1 B1BWBM 2002 CDALA1B3ADB1BWBOBLANAUAPD7A4B3A6 A3B3B0A0BHCAC0B0B8CEB6A7 2002A2 12CDA5 1 A0AQ 0 <b≤ a.CG (a ?b) 2 8a ≤ a + b 2 ? √ ab ≤ (a ?b) 2 8b . 2 A0CNDJA9C0B0 f(x)=ln(1+e ?x ),g(x)= √ c 2 + x 2 (c>0)BIBXBAC0B0A8 3 A0CNDJAOB8AVA9 (1) 1 > sin x x ≥ 2 π (0 <x≤ π 2 ); (2) sin x x > 3 √ cos x (0 <x≤ π 2 ); (3) (sin x) ?2 ≤ x ?2 +1? 4 π 2 (0 <x≤ π 2 ); (4) 2 2x +1 < ln(1 + 1 x ) < 1 √ x 2 + x (x>0); (5) ln x 1 ?x ≤ 1 √ x (x>0,xnegationslash= 1); (6) 4x π(1 ?x 2 ) ≤ tan( πx 2 ) ≤ πx 2(1?x 2 ) (0 ≤ x<1). 5 A0CNDJA9 (1 + x) α ≥ 1+αx (x ≥?1, 0 <α<1);(1 + x) α ≤ 1+αx (x ≥?1,α>1, or α<0). 6 A0CNDJA9AAD4BRB7 β C1D4B0B7α AUB6?n ∈ N + C5 (1 + 1 n ) n+α ≤ e ≤ (1 + 1 n ) n+β . 7 A0CNDJA9 lim n→∞ n summationdisplay k=1 ( √ k +2?2 √ k +1+ √ k) AZCFA2 8 A0CNDJB2BCCZ(Stolz) BDD0(CZAMBSB7DEAHB5AAC9BOBKCG). A3 1 A5 ( 0 0 BS) AQB0D8{a n }AXABC7DAB7A7B0D8{b n }AXBXBSBACIABC7DAB7A2AIBY lim n→∞ a n ?a n+1 b n ?b n+1 AZCFC6BI +∞, CG lim n→∞ a n b n C0AZCFC6BI +∞A7 lim n→∞ a n b n = lim n→∞ a n ?a n+1 b n ?b n+1 . A3 2 A5 ( ∞ ∞ BS) AQB0D8 {b n } BXBSB1BCCHCFA7A7 lim n→∞ b n =+∞. AIBY lim n→∞ a n ?a n+1 b n ?b n+1 AZCFC6BI +∞,CG lim n→∞ a n b n C0AZCFC6BI +∞A7 lim n→∞ a n b n = lim n→∞ a n ?a n+1 b n ?b n+1 . Hint of ProofA9Only prove that conclusions (1) and (2) hold for the cases with the finite limit, the proof of the other cases are similar. The proof of (1). Suppose that lim n→∞ a n ?a n+1 b n ?b n+1 = l ∈ R. Then, ?ε>0, there exists N 1 ∈ N + such that | a n ?a n+1 b n ?b n+1 ?l| <ε hold for all n ≥ N 1 . Further, the fact that sequence {b n } is strictly decreasing implies that ?ε(b m ?b m+1 ) <a m ?a m+1 ?l(b m ?b m+1 ) <ε(b m ?b m+1 ), 2 for all m ≥ N 1 . When n>N 1 , by adding the above inequalities in order from m = n to n + p,itiseasy to see that ?ε(b n ?b n+p ) <a n ?a n+p ?l(b n ?b n+p ) <ε(b n ?b n+p ), and hence ?ε(1 ? b n+p b n ) < a n b n ? a n+p b n ?l(1? b n+p b n ) <ε(1 ? b n+p b n ). (i) Since lim n→∞ a n =0, and lim n→∞ b n =0, then, by letting p →∞in (i), we get ?