0 ) > 0 A3CUA3ACB3CM x 0 BBC6BVDDCIAACMBSDDCIA7 f(x) ≥ q>0A3 27A0BBDGBQDCC2B4BBD8BYBVAAAOC9B0BGBUA3 (1) f(x)= ? ? ? ? ? ? x 2 ?4 x?2 ,xnegationslash=2, 4,x=2; (2) f(x)= ? ? ? ? ? | sin x x |,xnegationslash=0, 1,x=0; (3) f(x)= ? ? ? ? ? ? ? sin x |x| ,xnegationslash=0, 1,x=0; (4) f(x)=[x]. 28A0BBDGBQDCC2B4CMCXBGBEB1BBD8BYBVA3APAZCNBIBEAAB6A3BABBD2BTAC (1) f(x)= ? ? ? ? ? xsin 1 x ,x<0, 1,x≥ 0, x =0; (2) f(x)= ? ? ? ? ? ? ? 2 ? 1 x?3 ,xnegationslash=0, 2,x=0, x =3; 29A0BBDGBQDCC2B4BBD8BYBVA3APCCCNBIBEAAB6A3CNBIBEBBD2BTAC (1) f(x)= ? ? ? ? ? sin x x ,x<0, x 2 ?1,x≥ 0; (2) f(x)= ? ? ? ? ? ? ? e ? 1 x 2 ,xnegationslash=0, 2,x=0; (3) f(x)= ? ? ? ? ? ? ? 1 q ,x= p q (q>0,q,pBIC8D0BBCSB4), 0,xBIBKD4B4; (4) f(x)= ? ? ? ? ? cos πx 2 ,x≤ 1, |x?1|,x>1; (5) f(x)= ? ? ? ? ? sin πx, xBICCD4B4, 0,xBIBKD4B4; (6) f(x)= ? ? ? ? ? x, |x|≤1, 1, |x|≥1. ABD1A63A8B9ANAYBBC2B4CSD3DJA6RiemannA8C2B4AB 30A0B0ALBGAUB4AAAWBQDCC2B4CMx =0B1D8BYAC (1) f(x)= ? ? ? ? ? a + x, x ≤ 0, sin x, x > 0; (2) f(x)= ? ? ? ? ? arctan 1 x ,x<0, a + √ x, x ≥ 0; (3) f(x)= ? ? ? ? ? ? ? √ 1+x?1 3 √ 1+x?1 ,xnegationslash=0, c, x =0; (4) f(x)= ? ? ? ? ? sin x·sin 1 x ,xnegationslash=0, c, x =0; (5) f(x)= ? ? ? ? ? ? ? (1 + x) 1 x ,xnegationslash=0, c, x =0; (6) f(x)= ? ? ? ? ? tan x x ,xnegationslash=0, c, x =0. 6 31A0CUA3BQDCBWBDAC A61A8BLAX x2 x =1CM(0,1)A7CYATCCC6BVBYAD A62A8BLAX x 5 ?3x?1=0CM(1, 2.7)A7CYATCCC6BVBYAD A63A8AU f ∈ C[a,b]AAAP f CM[a,b]ASC7ARBIDEAACN f CM[a,b]ASC7BICTA6CCBRA8AD A64A8BLAXsin x + x +1=0CM[? π 2 , π 2 ]A7CYATCCC6BVBYAD A65A8BLAXa n x n + a n?1 x n?1 +···+ a 1 x + a 0 =0CMCYATCCC6BVBYAAABD1nAZACB4A0a j ∈ R(j = 0,1,2, ..., n)BIAUB4AA? a n negationslash=0A3 A66A8BLAXa n cos nx+ a n?1 cos(n?1)x+···+a 1 cos x +a 0 =0CM (0,2π)A7CYATCC2nBVBYAAAB D1 a 0 ,a 1 ,···,a n?1 ,a n DID3|a n | > |a 0 |+|a 1 |+···+|a n?1 |A3A6BCAYACCYDFBEx nk = kπ n ,k=0, ...,2n). 32A0AUC2B4f : R → RDID3CZCKBVAACHBJAMC5x 1 ,x 2 ∈ R,f(x 1 + x 2 )=f(x 1 )+f(x 2 )AAAO?f CM x =0B1D8BYAACUA3AC f CM R ASD8BYA3 33A0AU f ∈ C[a,+∞)AAAO? lim x→+∞ f(x) B3CM?CCBRAACUA3AC f CM[a,+∞)ASCCCUA3 34A0AU f ∈ C(a,b)AAAO? f(a +0)CGf(b?0)B3CMA6AID0CCBKAGCEBRA8?CAC3AACUA3ACCM(a,b) A7CYATB3CMC6BE ξ,AW f(ξ)=0. 