最新-北大版高等数学课后习题答案完整版-精品

习题1.12222222222222222223.33,,.3,3.3,,3132.961,9124,31.3,93,3,3.,,.,,,,pppqpqpqqppkpkpkkpkkppkkqqkqpqppaapabpapbbb证明为无理数若不是无理数,则为互素自然数除尽必除尽否则或除将余故类似得除尽与互素矛盾.设是正的素数证明是无理数设为互素自然数,则素证2.证1.2222222,,.,..,:(1)|||1|3.\;(2)|3|2.0,13,22,1,(1,0);01,13,13,(0,1);1,13,3/2,(1,3/2).(1,0)(0,1)papaapkpkpbpkbpbabxxxxxxxxxxxxxxxX数除尽故除尽类似得除尽此与为互素自然数矛盾.解下列不等式若则若则若则3.解(1)222(1,3/2).(2)232,15,1||5,1||5,(1,5)(5,1).,(1)||||||;(2)||1,||||1.(1)|||()|||||||||,||||||.(2)|||()||||||xxxxxabababababaabbabbabbababababbabb设为任意实数证明设证明证4.,|1.(1)|6|0.1;(2)||.60.160.1.5.96.1.(,6.1)(5.9,).(2)0,(,)(,);0,;0,(,).11,01,.1,1.11nnnnxxalxxxxXlXalallxalXaaannaabaa解下列不等式或或若若若若证明其中为自然数若显然解(1)证5.:6.120000(1)(1)(1).(,),(,).1/10.{|}.(,),,{|},10{|}./10,(1)/10,/10(1)/101/10nnnnnnnnnnnnnabbnaababnbamAAmAabABCBAxxbCAxxaBmmCbammZ设为任意一个开区间证明中必有有理数取自然数满足考虑有理数集合=若则中有最小数-=证7.(,),(,).1/10.{2|}.10nnnnababmnbaAmZ,此与的选取矛盾.设为任意一个开区间证明中必有无理数取自然数满足考虑无理数集合以下仿8题.8.证习题1.26426642642666613.1(1,)(1)(1)111(1).112113.(,).13||13,||1,3,11||3,(,).yxxxxxxyxxxxxxxxxxyxxxxxxxxxxxxxyyx证明函数在内是有界函数.研究函数在内是否有界时,时证解习题1.4221.-(1)lim(0);(2)lim;(3)lim;(4)limcoscos.|-||-||-|1)0,||,,||,||.,||,||,lim.(2)0xaxaxaxaxaxaxaaxaeexaxaxaxaxaxaxaaxaxaaaxaxaxaa直接用说法证明下列各极限等式:要使由于只需取则当时故证(222222,||1.||||||,|||||2|1|2|,1|2|)||,||.min{,1},||,1|2|1|2|||,lim(3)0,.||(1),01),1xaxaaxaxaaxaxaxaxaxaxaaaaxaxaxaaaxaxaxaeeeeee不妨设要使由于只需(取则当时故设要使即(.1,0ln1,min{,1},0,||,1|2|limlimlim0,|coscos|2sinsin2sinsin||,2222,|,|coscosxaaxaaxaxaxaxaxaxaeexaxaeeeaeeeeeexaxaxaxaxaxaxaxa取则当时故类似证故要使取则当|时...(4)20|,limcoscos.2.lim(),(,)(,),().1,0,0|-|,|()|1,|()||()||()|||1||.(1)1(1)limlim2xaxaxxxafxlaaaaaufxxafxlfxfxllfxlllMxx故设证明存在的一个空心邻域使得函数在该邻域内使有界函数对于存在使得当时从而求下列极限证3.:20022222000002212202lim(1)1.222sinsin1cos11122(2)limlimlim1.22221(3)limlim(0).()222(4)lim.22332(5)lim22xxxxxxxxxxxxxxxxxxxaaxaxxxaaaxxxxxxxx2.332010303030002232211121(23)(22)2(6)lim1.(21)2112(7)limlim1.(11)13132(8)limlimlim11(1)(1)(1)(1)(1)(2)limlim(1)(1)xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx214442100(2)31.