ε ≤ a n b n ?l ≤ ε, (ii) for any n>N 1 . The conclusion (1) follows (ii). The proof of (2). Suppose that lim n→∞ a n ?a n+1 b n ?b n+1 = l ∈ R. Then, ?ε>0, there exists N 1 ∈ N + such that | a n ?a n+1 b n ?b n+1 ?l| <ε hold for all n ≥ N 1 . Further, the fact that sequence {b n } is strictly increasing implies that ?ε(b m+1 ?b m ) <a m+1 ?a m ?l(b m+1 ?b m ) <ε(b m+1 ?b m ), for all m ≥ N 1 . When n>N 1 , by adding the above inequalities in order from m = N 1 to n,itiseasyto see that ?ε(b n ?b N 1 ) <a n ?a N 1 ?l(b n ?b N 1 ) <ε(b n ?b N 1 ), and hence ?ε(1? b N 1 b n ) < a n b n ? a N 1 b n ?l(1? b N 1 b n ) <ε(1 ? b N 1 b n ). (iii) Since lim n→∞ b n =+∞, then, there exists N 2 ∈ N + such that | a N 1 b n | <ε,and | b N 1 b n | <min{ε,1} hold for any n ≥ N 2 . Further, from (iii), we get ?2(1 +|l|)ε ≤ a n b n ?l ≤ 2(1 +|l|)ε, (iv) for any n>max(N 1 ,N 2 ). The conclusion (2) follows (iv). (3) DB x 1 ∈ (0, 1)CB x n+1 = x n (1 ?x n ),n=1, 2, 3,···.CNDJA9 lim n→∞ nx n =1 (4) AQC0B0D8 f 1 (x)=sinx, f n+1 (x)=sinf n (x),n=1, 2,···. AJ sin x>0, CNDJA9 lim n→∞ radicalbigg n 3 f n (x)=1. 9 ? A0(1) ( ∞ ∞ BSB2BCCZ(Stolz)BDD0B7BBBW)AQT BICMAQB0A8AJC0B0g(x),f(x),x∈ [a,+∞)DFD2A9 (i) g(x + T) >g(x),x∈ [a,+∞); (ii) lim x→+∞ g(x)=+∞,A7C0B0 f(x),g(x) CF [a,+∞)B7AHC2D0ACCHANC5COAA (iii) lim x→+∞ f(x + T)?f(x) g(x + T)?g(x) = l. CG lim x→+∞ f(x) g(x) = l. (2) ( 0 0 BSB2BCCZ(Stolz) BDD0B7BBBW) AQ T BICMAQB0A8AJC0B0 g(x),f(x),x∈ [a,+∞)DFD2A9 3 (i) g(x + T) >g(x) > 0,x∈ [a,+∞); (ii) lim x→+∞ g(x)=0,A7 lim x→+∞ f(x)=0; (iii) lim x→+∞ f(x + T)?f(x) g(x + T)?g(x) = l. CG lim x→+∞ f(x) g(x) = l. 10A0AQC0B0f DFD2A9(i) ?∞ <a≤ f(x) ≤ b<+∞, (a ≤ x ≤ b); (ii) |f(x)?f(y)|≤k|x?y|, (0 < k<1,x,y∈ [a,b]). AQ x 1 ∈ [a,b]A7ALBDC3BUD8{x n }: x n+1 = f(x n ),n=1, 2,···. AYCNDJA9 lim n→+∞ x n = x AZCFA7A7 x = f(x). 11A0AQC0B0f DFD2A9(i) ?