第一章 函数·极限·连续
一. 填空题
1. 已知f(x)?sinx,f[?(x)]?1?x2,则?(x)?__________, 定义域为___________. 解. f[?(x)]?sin?(x)?1?x2, ?(x)?arcsin(1?x2) ?1?1?x2?1, 0?x2?2,|x|??1?x?2．设lim??x??x??axa2
????ttedt, 则a = ________.
a解. 可得ea??a??ttedt=(te?e)tt?ae??a?e, 所以 a = 2.
a3. lim?n???21?n?n?11??2n?n?222?????=________. 2n?n?n?nn解. <
n?n?n122n?n?n222???n?n?nn22
12所以
n?n?1n?n?21?2???n1n?n?n2?n?n?nn?n?1n?n?1n?n?11?2???n2n?2???2<2< 2n?n?1n?n?2n?n?nn?n?1???
n(1?n)1?2???nn?n?n1?2???nn?n?122?12?, (n??) 2n?n?n2n(1?n)12?, (n??) 2n?n?121?2n?n?2|x|?1|x|?12?所以 lim?n???2?n?n?1?1????1?= 2n?n?n?2n4. 已知函数f(x)??
?0, 则f[f(x)] _______.
解. f[f(x)] = 1. 5. lim(n?3n?n??n?n)=_______.
(n?3n?n?n)(n?3n?n?nn?n)解. lim(n?3n?n??n?n)?limn??
n?3n? 1
=limn?3n?n?n?3n?n?nnn???2
6. 设当x?0时,f(x)?ex?1?ax1?ax1?bxx为x的3阶无穷小, 则a?_____,b?______.
e?x解. k?limx?0e?bxe?1?axe?bxe?1?ax1?bx ?lim?lim333x?0x?0xx(1?bx)xxxxx ?lime?bex?bxe2x?ax?03xe?2bexx2 ( 1 )
?lim?bxexx?06xx ( 2 )
x 由( 1 ): lim(e?be?bxex?0xxx?a)?1?b?a?0
x 由( 2 ): lim(e?2be?bxe)?1?2b?0
x?0 b??12,a?12
1??=______. x??limx?sinxx37. limcotx?x?0?1?sinx??解. limcosxsinxx?sinxxsinxx?0x?0?lim1?cosx3x2x?0?limsinx6xx?0?16
8. 已知limnk1990kn??n?(n?1)n1990k?A(? 0 ? ?), 则A = ______, k = _______.
解. limn??n?(n?1)k?limnkn1990n??k?1???A
1k11991所以 k－1=1990, k = 1991;
二. 选择题
1k?A，A??
1. 设f(x)和?(x)在(－?, +?)内有定义, f(x)为连续函数, 且f(x) ? 0, ?(x)有间断点, 则 (a) ?[f(x)]必有间断点 (b) [ ?(x)]2必有间断点 (c) f [?(x)]必有间断点 (d)
?(x)f(x)必有间断点
解. (a) 反例 ?(x)??
?0?1|x|?1|x|?1, f(x) = 1, 则?[f(x)]=1
2
(b) 反例 ?(x)???1??1
|x|?1|x|?1, [ ?(x)] = 1
2
(c) 反例 ?(x)??
?0?1|x|?1|x|?1, f(x) = 1, 则f [?(x)]=1
(d) 反设 g(x) = 以(d)是答案.
?(x)f(x)在(－?, +?)内连续, 则?(x) = g(x)f(x) 在(－?, +?)内连续, 矛盾. 所
2. 设函数f(x)?x?tanx?esinx, 则f(x)是
(a) 偶函数 (b) 无界函数 (c) 周期函数 (d) 单调函数 解. (b)是答案. 3. 函数f(x)?|x|sin(x?2)x(x?1)(x?2)2在下列哪个区间内有界
(a) (－1, 0) (b) (0, 1) (c) (1, 2) (d) (2, 3) 解. limf(x)??,x?1limf(x)??,x?0f(0?)?sin24,f(0?)??sin24
所以在(－1, 0)中有界, (a) 为答案. 4. 当x?1时,函数x?1x?121ex?1的极限
(a) 等于2 (b) 等于0 (c) 为? (d) 不存在, 但不为? 解. limx?1???ex?1?lim(x?1)ex?1??x?1x?1?0?352n?1?211x?1?0x?1?0. (d)为答案.