(1)3123(123)(2)(123)(9)limlim2(2)(2)(123)(28)(2)244lim.63(4)(123)(1)1(1)12(10)limlimlim.1(11)limxxxnnnxyyxxxxxxxxxxxxxxxxnnnyyyxynxyyx22221011001001010010142211lim0.11(12)lim(0)./,(13)lim(0)0,,.818(14)limlim1xmmmmnnnxnnmmmnnxnxxxxxaxaxaabbxbxbbabmnaxaxaabnmbxbxbmnxx42/1.11/xx33202233333322023333220333322033331312(15)lim(1312)(13131212)lim()(13131212)5lim(1)(13131212)55lim.3(1)(13131212)(16)0,lxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxa22220001imlim()()1lim()xaxaxaxaxaxaxaxaxaxaxaxaxaxaxa00()1lim()11lim.()2xaxaxaxaxaxaxaxaxaxaxaa000222200000sin14.lim1lim1sinsin(1)limlimlimcos.tansinsin(2)sin(2)2(2)limlimlim100323tan3sin2tan3sin2(3)limlimlimsin5sin5xxxxxxxxxxxxxexxxxxxxxxxxxxxxxxx利用及求下列极限:00()1/0321.sin5555(4)limlim2.1cos2sin2cossinsinsin22(5)limlimcos.2(6)lim1lim1lim1.(7)lim(15)xxxaxakxxxkkkkxxxyyxxxxxxxaxaxaaxaxakkkexxxy51/(5)50100100lim(15).111(8)lim1lim1lim1.5.lim()lim().lim():0,0,0|-|().lim(yyxxxxxxaxxaxyeexxxfxfxfxAxafxAfx给出及的严格定义对于任意给定的存在使得当时):0,0,().AxfxA对于任意给定的存在使得当时习题1.52222222222221.(1)10(2)sin5.(1)0,|110|.,1111,||,,|||110|,10555()(2)(1)0,|sin5sin5|2|cos||sin|.22xxxxaxxxxxxxxxxxxxaxaxa试用说法证明在连续在任意一点连续要使由于只需取则当时有故在连续.要使由于证000000555()2|cos||sin|5||,5||,||,225,|||sin5sin5|,sin55()()0,0||()0.(),()/2,0||(xaxaxaxaxaxaxaxxayfxxfxxxfxfxxfxxxfx只需取则当时有故在任意一点连续.2.设在处连续且证明存在使得当时由于在处连续对于存在存在使得当时证000000000000)()|()/2,()()()/2()/20.3.()(,),|()|(,),?(,),.0,0|||()()|,||()||()|||()()|,||.fxfxfxfxfxfxfxabfxabxabfxxxfxfxfxfxfxfxfx于是设在上连续证明在上也连续并且问其逆命题是否成立任取在连续任给存在使得当时此时故在连续其证220001,,(),()|11,ln(1),1,1,0,(1)()(2)()arccos,1.0;lim()lim11(0),lim()(0)xxxxfxfxxaxxxxfxfxaxxaxxfxxffxf逆命题是有理数不真例如处处不连续但是|处处连续.是无理数4.适当地选取,使下列函数处处连续:解(1)0111122sin2limsin301.(2)lim()limln(1)ln2(1),lim()limarccos(1)ln2,ln2.5.3:11(1)limcoscoslimcos01.(2)lim2.(3)limxxxxxxxxxxxxafxxffxaxafaxxxxxxxee利用初等函数的连续性及定理求下列极限sin22sin334422.88(4)limarctanarctanlimarctan1.114xxxxexxxx0000022222222()()(ln())()(5)lim(12)||lim(12)||3||33limlim.21211/12/6.lim