∞ <a≤ f(x) ≤ b<+∞, (a ≤ x ≤ b); (ii) |f(x)?f(y)|≤|x?y|, (x, y ∈ [a,b]). AQx 1 ∈ [a,b]A7ALBDC3BUD8{x n }: x n+1 = 1 2 parenleftbig x n + f(x n ) parenrightbig ,n=1, 2,···. AYCNDJA9 lim n→+∞ x n = x AZCFA7A7 x = f(x). 12. AQ a 0 =0,BDC3 a n+1 =1+sin(a n ?1),n=0, 1, 2,···.AYAAA9 lim n→+∞ n summationdisplay k=0 a k n . (Hint: Let b n = a n ?1. Then, it is easy to prove that b n < 0andb n <b n+1 for any n ∈ N + .) 13 ? . AQC0B0 f D3BVAOCIA7?BD x 0 A7C4AXBDC3 x n = f(x n?1 ),n=1, 2,···A2AJ x 0 = a ARA7{x n } AZD4A7AYCNDJB2min(a,f(a)) ≤ x 0 ≤ max(a,f(a)) ARA7{x n }C0AZD4A2 14 ? A0AJ ε 0 ,ε 1 ,ε 2 ,···CUB7DHC1BTCU?D1?1, 0, 1AKB0COC1A7CNDJ a n =: ε 0 radicalbigg 2+ε 1 radicalBig 2+···+ ε n √ 2=2sin bracketleftBigg π 4 n summationdisplay k=0 ε 0 ε 1 ···ε k 2 k bracketrightBigg =: b n BGDHBTn ASD2A2 15 ? A0BDC3B0D8 x n = y n e ? 1 12n ,y n = n!n ?n? 1 2 e n ,n =1, 2,···. AYCNACCH (x k ,y k ) AFBZD6ACCH (x k+1 ,y k+1 ). 16 ? A0A3 1 A5B0D8 braceleftBig a n = parenleftBig 1+ 1 n parenrightBig n (1 + x n ) bracerightBig B1BCCIAOB7ATBNAHBZB8CJAX x ≥ 1 2 . A3 2 A5B0D8 braceleftBig a n = parenleftBig 1+ x n parenrightBig 1+n bracerightBig B1BCCIAOB7ATBNAHBZB8CJAX 0 <x≤ 2. 17A0AQ f(x) CF [a,b]ANBJCNCWB3A7f(a) < 0,f(b) > 0,BGC1A6x ∈ [a,b],f prime (x) ≥ δ>0,f primeprime (x) ≥ 0. (1)DBx 1 = b? f(b) f prime (b) ,x n+1 = x n ? f(x n ) f prime (x n ) ,n=1, 2, 3,···.CNDJ{x n }AZD4C7f CF[a,b]ANB7DABB ξ. (2)DBy 1 = b? (b?a)f(b) f(b)?f(a) ,y n+1 = y n ? (b?y n )f(y n ) f(b)?f(y n ) ,n=1, 2, 3,···.CNDJ{y n }C0AZD4C7f CF [a,b]ANB7DABBξ. (3)CNDJC4(1)A0(2)AAf B7DABBB8CGC7AA g(x)=x? f(x) f prime (x) A7h(x)=x? (b ?x)f(x) f(b) ?f(x) B7AOBEBBA2 (Hint of (2): Let y = f(b)+ f(b)?f(a) b?a (x?b). Find out the point of intersection x 1 of the straight line with x?axes. After that, find out the point of intersection x 2 of the new straight line with x?axes by using x 1 replecing a, and repeat continuously the program. What is the signification of geometry?) 