(2012版)ZYM第二章极限与连续习题参考答案
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第二章极限与连续习题参考答案(2012)
练习 2.1
1.根据函数f(x)的图形,求下列极限或解释它们为什么不存在.
(1)limf(x) (2)limf(x) (3)limf(x) (4)limf(x)
x??1x?0x?1x?1.5
?1yy?f(x)?21.50.51?11.50xf(x)?1,f(x)?2,f(x)不存在, 解:(1)xlim(2)lim(3)lim??1x?0x?1因为limf(x)?0.5,limf(x)?1.5,limf(x)?limf(x);
x?1?x?1?x?1?x?1?f(x)不存在, 因为limf(x)???. (4)xlim?1.5x?1.5?x2?3x?22.设函数f(x)?
x?2(1)列表计算f(x)在点x?1.9,1.99,1.999,?,和x?2.1,2.01,2.001,?,的
f(x). 函数值,并估计极限limx?2(2)画出函数f(x)的图形,并根据函数图形检验(1)所得的极限.
f(x). (3) 根据函数解析式求极限limx?2x2?3x?2(x?2)(x?1)??x?1(x?2) 解:f(x)?x?2x?2limf(x)?1. (图、表略).
x?2 1
f(x)?2,3.如果有极限lim则函数f(x)在x?0处一定要有定义吗? x?0请归纳f(x)在点x?0处可能出现的所有情况,并用图形表示.
limf(x)?2 解:不一定.如:①f(x)?x?2,在点x?0处有定义且连续,x?0?x?2,x?0f(x)?在点x?0处无定义,但limf(x)?2 ②?2x?0?x?2,x?0?x?2,x?0?limf(x)?2 ③f(x)??3x?0在点x?0处有定义但断开,x?0?x?2,x?0?(图形略)
f(x)一定存在吗? 如果极限limf(x)存4.如果f(0)?2,则极限limx?0x?0f(x)?2?请归纳极限limf(x)可能出现的所有情况,在,是否必须limx?0x?0并用图形表示.
?x?1,x?0?解:(1)不一定,如:f(x)??2x?0,f(0)?2
?x?1,x?0?f(x)不存在. 但limf(x)?lim(x?1)?1?limf(x)?lim(x?1)??1,limx?0x?0?x?0?x?0?x?0?f(x)?2; (2)不一定,如:①f(x)?x?2 f(0)?2,有limx?0②f(x)???x,x?0f(x)?limx?0?2 ,f(0)?2,但limx?0x?0?2,x?0(图形略) 5.设f(x)???x,x?1,画出函数f(x)的图形;求x?1时,f(x)的
?3x?1,x?1f(x)是否存在. 左右极限,并判定极限limx?1解:图形略.
f(x)?lim(3x?1)?2 ,limf(x)?limx?1 因为 limx?1x?1??x?1?x?1? 2
x?1?limf(x)?limf(x) 所以limf(x)不存在. ?x?1x?16.(不作要求略之) 用下列函数的图形求??0,使当0?x?x0??时,不等式f(x)?A??成立. (1)
y4.34f(x)?x?2x0?6??0.3A?405.73.766.3x
(2)
y431230f(x)?xx0?113A?1??491691x 5 解:(1)??0.3(2)??
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7.(不作要求略之)设函数f(x)?2x?3,
(1) 求一个??0,使当x?1??时, f(x)?5?0.1; (2) 求一个??0,使当x?1??时, f(x)?5?0.01;
(3)设?是一个任意给定的正数,求一个??0,使当x?1??时,
f(x)?5??.
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解:(1)??0.05(2)??0.005 (3)??.(解略)
2?8. (不作要求略之)用???语言证明
x?1)?8; (2)lim (1)lim(3x?3x?3x?0?x?31?; x2?96 (3)limx?0; (4)lim1?x?0;
x?1?9.求下列函数在指定点的极限,如果不存在,请说明理由. (1)f(x)?x,在x?0处; x(2)f(x)??(3)f(x)??x?4,?2x?1,x?1,在x?0,x?1,x?2处; x?11,在x?2处; x?2?x?2,x??1(4)f(x)??,在x??1处.
