解 limsin?x??limsin?x???
x?0x?0?xx (2)limtan3x?
x?0x 解 limtan3x?3limsin3x?1?3?
x?0x?03xxcos3x (3)limsin2x?
x?0sin5x 解 limsin2x?limsin2x?5x?2?2?
x?0sin5xx?02xsin5x55 (4)limxcotx?
x?0 解 limxcotx?limx?cosx?limx?limcosx?1? x?0x?0sinxx?0sinxx?0 (5)lim1?cos2x?
x?0xsinx21?cos2x1?cos2x2sinx?2lim(sinx)2?2? 解 lim?lim?limx?0xsinxx?0x?0x?0xx2x221?cos2x2sinx?2limsinx?2? 或 lim?limx?0xsinxx?0xsinxx?0x (6)lim2nsinx(x为不等于零的常数)?
n??2nsinx2n?x?x? 解 lim2nsinx?limn??2nn??x2n 2? 计算下列极限?
1(1)lim(1?x)x? x?0 解
1lim(1?x)xx?01(?1)(??lim[1?(?x)]x)x?01(??{lim[1?(?x)]x)}?1?e?1?
x?0
1(2)lim(1?2x)x? x?011?21解 lim(1?2x)x?lim(1?2x)2x?[lim(1?2x)2x]2?e2?
x?0x?0x?0
(3)lim(1?x)2x? x??x 解 lim(1?x)2x?[lim(1?1)x]2?e2?
x??xx??x (4)lim(1?1)kx(k为正整数)?
x??x 解 lim(1?1)kx?lim(1?1)(?x)(?k)?e?k?
x??x??x?x 3? 根据函数极限的定义? 证明极限存在的准则I?? 证明 仅对x?x0的情形加以证明?
设?为任一给定的正数? 由于limg(x)?A? 故由定义知? 对??0? 存在?1?0? 使
x?x0得当0?|x?x0|??1时? 恒有|g(x)?A|??? 即
A???g(x)?A???
由于limh(x)?A? 故由定义知? 对??0? 存在?2?0? 使得当0?|x?x0|??2时? 恒有
x?x0|h(x)?A|??? 即
A???h(x)?A???
取??min{?1? ?2}? 则当0?|x?x0|??时? A???g(x)?A??与A???h(x)?A?? 同时成立? 又因为
g(x)?f(x)?h(x)? 所以 A???f(x)?A??? 即 |f(x)?A|??? 因此limf(x)?A?
x?x0
证明 仅对x?x0的情形加以证明? 因为
limg(x)?A? limh(x)?A?
x?x0x?x0所以对任一给定的??0? 存在??0? 使得当0?|x?x0|??时? 恒有 |g(x)?A|??及|h(x)?A|???
即 A???g(x)?A??及A???h(x)?A???
又因为 g(x)?f(x)?h(x)? 所以 A???f(x)?A??? 即 |f(x)?A|??? 因此limf(x)?A?
x?x0
4? 利用极限存在准则证明? (1)lim1?1?1?
n??n 证明 因为1?1?1?1?1?
nn而 lim1?1且lim(1?1)?1?
n??n??n由极限存在准则I? lim1?1?1?
n??n (2)limn(21?21? ? ? ? ?21)?1?
n??n??n?2?n?n? 证明 因为
22n111n 2? ?n(2?? ? ? ? ?2)?n?n?n??n2?2?n?n?n2??22nn而 lim2?1? lim2?1? n??n?n?n??n??所以 limn(21?21? ? ? ? ?21)?1?
n??n??n?2?n?n? (3)数列2?
2?2? 2?2?2? ? ? ? 的极限存在?
证明 x1?2? xn?1?2?xn(n?1? 2? 3? ? ? ?)? 先证明数列{xn}有界?
当n?1时x1?2?2? 假定n?k时xk?2? 则当n?k?1时? xk?1?2?xk?2?2?2? 所以xn?2(n?1? 2? 3? ? ? ?)? 即数列{xn}有界?
再证明数列单调增? 因为
22?xn?xn?(xn?2)(xn?1) xn?1?xn?2?xn?xn?? ?2?xn?xn2?xn?xn而xn?2?0? xn?1?0? 所以xn?1?xn?0? 即数列{xn}单调增?
因为数列{xn}单调增加有上界? 所以此数列是有极限的? (4)limn1?x?1?
x?0 证明 当|x|?1时? 则有 1?x?1?|x|?(1?|x|)n ? 1?x?1?|x|?(1?|x|)n ? 从而有 1?|x|?n1?x?1?|x|? 因为 lim(1?|x|)?lim(1?|x|)?1?
x?0x?0根据夹逼准则? 有 limn1?x?1?
x?0 (5)lim?x[1]?1?
x?0x 证明 因为1?1?[1]?1? 所以1?x?x[1]?1?
xxxx 又因为lim?(1?x)?lim?1?1? 根据夹逼准则? 有lim?x[1]?1? x?0x?0x?0x
习题 1?7
1? 当x?0时? 2x?x2 与x2?x3相比? 哪一个是高阶无穷小？
232x?xx?x 解 因为lim?lim?0? x?02x?x2x?02?x所以当x?0时? x2?x3是高阶无穷小? 即x2?x3?o(2x?x2)?
2? 当x?1时? 无穷小1?x和(1)1?x3? (2)1(1?x2)是否同阶？是否等价？
23(1?x)(1?x?x2)1?x 解 (1)因为lim?lim?lim(1?x?x2)?3? x?11?xx?1x?11?x所以当x?1时? 1?x和1?x3是同阶的无穷小? 但不是等价无穷小?
1(1?x2)?1lim(1?x)?1? (2)因为lim2x?11?x2x?1所以当x?1时? 1?x和1(1?x2)是同阶的无穷小? 而且是等价无穷小?
2 3? 证明? 当x?0时? 有? (1) arctan x~x?
2 (2)secx?1~x?
2 证明 (1)因为limarctanx?limy?1(提示? 令y?arctan x? 则当x?0时?
x?0y?0tanyxy?0)?
所以当x?0时? arctanx~x?
2sin2x2sinxcosx?lim2?lim(2)2?1? (2)因为limsecx?1?2lim1?22x?012x?0xcosxx?0x?0xxx2222x所以当x?0时? secx?1~? 2 4? 利用等价无穷小的性质? 求下列极限? (1)limtan3x?
x?02xsin(xn) (2)lim(n? m为正整数)?
x?0(sinx)msinx? (3)limtanx?x?0sin3x (4)limsinx?tanx? x?0(31?x2?1)(1?sinx?1) 解 (1)limtan3x?lim3x?3?
x?02xx?02x21 n?mn??sin(xn)?limxm??0 n?m? (2)limmx?0(sinx)x?0x??? n?m1x2sinx(1?1)sinx?lim1?cosx?lim21? cosx?lim? (3)limtanx?x?0x?0x?0cosxsin2xx?0x2cosx2sin3xsin3x (4)因为
2x sinx?tanx?tanx(coxs?1)??2tanxsin~?2x?(x)2??1x3(x?0)? 2222x1x2(x?0)? 1?x?1?~3(1?x2)2?31?x2?1332x?1? 1?sinsinx~sinx~x(x?0)?
1?sinx?1
高等数学上册第六版课后习题答案 - 图文
