第一章 函数·极限·连续
一. 填空题
a?1?x?1．设lim???tetdt, 则a = _2._______. ???x???x?ax2. =___
112n??_____.lim?2?2???2?
n??2n?n?n??n?n?1n?n?23. 已知函数f(x)??
?1?0|x|?1|x|?1, 则f[f(x)] ___1____.
4. lim(n?3n?n?n)=___limn??n??n?3n?n?nn?3n?n?n?2____.
5. limcotx?x?01??1??=______.
?sinxx?解. limcosxx?sinxx?sinx1?cosxsinx1??lim?lim?lim?
x?0sinxx?0x?0x?06xxsinxx33x26n19906. 已知limk?A(? 0 ? ?), 则A = ______, k = _______.
n??n?(n?1)k所以 k－1=1990, k = 1991;
111?A，A?? kk1991
二. 选择题
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) 所以(d)是答案.
2. 设函数f(x)?x?tanx?esinx?(x)必有间断点 f(x), 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. 极限lim?352n?1?的值是 ????2222?n??12?222?3n?(n?1)???(a) 0 (b) 1 (c) 2 (d) 不存在
解., 所以(b)为答案.
(x?1)95(ax?1)54. 设lim?8, 则a的值为 250x??(x?1) 1
(a) 1 (b) 2 (c) 解.所以(c)为答案. 5. 设lim58 (d) 均不对
(x?1)(x?2)(x?3)(x?4)(x?5)??, 则?, ?的数值为
x??(3x?2)?111 (b) ? = 5, ? = (c) ? = 5, ? = 5 (d) 均不对 333(a) ? = 1, ? = 解. (c)为答案.
6. 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. 所以(b)为答案. 7. 设lim(1?x)(1?2x)(1?3x)?a?6, 则a的值为
x?0x(a) －1 (b) 1 (c) 2 (d) 3 解.所以(a)为答案. 8. 设limx?0atanx?b(1?cosx)cln(1?2x)?d(1?e?x)2?2，其中a2?c2?0, 则必有
(a) b = 4d (b) b =－4d (c) a = 4c (d) a =－4c 解.所以(d)为答案.
三. 计算题 1. 求下列极限 (1) lim(x?e)
x???x1x解. lim(x?e)?limex???x???x1xln(x?ex)x?eln(x?ex)x???xlim?ex???x?e?e1?e
xlim1?ex21?cos)x
x??xx1解. 令y?
x(2) lim(sinlim21xy?0ylim(sin?cos)?lim(sin2y?cosy)=ex??y?0xx1ln(sin2y?cosy)y?e2cos2y?sinyy?0sin2y?cosylim?e2
?1?tanx?(3) lim??x?01?sinx???1?tanx?解. lim??x?01?sinx??
1x3
11x3tanx?sinx???lim?1??x?01?sinx??2
x3
taxn?sixn?tanx?sinx?? ?lim??1??x?0?1?sinx???limsinx(1?cosx)x31?sixntaxn?sixnn)x3?(1?sixtanx?sinxlim3?=ex?0x ??sinx?2sin2 =ex?0=ex?0limx3x2?e.
122. 求下列极限 (1) limx?1ln(1?3x?1)arcsin2x?132
解. 当x?1时, ln(1?3x?1)~3x?1, arcsin23x2?1~23x2?1. 按照等价无穷小代换 limx?1ln(1?3x?1)arcsin23x2?1?x? ?3?limx?1x?123x2?1?lim11 ?3x?123x?122(2) lim??12?cotx?0x2?解. 方法1：
?1lim?2?cot2x?0x??sin2x?x2cos2?1cos2x??x?=lim?=lim???22??x?0x2sin2xsinx?x?0????xx??? ??1?(x2?1)cos2 =lim?x?0?x4???2xcos2x?2(x2?1)cosxsinx?x? ???3?=lim?x?0?4x????2xcos2x?sin2x2x2cosxsinx?lim =lim 33x?0x?04x4x?2cos2x?4xcosxsinx?2cos2x1? =limx?012x22?2cos2x?2cos2x114cosxsinx?4sin2x11???lim?? =lim2x?0x?012x3224x32 =lim?2sin2x111112???????
x?024x3263233. 求下列极限 (1) limnn(n?1)
n??lnnnxnnn?1?1 解. lim(n?1)?limn 令nn?1?x limx?0ln(n??lnnn??lnn1?x) 3
1?e?nx(2) lim
n??1?e?nx1?en??1?e?nx?nx解. limx?0?1???0 x?0 ??1x?0?n?na?nb??, 其中a > 0, b > 0 (3) lim??n???2??解. lim??a?b??1?c? x?1/n,c?b/a alim????n???x?02?2??nnnx????ae?1xln(1?cx)?ln2xx?0?lim
=aeln(1?cx)?ln2xx?0?lim?aex?0?1?c?ac?alimcxlncxb?ab a
4. 求下列函数的间断点并判别类型
(1) f(x)?2?12?11x1x
解. f(0)?lim?x?0?2?12?11x1x?1, f(0?)?lim?x?02?12?11x1x??1
所以x = 0为第一类间断点.
?x(2x??)x?0??2cosx( 2 ) f(x)??
?sin1x?0??x2?1解. f(+0) =－sin1, f(－0) = 0. 所以x = 0为第一类跳跃间断点;
sin limf(x)?limx?1x?11不存在. 所以x = 1为第二类间断点; x2?1x?? f(??2)不存在, 而lim?2x(2x??)??,所以x = 0为第一类可去间断点;
2cosx2
x??k??lim?2?x(2x??)??, (k = 1, 2, …) 所以x =?k??为第二类无穷间断点.
22cosx1??xsinx?0?5. 讨论函数f(x)?? 在x = 0处的连续性. x
x?0x?e???
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1)不存在, 所以x = 0为第二类间断点;
x?0x1?(xsin)?0, 所以 当??0, limx?0?x(x?sin解. 当??0时lim? ???1时,在 x = 0连续, ???1时, x = 0为第一类跳跃间断点.
6. 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)???cn.
c1?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)???cn? M
c1?c2???cnc1f(x1)?c2f(x2)???cn
c1?c2???cn所以存在?( a < x1 ? ? ? xn < b), 使得f(?)?7. 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(?) = ?.
8. 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(?) = ?.
9. 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(?) = ?. 10. 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.
?2?x2(1?cosx)?11. 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.
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