2x?3,x??1?解:(1)因为f(x)?|x|?1x?0 ??x??1x?0x?0x?0x?0?limf(x)?lim1?1?limf(x)?lim(?1)??1 , ???x?0
所以f(x)?|x|在x?0处极限不存在. xf(x)?lim(x?4)?4, (2)因为 limx?0x?0所以f(x)??x?1?x?1?x?4 在x?0处极限为4. x?12x?1?x?1?x?1?x?1?因为 limf(x)?lim(2x?1)?1?limf(x)?lim(x?4)?5, 所以f(x)??x?1?x?4 在x?1处极限不存在.
?2x?1x?1f(x)?lim(2x?1)?3, 因为 limx?2x?2所以f(x)??x?1?x?4 在x?2处极限为3.
?2x?1x?1 4
x?2)?0,所以 lim(3)因为 lim(x?2x?2x??1?x??1?1?? x?2(4)因为 lim(2x?3)?1,lim(x?2)?1
f(x)?1 所以 xlim??110.用图形表示函数f(x),使f(x)满足条件: (1)f(0)?2,f(2)??1
(2) limf(x)???, limf(x)???..
x?1?x?1?解:(1)例如:设y?f(x)?ax?b,f(0)?b,而f(0)?2,所以b?2
3于是y?f(x)?ax?2,f(2)?2a?2,而f(2)??1,所以a??
23故y??x?2.(图略).
211f(x)???. ,有 limf(x)?lim???,lim(2)例如:取y?x?1?x?1?x?1?x?1x?1(图略).
11.下列函数在什么情况下是无穷大量,什么情况下是无穷小量? (1)y?1; (2)y?lnx; x?1(3)y?x2; (4)y?ex. 解:(1)因为
11??(x?1),?0(x??); x?1x?11所以 当x?1时,y?是无穷大量,
x?11当x??时,y?是无穷小量.
x?1(2)因为 lnx???(x?0?),lnx???(x???), 所以 当x???时,y?lnx是正无穷大量, 当x?0?时,y?lnx是负无穷大量;
又因为 lnx?0(x?1),所以 当x?1时,y?lnx是无穷小量. (3)因为 x2???(x??),x2?0(x?0)
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所以当x??时, y?x2是正无穷大量, 当x?0时,y?x2是无穷小量.
(4)因为 ex???(x???),ex?0(x???).
所以当x???时,y?ex是正无穷大量,当x???时,y?ex是无穷小量. 12.下列变量哪些是无穷小量?哪些是无穷大量? (1)x?0,cotx; (2)x?0,2?x?1; (3)x?0?,lgx; (4)??0,tan?.
解:(1)因为 x?0时,cotx??,所以x?0,cotx是无穷大量;
2?x?1?0,所以x?0,2?x?1是无穷小量; (2)因为 x?0时,(3)因为 x?0?时,lgx???,所以x?0?,lgx是负无穷大量; (4)因为 ??0时,tan??0,所以??0,tan?是无穷小量. 13.下列说法是否正确?
(1)无穷大量是极限为无穷大的变量;
(2)无穷大量是无界变量,无界变量也是无穷大量; (3)无极限的数列一定无界.
解:(1)不正确. 无穷大量是绝对值无限增大的变量;
(2)不正确. 例如:2,0,4,0,6,0, ??无界但不是无穷大量; (3)不正确.例如:1,-1,1,-1,??有界但它没有极限. 14.用图形表示一个函数f(x),使f(x)满足条件:
(1)f(0)?1,f(1)?3
f(x)?0 ,limf(x)?2. (2) xlim???x???解:(1)如:f(x)?2x?1,(图略).