18A0AQf AXBDC3CF[a,b]ANB7ATCQC0B0A7C6AQf CFx 0 AWCWBGBNA7A4CUa<x 0 <bA7ALAQB0D8{a n }, 4 {b n }DFD2B8CJA9a<a n <x 0 <b n <bA7 lim n→+∞ a n = lim n→+∞ b n = x 0 .AYCNDJA9 lim n→+∞ f(b n ) ?f(a n ) b n ?a n = f prime (x 0 ). 19A0AQC0B0f CFC5A8C6BKA8B7ACCH(a,b)CUB7AHC2C1BBxAWC5C5BOB7B3B0f prime (x)A7A7 lim x→a + f(x)= lim x→b ? f(x). AYCNDJCF (a,b)CUAZCFBB c A7AUB6 f prime (c)=0. 20A0AQf(x)=a 1 sin x + a 2 sin 2x +···+ a n sin nx,ALA7|f(x)|≤|sin x|,?x ∈ R, a 1 ,a 2 ,···,a n BI ATAQB0A2AACNA9 |a 1 +2a 2 +···+ na n |≤1. 21A0AAB0D8 braceleftBig n parenleftbig e? parenleftbig 1+ 1 n parenrightbig n parenrightbig bracerightBig B7C9BOA2 22A0CNDJB0D8{ 3 √ n 2 (2n?5)}(n ≥ 3)AXBACHB0D8A2 23A0AAB0D8 braceleftBig n 2 parenleftBig 1 2 n+1 parenrightBigbracerightBig CU?D4B0CQB7BQA2 24A0CF?CPBI a(a>0)B7A9B7A0CMCCCXCUA7AAB7C7D4B0B7CCCXA2 25A0CNDJA9AJ f CFCLBTBDC3C9ANAXBXBAA3C6BXACA5B7A7CG f CFA4BDC3C9A0CSBHC5C1BTC9CQA2 26A0AA f(x)=x 2 ln(ax)(a>0)B7BVBBA7B2a AIBEARBVBBB7BXC8AXASDGAB 27A0CNDJC4B8AVA9 arctan x =arcsin x √ 1+x 2 ,?∞ <x+∞. 28A0AQf CF[a,+∞)A0CWBGA7A7 lim x→+∞ f(x) x =0A7CNDJAHC5BBD8{ξ n },ξ n → +∞(n →∞),AUB6 lim n→∞ f prime (ξ n )=0. 29A0AAC0B0 y = x 3 √ 1+x B7 nCNB3B0A2 CYA9CF R n CUCVACC9C1AGACC9B7BDC3A7AIBYC1BTCAC3AXB5DCD3B9A3CCA7A4CUAHC2D5BBA7BFC5BFAF DDCFBRCAC3B7D3BVADBPCKCID5BBD3CMA5CXA5B7CVCACLCVACC9A7CVACC9B7AGAFCLAGACC9A2CFR n CUBACA C3B7BDC3A7AIBYC1BTCAC3E ? R n CUB7AHC2D5BBx 1 ,x 2 A7BFC5 x = tx 1 +(1?t)x 2 ∈ E,?t ∈ [0, 1]A7CQ ARCAC3E AXBACAA2 30A0AQ G 1 ,G 2 AX R n AHC2CVCAA7A7 G 1 ∩ G 2 = ?, AYCNDJA9 G 1 ∩ G 2 = ?. 31A0AQE ? R n BIAHC2CAC3A7E prime AJAWE B7AFB7CSBBD3ASB7CAC3A7ARBIE B7B3CAA7AYCNE prime BIAG CAA2 32A0AQ A, B ? R n BICVCAA7A∩ B = ?.AYCNA9 ?(A ∪ B)=?A∪ ?B. 33 A0AQ A, B ? R n BIC5COAGCAA7 A ∩ B = ? A2AYCNA9 ? CVCA W, V AUB6 A ? W, B ? V , A7 W ∩ V = ? 34A0AGBDBND8C0B0B7BDC3C9A2 (1) u = √ 1 ?x 2 + radicalbig 1 ?y 2 ;(2)u = radicalBigg 2x?x 2 ?y 2 x 2 + y 2 ?x ; (3) u =arcsin y x 4u =ln(?1 ?x 2 ?y 2 + z 2 ). 