x?1?2???25. 极限lim?2的值是 222?n??1?22?3n?(n?1)??(a) 0 (b) 1 (c) 2 (d) 不存在
?2???2解. lim?2 222?n??1?22?3n?(n?1)???lim?1??1, 所以(b)为答案. =lim?2?2?2?2???2?2?2?n??1n??223n(n?1)?(n?1)????111111??1??352n?1?6. 设lim(x?1)(ax?1)(x?1)250955x???8, 则a的值为
(a) 1 (b) 2 (c)
58 (d) 均不对
3
解. 8 = lim(x?1)(ax?1)(x?1)95250955x??=lim(x?1)952/x(ax?1)/x509555x??(x?1)5/x100
=lim(1?1/x)(a?1/x)(1?1/x)25055?a, a?8, 所以(c)为答案.
x??7. 设lim(x?1)(x?2)(x?3)(x?4)(x?5)(3x?2)13?x????, 则?, ?的数值为
135(a) ? = 1, ? = 解. (c)为答案.
(b) ? = 5, ? =
13 (c) ? = 5, ? = (d) 均不对
8. 设f(x)?2x?3x?2, 则当x?0时
(a) f(x)是x的等价无穷小 (b) f(x)是x的同阶但非等价无穷小 (c) f(x)比x较低价无穷小 (d) f(x)比x较高价无穷小 解. lim2?3?2xxxx?0=lim2ln2?3ln31xxx?0?ln2?ln3, 所以(b)为答案.
9. 设lim(1?x)(1?2x)(1?3x)?axx?0?6, 则a的值为
(a) －1 (b) 1 (c) 2 (d) 3
解. lim(1?x)(1?2x)(1?3x)?a?0, 1 + a = 0, a = －1, 所以(a)为答案.
x?010. 设limatanx?b(1?cosx)cln(1?2x)?d(1?e?x2x?0?2，其中a?c?0, 则必有 )22(a) b = 4d (b) b =－4d (c) a = 4c (d) a =－4c
aatanx?b(1?cosx)cln(1?2x)?d(1?e?x2解. 2 =limx?0=limcosx)x?0?2c?2xde1?2x2?bsinx???x2a2c, 所以a =－4c, 所以(d)
为答案.
三. 计算题 1. 求下列极限
1(1) lim(x?e)x
x???1ln(x?e)xx???xx解. lim(x?e)x?limex???x?ex???limln(x?e)xx?ex???x?e?e?e
lim1?exx1(2) lim(sinx??2x?cos1x)
x 4
解. 令y?2x1x
1x1lim(sinx???cos)?lim(sin2y?cosy)=ey?0xylimy?0ln(sin2y?cosy)ylim2cos2y?sinysin2y?cosy2?e
?ey?0?1?tanx?(3) lim??x?01?sinx???1?tanx?解. lim??x?01?sinx??1x3
1x1x3tanx?sinx???lim?1??x?01?sinx??tanx?sinx1?sinx33
x)x??(1?sintanx?sinxlimtanx?sinxtanx?sinx3??x?0x? ?lim??1?=e ?x?0??1?sinx???? =elimx?0sinx(1?cosx)x3sinx?2sin2x21=elimx?0x3?e2.
2. 求下列极限 (1) limln(1?33x?1)2
x?1arcsin2x?13解. 当x?1时, ln(1?代换 limln(1?33x?1)~3x?1, arcsin233x?1~2x?1. 按照等价无穷小12x?13232x?1)2x?1?limx?12arcsin2x?1x?12x?13?limx?1?1223
?1?2(2) lim?2?cotx?
x?0?x?解. 方法1：
2?sin2x?x2cos2x??1cosx??1?2???lim?2?cotx?=lim??2222?=lim? x?0?xx?0?x?0xsinxsinxx???????1?(x2?1)cos2x???2xcos2x?2(x2?1)cosxsinx??? =lim?=lim? 43???x?0?x?0x4x???? =lim?2xcosx?sin2x4x?2cos232x?0?lim2xcosxsinx4x32x?0
=limx?4xcosxsinx?2cos2x12x2x?0?12
5