习题1.1
1.证明3为无理数.pp22证 若3不是无理数,则3?,p,q为互素自然数.3?2,p?3q2.3除尽p2,qq必除尽p,否则p?3k?1或p?3k?2.p2?9k2?6k?1,p2?9k2?12k?4,3除p2将余1.故p?3k,9k2?3q2,q2?3k2,类似得3除尽q.与p,q互素矛盾.2.设p是正的素数,证明p是无理数.aa22证 设p?,a,b为互素自然数,则p?2,a?pb2,素数p除尽a2,故p除尽a,bba?pk.p2k2?pb2,pk2?b2.类似得p除尽b.此与a,b为互素自然数矛盾.3.解下列不等式:(1)|x|?|x?1|?3.\\;(2)|x2?3|?2.解 (1)若x?0,则?x?1?x?3,2x??2,x??1,(?1,0);若0?x?1,则x?1?x?3,1?3,(0,1);若x?1,则x?x?1?3,x?3/2,(1,3/2).X?(?1,0)?(0,1)?(1,3/2).(2)?2?x2?3?2,1?x2?5,1?|x|2?5,1?|x|?5,x?(1,5)?(?5,?1).4.设a,b为任意实数,(1)证明|a?b|?|a|?|b|;(2)设|a?b|?1,证明|a|?|b|?1.证(1)|a|?|a?b?(?b)|?|a?b|?|?b|?|a?b|?|b|,|a?b|?|a|?|b|.(2)|a|?|b?(a?b)|?|b|?|a?b|?|b|?1.5.解下列不等式:(1)|x?6|?0.1;(2)|x?a|?l.解(1)x?6?0.1或x?6??0.1.x??5.9或x??6.1.X?(??,?6.1)?(?5.9,??).(2)若l?0,X?(a?l,??)?(??,a?l);若l?0,x?a;若l?0,X?(??,??).a?16.若a?1,证明0?na?1?,其中n为自然数.n证若a?1,显然na?b?1.a?1?na?1?(na?1)(bn?1?bn?2?7.设(a,b)为任意一个开区间,证明(a,b)中必有有理数.证取自然数n 满足1/10n?b?a.考虑有理数集合mA=An?{n|m?Z}. 若An?(a,b)??,则A?B?C,B?A?{x|x?b},10C?A?{x|x?a}.B中有最小数m0/10n,(m0?1)/10n?C,b?a?m0/10n-(m0?1)/10n=1/10n,此与n的选取矛盾. 8.设(a,b)为任意一个开区间,证明(a,b)中必有无理数.证取自然数n 满足1/10n?b?a.考虑无理数集合An?{2?m|m?Z}. 以下仿8题.10nn?1)?n(na?1).习题1.2
13.证明函数y?1?x?x在(1,??)内是有界函数.(1?x?x)(1?x?x)11??(x?1).1?x?x1?x?x2?1x6?x4?x213.研究函数y?在(??,??)内是否有界. 61?xx6?x4?x2x6?x4?x23x6解|x|?1时,?3,|x|?1时,?6?3,1?x61?x6x|y|?y?3,x?(??,??).证y?1?x?x?习题1.4
1.直接用?-?说法证明下列各极限等式:(1)limx?ax?a(a?0);(2)limx2?a2;(3)limex?ea;(4)limcosx?cosa.x?ax?ax?a证(1)???0,要使|只需x?a|?|x-a||x-a||x-a|??,由于?,x?ax?aax?a. |x?a|??,|x?a|?a?.取??a?,则当|x?a|??时,|x?a|??,故limx?aa(2)???0,不妨设|x?a|?1.要使|x2?a2|?|x?a||x?a|??,由于|x?a|?|x?a|?|2a|?1?|2a|,只需(1?|2a|)|x?a|??,|x?a|?|x2?a2|??,故limx2?a2.x?a.取??min{,1},则当|x?a|??时,1?|2a|1?|2a|??(3) ???0,设x?a.要使|ex?ea|?ea(ex?a?1)??,即0?(ex?a?1)??ea,1?ex?a?1??ea,????0?x?a?ln?1?a?,取??min{,1},则当0?x?a??时,|ex?ea|??,e?1?|2a|?故limex?ea. 类似证limex?ea.故limex?ea.x?a?x?a?x?a(4)???0,要使|cosx?cosa|?2sinx?ax?ax?ax?asin?2sinsin?|x?a|,2222x?a取???,则当|x?a|??时,|cosx?cosa|??,故limcosx?cosa.2.设limf(x)?l,证明存在a的一个空心邻域(a??,a)?(a,a??),使得函数u?f(x)在x?a该邻域内使有界函数.证对于??1,存在??0,使得当 0?|x-a|??时,|f(x)?l|?1,从而|f(x)|?|f(x)?l?l|?|f(x)?l|?|l|?1?|l|?M. 3.求下列极限:(1)limx?0(1?x)2?12x?x2x?lim?lim(1?)?1.x?0x?02x2x222??x??x??2sin??sin????1?cosx?2??1lim??2???112?1.(2)lim?lim22x?0x?0xxx2x?0?22???2??(3)limx?0
x?a?xa?limx?0xx(x?a?a)?12a(a?0).x2?x?2?x?12x2?2x?3x2?x?2(5)lim?x?02x2?2x?3(4)lim?2.?3?2.?3
(2x?3)20(2x?2)10230(6)lim?30?1.x??(2x?1)302(7)limx?01?x?1?x2x?lim?1.x?0x(1?x?1?x)x3?x2?x?1?3x2?x?2?1(8)lim??3??lim?lim2x??1x?1x??1x?1?(x?1)(x?x?1)x??1(x?1)(x2?x?1)?(x?1)(x?2)(x?2)?3?lim?lim???1.x??1(x?1)(x2?x?1)x??1(x2?x?1)3(9)limx?41?2x?3(1?2x?3)(x?2)(1?2x?3)?limx?4x?2(x?2)(x?2)(1?2x?3)?lim(2x?8)(x?2)244??.x?4(x?4)(1?2x?3)63nnn(n?1)2y?x?1(1?y)?12(10)lim?lim?limx?1x?1y?0y?0yy2(11)limx2?1?x2?1?lim?0.22x??x??x?1?x?1a0xm?a1xm?1??amam(12)lim(b?0)?.nx?0bxn?bxn?1??bnbn01ny??yn?n.??
(13)lima0x?a1x?x??bxn?bxn?1?01mm?1?a0/b0,m?n?am?(a0b0?0)??0, n?m?bn??, m?n.?x4?81?8/x4(14)lim2?lim?1.x??x?1x??1?1/x21?3x?31?2x(15)limx?0x?x2?limx?03(1?3x?1?2x)(1?3x?1?3x1?2x?1?2x)(x?x2)(31?3x?31?3x31?2x?31?2x)5x222233323332?limx?0x(1?x)(31?3x?31?3x31?2x?31?2x)55?lim?.22x?033333(1?x)(1?3x?1?3x1?2x?1?2x)(16)a?0,lim??lim?x?a?0???x?a1??lim????2222x?a?0x?a?0?x?ax?a?x?a?(x?a)(x?a)1???x?ax?a(x?a)x?a??x?a?x?a