5 35A0AABND8C0B0B7C9BO (1) lim (x,y)→(0,0) e x + e y cos x +siny ; (2) lim (x,y)→(0,0) x 2 y 3 2 x 4 + y 4 ; (3) lim (x,y)→(+∞,+∞) (x 2 + y 2 )e x+y ; (4) lim (x,y)→(0,0) sin(x 3 + y 3 ) x 2 + y 2 . 35A0BGBND8C0B0 f(x, y), CNDJ lim (x,y)→(0,0) f(x, y) AOAZCFA8 (1) f(x, y)= x 2 x 2 + y 2 ;(2)f(x, y)= x 2 y 2 x 3 + y 3 . 36A0BJBND8C0B0AXBPCFAFA3DID3BVA7BIASDGAB (1) f(x, y)= ? ? ? ? ? ? ? x 2 ?y 2 x 2 + y 2 ,x 2 + y 2 negationslash=0, 0,x 2 + y 2 =0; (2) f(x, y)= ? ? ? ? ? | sin(xy) x |,xnegationslash=0, y, x =0; (3) f(x, y)= ? ? ? ? ? ? ? ? x 2 y 2 e ? x 4 y 2 ,ynegationslash=0, 0,y=0; (4) f(x, y)= ? ? ? ? ? y 2 ln(x 2 + y 2 ),x 2 + y 2 negationslash=0, 0,x 2 + y 2 =0. 37A0AQC0B0f(x, y)CFCV?A3DIx>0AND3BVA7A7BG?y 0 ,C9BO lim (x,y)→(0 + ,y 0 ) f(x, y)=φ(y 0 )AZCFA2 B2C0B0 f CF y CWANAMATBDC3 φ(y) C5A7CNDJC0B0 f CFAG?A3DI x ≥ 0 AND3BVA2 38A0AQC0B0f(x, y)CFCV?A3DIx>0ANC1CTD3BVA2CNDJA9(1) ?y 0 ,C9BO lim (x,y)→(0 + ,y 0 ) f(x, y)=φ(y 0 ) AZCFAA (2) C0B0 f CF y CWANAMATBDC3 φ(y) C5B4B6C0B0CFx ≥ 0 ANC1CTD3BVA2 39 A0AQ u = f(x) CFBB x 0 ∈ R n D3BVA7A7 f(x 0 ) > 0, CNDJA9AZCF x 0 B7C1BTD9C9 U(x 0 ,δ)(δ>0) AUB6 f CF U(x 0 ,δ) AN?CMCQA2 40A0AQ E AX R n CUAHC2BBCAA2AACNA9 d(x, E)(x B4CAC3E B7CTCZ) CFR n ANC1CTD3BVA2 41 A0AQC0B0 f CF R n AND3BVA7BGAHC2ATB0α A7D5CAC3G = {x vextendsingle vextendsingle f(x) >α},F= {x vextendsingle vextendsingle f(x) ≥ α}AA CNA9 G AX R n CUB7CVCAA7F AXR n CUB7AGCAA2 41A0AQ x ∈ R n , x =(x 1 ,x 2 ,···,x n ) A2AACNA9 (1) ?a>0,b>0,AUB6 a|x|≤ n summationdisplay j=1 |x j |≤b|x|; (2) ?a>0,b>0,AUB6 a|x|≤ max 1≤j≤n |x j |≤b|x|. 42A0AQ ? ? R n BIC5COAGACC9A7 f AX ?B4 R n A0B7B1APA7D3BVA2AACNA9 f ?1 CF f(?) AND3BVA2 43A0AQ f(x, y) AVCPBPx = aC8 y = bBEC5BDC3A2DFD2 (1) lim y→b f(x, y)=φ(x) AZCFAA (2) lim x→a f(x, y)=ψ(y) C1CTAZCF (CCA7 ?ε>0,?δ(ε) > 0, B2 0 < |x ? a| <δARA7 ?y negationslash= b, C5 |f(x, y) ?