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?1?2x?0(2)如:f(x)?? (图略). ?x?x?0?015.用图形表示一个函数f(x),使f(x)满足条件:
(1)f(?1)?0,f(2)?3
f(x)?2. (2) limx??解:(1)如:f(x)?x?1,(图略). (2)如:f(x)??2 (图略). 16. (不作要求略之) 设函数f(x)?1, x2?21x(1)求一个M?0,使当x?M时, f(x)?0?0.01; (2) 求一个M?0,使当x?M时, f(x)?0?0.0001;
(3) 设?是一个任意给定的正数,求一个M?0,使当x?M时,
f(x)?0??.
解:(1)M?10(2)M?100 (3)M?17. (不作要求略之)用??M语言证明
(1)limx??1?. 2x?3?2; (2)lim2x?0. x???x18.写出下列数列的前5项
2n?11?(?1)n(1)an?; (2)an? 33n?2n1n(?1)n?1x2n?1(3)an?(1?); (4)an?.
n(2n?1)!解:(1)由an?2n?1(n?1,2,3,?) 得数列的前五项为 3n?213579,,,,. 58111417 7
1?(?1)n(n?1,2,3,?) 得数列的前五项为 ?2?由an?n32,0,
n22,0,. 3335(n?1,2,3,?)得数列的前五项为
?3?由an?(1?1)n322,()2,()3,()4,()5.
(?1)n?1x2n?1(n?1,2,3,?)得数列的前五项为 ?4?由an?(2n?1)!xx3x5x7x9 ,?,,?,.
1!3!5!7!9!43546519.画出下列数列的点图,并指出哪些数列收敛,哪些数列发散.
1n; (2)a?(?1)n; nn21n(3)an?(?1)n; (4)an?;
n?3n?11?(5)an?sin; (6)an?[1?(?1)n]n.
nn(1)an?解:作图略.
(1)收敛于0 (2)发散 (3)收敛于0 (4)收敛于1 (5)收敛于0 (6)发散.. 20. (不作要求略之)设an?3n?2; n?1(1)求a1?3,a10?3,a100?3的值;
(2)求正整数N,使当n?N时,不等式an?3?10?4成立; (3)求正整数N,使当n?N时,不等式an?3??成立. 解:(1),
12111, (2)N?104?1 (3) N?[?1] 11101?21. (不作要求略之)用??N语言证明
8
(1)limn??13n?23?0lim?. ; (2)n??2n?1n22练习2.2
1. 下列说法是否正确,如不正确,请说明理由.
f(x)存在,则函数f(x)是有界函数. (1)若极限limx?2f(x)存在且大于0,则函数f(x)是非负函数. (2)若极限limx?2f(x)存在,只能说明f(x)在点x?2附解:(1)不正确,因为limx?2近局部有界,并非整体有界;
f(x)存在且大于0,只能说明f(x)在点(2)不正确, 因为limx?2x?2附近局部非负,并非整体非负.
2.根据下列函数f(x)和g(x)的图形,求下列极限或解释它们为什么不存在.
f(x)?g(x)] (2)lim[f(x)?g(x)] (1)xlim[??1x?1yy632g(x)4.54f(x)??101x?101x 解:(1)根据图形可得
x??1lim[f(x)?g(x)]?2?4?6
?f(x)?g(x)]不存在.因为根据图形可得limg(x)?4.5 (2)lim[x?1x?1
x?1?g(x)?limg(x),limg(x)不存在, limg(x)?6,lim??x?1x?1x?1 9
f(x)?3,若lim[f(x)?g(x)]存在,反证法,矛盾. 而limx?1x?1f(x)?1,limg(x)?3,请指出下列解题过程 3.设极限limx?2x?2(a),(b),(c)所依据的极限运算法则.
f(x)?g(x)]2f(x)?g(x)lim[2x?2lim? (a) 4x?2[f(x)g(x)]4lim[f(x)g(x)]x?2?2limf(x)?limg(x)x?2x?2?lim[f(x)g(x)]?x?24 (b)
?2limf(x)?limg(x)x?2x?2[limf(x)?limg(x)]x?2x?24 (c)
?2?1?31?? (1?3)481 (a)商的运算法则;(b)代数和的运算法则,常数因子提到极限号外,幂的运算法则;(c)积的运算法则.
