考研数学三 经济ch1 各章复习题目及答案

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

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=lim =lim方法2：

?2cosx?2cos2x12x?2sin2x24x?22x?0?1316?1213?lim124cosxsinx?4sin2x24x23x?0?13?12

13x?0?12?????

2?sin2x?x2cos2x??1cosx??1?2 lim?2?cotx?=lim?????2222?=lim? x?0?xx?0?x?0xsinxsinxx?????? =lim??x?0?1?(x?1)cos?x422x????1??21?(x?1)(cos2x?1)??2? =lim?4x?0x??????24?1(2x)(2x)24??1?(x?1)(1?1???0(x)?22!4!? =lim?4x?0??x????1164?2424x?0(x))?1?(2x?2x?2?2x?224 =lim?4x?0x?????? ???22 =lim34?

x?0x33. 求下列极限

x4(1) limnlnnn??(nn?1)

n解. limnlnnn??(n?1)?limnn?1nn??lnn 令nn?1?x limxln(1?x)x?0?1

(2) lim1?e1?e?nx?nxn??

x?0?1???0 x?0 ??1x?0?b??, 其中a > 0, b > 0 ??n解. lim1?e1?e?nx?nxn???(3) lim?n?????解. lim?n????na?2nna?2n?1?cb?? x?1/n,c?b/a alim????x?0?2?n1xlim?x?x?0??ae??ln(1?c)?ln2xx

6

=aelimx?0ln(1?c)?ln2?xxlimclnc?x?aex?01?cx?ac?aba?ab

?2?x2(1?cosx)?4. 设f(x)??1?1x??cost2dt?x0x?0x?0 x?0试讨论f(x)在x?0处的连续性与可导性.

1解. f?'(0)?limf(x)?f(0)?x?0x?lim?x?0?xxx0costdt?1x?lim?x?02?x0costdt?xx22

?limx?0cosx?1?2??lim?x?0122x22x2x2?0

(1?cosx)?1?lim?x?0 f?'(0)?limf(x)?f(0)?x?0 ?limx?0x2sinx?2x??lim?x?02(1?cosx)?xx323x2?lim?x?0x2(cosx?1)6x

?0

所以 f'(0)?0, f(x)在x?0处连续可导. 5. 求下列函数的间断点并判别类型

1(1) f(x)?2x?11

2x?111解. f(0)?limx?0?2?x1?1?1, f(0)?lim?x?0?2x?11??1

2x?12x?1所以x = 0为第一类间断点. ?x(2x??)x?0??2cosx(2) f(x)??

1?sinx?02?x?1?解. f(+0) =－sin1, f(－0) = 0. 所以x = 0为第一类跳跃间断点; limf(x)?limsinx?1x?11x?12不存在. 所以x = 1为第二类间断点;

f(??2)不存在, 而limx??x(2x??)?22cosx??2,所以x = 0为第一类可去间断点;

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limx??k??x(2x??)?22cosx??, (k = 1, 2, …) 所以x =?k???2为第二类无穷间断点.

1??x?0?xsin6. 讨论函数f(x)?? 在x = 0处的连续性. x

x?0?ex???解. 当??0时lim(x?sinx?0?1x1x)不存在, 所以x = 0为第二类间断点;

当??0, lim(x?sinx?0?)?0, 所以

???1时,在 x = 0连续, ???1时, x = 0为第一类跳跃间断点.

7. 设f(x)在[a, b]上连续, 且a < x1 < x2 < … < xn < b, ci (I = 1, 2, 3, …, n)为任意正数, 则在(a, b)内至少存在一个?, 使 f(?)?c1f(x1)?c2f(x2)???cnc1?c2???cn.

证明: 令M =max{f(xi)}, m =min{f(xi)}

1?i?n1?i?n所以 m ?

c1f(x1)?c2f(x2)???cnc1?c2???cn? M

所以存在?( a < x1 ? ? ? xn < b), 使得f(?)?c1f(x1)?c2f(x2)???cnc1?c2???cn

8. 设f(x)在[a, b]上连续, 且f(a) < a, f(b) > b, 试证在(a, b)内至少存在一个?, 使f(?) = ?. 证明: 假设F(x) = f(x)－x, 则F(a) = f(a)－a < 0, F(b) = f(b)－b > 0

于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?.

9. 设f(x)在[0, 1]上连续, 且0 ? f(x) ? 1, 试证在[0, 1]内至少存在一个?, 使f(?) = ?. 证明: (反证法) 反设?x?[0,1],?(x)?f(x)?x?0. 所以?(x)?f(x)?x恒大于0或恒小于0. 不妨设?x?[0,1],?(x)?f(x)?x?0. 令m?min?(x), 则m?0.

0?x?1因此?x?[0,1],?(x)?f(x)?x?m. 于是f(1)?1?m?0, 矛盾. 所以在[0, 1]内至少存在一个?, 使f(?) = ?.

10. 设f(x), g(x)在[a, b]上连续, 且f(a) < g(a), f(b) > g(b), 试证在(a, b)内至少存在一个?, 使 f(?) = g(?).

证明: 假设F(x) = f(x)－g(x), 则F(a) = f(a)－g(a) < 0, F(b) = f(b)－g(b) > 0 于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?. 11. 证明方程x5－3x－2 = 0在(1, 2)内至少有一个实根. 证明: 令F(x) = x5－3x－2, 则F(1) =－4 < 0, F(2) = 24 > 0 所以 在(1, 2)内至少有一个?, 满足F(?) = 0.

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f(x)??sin3x12. 设f(x)在x = 0的某领域内二阶可导, 且lim????0, 求32x?0x??xf(0),f'(0),f''(0)及limf(x)?3x2x?0.

sin3xf(x)?sin3x?xf(x)?sin3x解. lim???lim?lim?323x?0x?0x?0x?x?xxx?f(x)2?0. 所以

?sin3x? lim??f(x)??0. f(x)在x = 0的某领域内二阶可导, 所以f(x),f'(x)在x = 0

x?0?x?连续. 所以f(0) = －3. 因为

sin3xxx?f(x)2sin3x limx?0?0, 所以lim3??limx?0x?3?f(x)?3x2x?0?0, 所以

sin3xxx2 limf(x)?3x2x?0?lim3x?sin3xx3x?0?lim3?3cos3x3x2x?0

=lim3sin3x2xx?0?92

?limf(x)?3x?limx?x?0 f'(0)?lim由limf(x)?3x2f(x)?f(0)x?0f(x)?3x2x?0x?0?0?92?0

x?0?92, 将f(x)台劳展开, 得

12!xf''(0)x?0(x)?3222f(0)?f'(0)x? limx?0?92, 所以

12f''(0)?92, 于是

f''(0)?9.

(本题为2005年教材中的习题, 2008年教材中没有选入. 笔者认为该题很好, 故在题解中加

入此题)

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