高等数学 上册 答案

70.当n?1时，x1?2?2

设当n?k时，xk?2成立则当n?k?1时，xk?1?2?xk?2?2?2

由数学归纳法原理知：数列?xn?有上界为2

下面再用数学归纳法证明数列?xn?为单调增当n?1时，x2?2?x1?2?2?2?x1设n?k时，xk?1?xk成立则当n?k?1时，xk?1?1?xk?2?2?xk?1?2?xk?xk?1所以数列?xn?单调增

根据单调有界数列必有极限，知limxn存在

x??可设limxn?A，则A?2?0n??由limxn?1?lim2?xn，得A?2?An??n??

解得：A1?2，A2??1(舍去），因此limxn?2

n??1?xsinx?cosx x?0xtanx(1?sinx?cosx)1131xsinx1?cosx(?) ?(1?)? ?limx?02xtanxxtanx22471. 原式?limsinxcos??sin?cosx x??(x??)cosxcosxsin(x??)1?lim? x??x??cosxcosx12?sec? ?2cos?72. 原式?lim73. 原式?lim(1?tanx)?(sinx?1) x?0x3(1?tanx?1?sinx)1sinx(1?cosx)?lim 2x?0x31sinx1?cosx?lim? 2x?0xx21? 4 74. f(x)?(ax?1)(2x?1)

(ax?1)(x?a)x?1(1)当a?1时，limf(x)?lim2x?1??

x?1x?1(2)limf(x)?limx?12x?111??x?1x?a1?a2

得a??1(3)lim(ax?1)(2x?1)?01x? 1故欲使limf(x)?0，必须lim(x?a)??a?0112x?x?222即a?1 21(x?1)(2x?1)2limf(x)?lim?2

1111x?x?x?1)(x?)22(222e3x?3e?2x2?3e?5x1?lim? 75. limx???4e3x?e?2xx???4?e?5x22e3x?3e?2x2e5x?3而lim?lim??3 x?－?4e3x?e?2xx???4e5x?12e3x?3e?2x因此lim不存在. x??4e3x?e?2x1?cosx2x76. 原式?lim ? x??133?sinxx2?3n1077. 原式?lim??15 n??32?n?110102?78. 原式?lim3 n(n??1n2)?()n?1 ?3 331k?1k?1 )??2kkk1324n?1n?1原式?lim(?)(?)?(?) n??2233nn1n?1 ?lim n??2n1 ? 279. 由(1?an?0 所以limnnn??因为lima?02?an??80.当a?1时，当a?1时，an1?lim?1 所以lim1nn??2?ann??1n因为lim()?02()?1n??aa1?n(n?1)n2??? 81. 原式?lim?2n??n?22??1?n ?lim ?? n??2(n?2)2 82.B 83.B 84.C 85.C 86.A 87.D 88.D 89.D 90.B 91.B

92.A 93.C 94.C 95.B 96.C 97.C 98.D 99C 100.C 101.D

102.B 103.D 104.A 105.C 106.D 107.C 108.B 109.D

110.解法一：若x?0是f(x)的可去间断点，则limf(x)存在x?0

从而limf(x)?sin2x?lim(1?sinx?sin2x?a?bsinx)?0x?0x?0

故a?lim(1?sinx?sin2x?bsinx)?1

x?0再由limf(x)?sinx?0得x?01?sinx?sin2x?1lim(?b)?0x?0sinx即b?limx?0

1?sinx1?sinx?sin2x?1?1 2故当a?1，b?1时，x?0是f(x)的可去间断点 2

解法二：若x?0是f(x)的可去断点，则必极限limf(x)存在x?0而limsin2x?0　所以必须x?0lim(1?sinx?sinx?(a?bsinx))?0x?02

求得：a?1，此时1?sinx?sin2x?(1?bsinx) limf(x)?lim2x?0x?0sinx(1?2b)?(1?b2)sinx　　　?limx?0sinx1?sinx?sin2x?(1?bsinx)??仅当1?2b?0，即b?综上述，a?1，b?111. f(x)?1时，上面极限存在 21时，x?0是f(x)的可去断点 2(x?1)(x?1)，x?0与x?1是f(x)的间断点

x(x?1)因为：lim(x?1)(x?1)??

x?0x(x?1)