4.设极限limx?1f(x)?2,求极限 2x?13f(x) x2f(x) (2)lim(1)limx?1x?1f(x)?2,而lim(2x?1)?3,所以limf(x)存在 x?1x?12x?1limf(x)limf(x)f(x)x?1??x?1?2, limf(x)?6 于是(1)limx?1x?12x?1lim(2x?1)3解:因为limx?1x?1f(x)3?63f(x)3limx?1???18 (2)lim22x?1xlimx1x?15.设极限limx?1f(x)?3?3,求极限 x?1x?2f(x)f(x) (2)lim2(1)lim x?1x?1x?1)?0,而lim解:(1)因为lim(x?1x?1f(x)?3?3, x?1所以limx?1
f(x)?30为()型lim[f(x)?3]?0,从而 limf(x)?3 x?1x?1x?1010
limf(x)f(x)3x?1lim???1 (2)x?122x?2lim(x?2)1?2x?16.下列解题过程是否正确,如不正确,请说明理由,并给出正确的解题过程.
(x?1?x)?limx?1?limx????(??)?0; (1)xlim???x???x???x?(2)lim(x?011)?limx?lim?0?(??)???; 22x?0x?0xx(3)limx?2x?2x?26????; x?2lim(x?2)0x?22lim(x2?2)解:(1)不正确,不能直接用代数和的极限运算法则,而应乘有理化因子后,再求极限.
lim(x?1?x)?lim1?0
x?1?xx???x???(2)不正确,不能直接用代数和的极限运算法则,而应通分再利用无穷大量与无穷小量互为倒数的关系求之.
1x3?1lim(x?2)?lim2??? x?0x?0xx(3)不正确,不能直接用商的极限运算法则,应利用无穷大量与无穷小量互为倒数的关系求之.
x2?2lim??. x?2x?27. 求下列极限:
x5?2x?3); (2)lim(1)lim(3x?1x?12x?1; 2x?xx2?2xxtanx?1); (4)lim2(3)lim(; ?x??23x?x?10x?4(5)limx?4
y?1x?2; (6)lim; y?1x?4y?3?211
5x2?2x?34x3?2x2?1(7)lim; (8)lim; x??x??3x4?16x2?1312x3?1lim(?). (9)lim; (10)x?11?x3x??8x2?7x1?xx5?2x?3)?lim3x5?lim2x?lim3?3?2?3?8. 解:(1)lim(3x?1x?1x?1x?1(2)limx?1x?2x?12?13??. 2x?x1?12(3)lim(xtanx?1)?limxtanx?lim1??1. ???4x?4x?4?4x2?2xx(x?2)x?22(4)lim2?lim?lim??. x??23x?x?10x??2(3x?5)(x?2)x??23x?5?6?511(5)limx?4x?2x?411?lim?lim?. x?4x?4(x?4)(x?2)x?4x?24(y?1)(y?3?2)y?1?lim?lim(y?3?2)?4 y?1y?1y?1y?3?22(6)limy?1235??25x?2x?3xx?5. (7)lim?limx??x??16x2?166?2x421?2?4324x?2x?1x?0. (8)lim=limxx4x??x??13x?13?4x87?228x?7xxx?0?0 (9)因为lim=limx??2x3?1x??12?32x2x3?1所以 lim=?. x??8x2?7x313?(1?x?x2)2?x?x2?)?lim?lim(10)lim( 33x?11?x3x?1x?11?x1?x1?x?lim(1?x)(2?x)2?x?lim?1 x?1(1?x)(1?x?x2)x?11?x?x2
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8. 求下列数列的极限:
2n2?2n?37n3(1)lim; (2)lim; n??n??n3?3n2?6n3n2?132n?n?1n?5n(3)lim; (4)lim3. 5n??n??3n?7n?n解:(1)lim2n?2n?3?lim2n??n??3n?122?23?2nn?2. 133?2n7n37?lim?7. (2)limn??n3?3n2?6nn??361??2nn21?22n?n0n(3)lim?limn??0. n??n??73n?733?n?11n1?31523n?5nnn?1. (4)lim?lim?lim5n??3n?5nn??nn??1?11?3152nn51?9. (不作要求略之) 用???语言证明
f(x)?A,limg(x)?B, 若极限xlim?xx?x00f(x)?limg(x)=AB. 则 xlim?f(x)?g(x)?=xlim?xx?x?x000练习2.3
1.利用夹逼定理证明
cosx?0. x??x2?1cosx1证明 :因为 ?1?cosx?1, 所以 2?2?2x?(??,??)