35A0AUf ∈ C[a,b]AO??x ∈ [a,b],?y ∈ [a,b],AWf(y)= 1 2 |f(x)|A3CUA3ACB3CMξ ∈ [a,b]AWf(ξ)=0. 36 A0AU f ∈ [a,b],{x n }?C[a,b] AZAMC6B4DCA3AP lim n→∞ f(x n )=A ∈ R AACUA3BLAXf(x)=A CM [a,b]A7ALCCC6BVBYA3 37A0AU f ∈ C(?∞,+∞)AAAO? f AZACC2B4AACUA3BLAXf(x)=0CYATCCC6BVBYA3APAZf C1BUB6 BFBBAACN x =0AZBABBBHC6BYA3 38A0CBC6CZD8BYBGC9C3CUAC (1) f(x)= 3 √ x CM[0,1]ASAZC6CZD8BYBBAD (2) f(x)=sinx CM(?∞,+∞)ASAZC6CZD8BYBBAD (3) f(x)=sinx 2 CM(?∞,+∞)ASAZARC6CZD8BYBBA3 39A0CWB0BQC6BQDCC4AHBB f: (1) f(+0) = 0,f(?0) = 1; (2) f(+0)ARB3CMAA C5BM∞,f(?0) = 0; (3) lim x→+∞ f(x) = 0, lim x→?∞ f(x)ARB3CMAD (4) lim x→+∞ f(x) = lim x→?∞ f(x)=AA6AUB4A8AD (5) f(x 0 +0)C4 f(x 0 ?0)BHARB3CMAD (6) f(x 0 +0)=+∞, f(x 0 ?0) = ?∞; (7) f(x 0 +0)=1,f(x 0 ?0) = +∞; (8) lim x→+∞ f(x) ARB3CMAAC5BM∞, lim x→?∞ f(x)=?∞. 40A0A61A8CUA3ACB6BFCCCUC2B4B3CMD4CDCEBRAD A62A8CUA3ACB6BFCCCUC2B4BBC6ADARD8BYBEBHBIBDC6D2ARD8BYBEA3 7 41 A0CUA3AC lim x→+∞ f(x) B3CM?CCBRBBAYBNALC4BFCQAZACBJAMC8BXBG ε>0 AAB3CM X>0, B8 x prime ,x primeprime >XAVAAC7CC|f(x prime )?f(x primeprime )| <εA3 42 A0CUA3AC lim x→x 0 f(x) B3CM?CCBRBBAYBNALC4BFCQAZACBJAMC8BXBG ε>0 AAB3CM δ>0, B8 0 < |x prime ?x 0 | <δ, 0 < |x primeprime ?x 0 | <δ,AVAAC7CC|f(x prime )?f(x primeprime )| <εA3 43 A0CUA3AC f CM x 0 BED8BYBBAYBNALC4BFCQAZACBJAMC8BXBG ε>0 AAB3CM δ>0, B8|x prime ? x 0 | < δ, |x”?x 0 | <δ,AVAAC7CC|f(x prime )?f(x”)| <εA3 44A0CUA3ACAPB6BFCCCUC2B4 f CZAJB9 f(a), f(b)CVCNBBC6ADCWAACN f CM[a,b]D8BYA3 45A0CUA3ACC2B4f CM(a,b)D8BYAAAO?f(a +0),f(b?0)B3CMAACNf CZAJB9 f(a +0)C4f(b?0) CVCNA6B7CZA8ARBCCF f(a +0),f(b?0))C6ADCWA3 46A0CUA3AC (a,b)ASBBD8BYC2B4 f BIC6CZD8BYBBAYBNALC4BFCQAZAC f(a +0),f(b?0)B3CMA3 47A0APC2B4f CM(?∞,+∞)ASBBAMC6CCBRAKAICNASD8BYAACNBACM(?∞,+∞)ASBBAMC6CCBRCXAI CNASC5C6CZD8BYA3 48A0C2B4 f(x)=x 2 CM(?∞,+∞)CG(?l,l)AS(l ∈ R,l > 0)AZBPC6CZD8BY? 49A0AP f CM(a,b)A7CCBGC9AAAO?BJ(a,b)A7AMC5 xAAB3CM xBBA5BVDDCI O x AAAWBA f CMO x A7 CCCUA3BJ f CM(a,b)AZBPCCCU?CEAPCR (a,b)BTBI[a,b]AAAOC5? 50A0CUA3AC (a,b)ASBBC6CZD8BYC2B4ALCCCUA3 51A0AFCQBGC9CUA3ACDABVC6CZD8BYC2B4BBC4ANC6CZD8BYA3CEBJACDABVC6CZD8BYC2B4BBCDAOC5?