ψ(y)| <ε); 6 (3) lim y→b ψ(y)=cAZCFA8 CNDJA9 (i) CYAYC9BO lim x→a lim y→b f(x, y) = lim x→a φ(x)=c AZCFAA (ii) CYAYC9BO lim y→b lim x→a f(x, y) = lim y→b ψ(y)=cAZCFAA (iii) BJCVC9BO lim (x,y)→(a,b) f(x, y)=c. 44 A0AQC0B0 u = f(x, y, z) CFAGD2BLB7 a ≤ x ≤ b,a ≤ y ≤ b,a ≤ z ≤ b AND3BVA7DB φ(x)= max a≤y≤b { min a≤z≤b f(x, y, z)}. AYCNDJA9 φ CF[a,b] AND3BVA2 45 A0AJC0B0 f(x, y) CFCRBT D : ?a ≤ x ≤ a,?b ≤ y ≤ b (a>0,b>0) ANBNAKBI x C1 y B7D3BVC0 B0A7BIA7f(0, 0) = 0A2B2xBUBDARA7f(x, y)AXy B7BXBSBACIC0B0A7CGC5δ>0A7AUBGDHBTx ∈ (?δ, δ) C5 y ∈ (?b,b)DFD2 f(x, y)=0. Hint: Because the function f(0,y) is strictly decreasing on the interval [?b,b], and f(0, 0) = 0, then f(0,b) < 0, f(0,?b) > 0. Further, consider to use the continuity of the function f. 46A0AJ f(x, y) C0B0CFxy A3DIANAWAWD3BVA7AYCNA3DIBBCAE = {(x, y)|f(x, y)=0}BIAGCAA2 47A0AQf CFR n AND3BVA7x negationslash= O(R n CUB7CBBB)ARA7f(x) > 0,A7?x ∈ R n ,CBc>0C5f(cx)=cf(x). AYCNDJA9?a,b > 0,AUB6 a|x|≤f(x) ≤ b|x|, (?x ∈ R n ). 48A0AQAAXn×nCRCKA7detA negationslash=0.AYCNDJA9?α>0,β>0AUB6?x ∈ R n C5β|x|≤|Ax|≤α|x|. 49 ? A0AQ f : R n → R,DFD2AIBNAKB8CJA9 (1)?x ∈ R n ,f(x) ≥ 0,A7 x = O ?? f(x)=0; (2)?λ ∈ R,C5 f(λx)=|λ|f(x); (3)?x, y ∈ R n ,f(x + y) ≤ f(x)+f(y). AYCNDJA9 (i)?M>0, AUB6?x ∈ R n , C5 f(x) ≤ M|x|; (ii) f DFD2 LipschitzB8CJA9CC?L>0,AUB6|f(x) ?f(y)|≤L|x?y|(?x, y ∈ R n ); (iii)?AQB0 α>0 AUB6?x ∈ R n C5 α|x|≤f(x). 50 A0AQ f CF R n CUBB x 0 B7D9C9D1C5COA7CE M f (x 0 ,δ)=sup{f(x) vextendsingle vextendsingle x ∈ R n , |x ? x 0 | <δ}, m f (x 0 ,δ)=inf{f(x) vextendsingle vextendsingle x ∈ R n , |x?x 0 | <δ}A2CGA31 A5C9BOω f (x 0 ) =: lim δ→0 + bracketleftbig M f (x 0 ,δ)?m f (x 0 ,δ) bracketrightbig AZ CFA7ALARCOBIf CF x 0 AWB7CJBQAAA32 A5 f(x) CF x 0 D3BVB7ATBNAHBZB8CJAX ω f (x 0 )=0. 51A0AQCAC3E ? R n AXBMAGB7A7C0B0f CFE ANC1CTD3BVA2AYCNDJA9f CRA1BHC1B9D3BVBYBDB4E ANA7CCBHC1B9AZCF E ANB7C0B0 F DFD2 F(x)=f(x),x∈ E, AUB6 F CF E ANC1CTD3BVA2 52A0AQ f CF R n AND3BVA7 lim |x|→+∞ f(x) AZCFA2AYCNDJ f CF R n ANC1CTD3BVA2