xxx11cosx?2)?0,lim2?0,由夹逼定理 lim2?0 而lim(x??x??xx??xxlim证毕.
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2.求下列极限 (1)limx?0(3)limx?0tan2xsin?x; (2)lim(??0); x?0sin?xsin5x1?cosxarcsinxlim; (4); x?0tan2x2xx?sinxsin(x2?1)lim(5)lim; (6). x?0x?sinxx?1x?1tan2x?2x22x解:(1)原式=lim?.
x?0sin5x?5x55xsin?x??x??x(2)原式=lim?.
x?0sin?x??x??xxx2sin2sinx2cos2x1222(3)原式=lim?lim()?()??. x?0tan2xx?0xsinx222(1?cosx)(1?cosx)cos2x1或原式=lim ?lim?2x?0x?0tanx(1?cosx)1?cosx2arcsinxt?arcsinxt1??lim? (4)原式=.limx?0x?02sint2x2sin(x2?1)(x?1)?2. (5)原式=limx?1x2?1x?1sinx(6)原式=lim?0. x?0x?1sinx3. 求下列极限
x?11x22lim(1?)?); (1)x??; (2)lim(1x??2xx(3)lim(x??x?axn?2n?1); (4)lim(); n??x?an2x11??1?1(1?)x?[lim(1?)?1]2?e2. 解:(1)原式=limx??x??2x2x 14
?2(?1)?2(2)原式=lim(1?)?2lim(1?)?1?e?1.
x??x??xx?2a?2a(?2a)?a?2a?2a(?2a)?2a?a(3)原式=lim(1?)?lim(1?)lim(1?) x??x??x??x?ax?ax?ax?ax?ax=e?2a.
aa?a?(?a)1?(1?)e?axxx)?lim?a?e?2a. 或原式?lim(xx??x??aea?a1?(1?)axx?22n2?)2(1?)?e2. (4)原式=lim(1n??nnx练习2.4
1.求下列极限
x3?x2?x); (2)lim(x2?1)cos(1)lim(2x?0x?131; x?1n2xcosxarctann. (3)lim; (4)limn??n3?2n?1x??x?1x3?x2?x)?0 解(1)lim(2x?0x2?1)?0,cos(2)因为lim(x?11有界变量 x?1x2?1)cos所以 lim(x?11?0 x?13132xx(3)因为lim?lim?0,cosx有界变量 x??x?1x??11?x所以limx??3xcosx?0 x?1n2arctann?0 (4)同理:limn??n3?2n?12. 当x?1时,下列各对无穷小量是否等价?
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(1)1?x与1?x3; (2)1?x与(1?x3); (3)
1?x与1?x (4)arctan(1?x)与1?e1?x. 1?x1?x11?lim?解:(1)否.因为lim , x?11?x3x?11?x?x2313所以x?1时,1?x与1?x3是同阶无穷小量非等价. (2)是. 因为limx?11?x1(1?x3)3?limx?111(1?x?x2)3?1,
所以x?1时,(1?x)~(1?x3).
1?x1?x?lim1?x?1 , (3)是. 因为limx?11?xx?11?x13所以x?1时,
1?x~(1?x). 1?x(4)否. 因为x?1时,arctan(1?x)~(1?x),e1?x?1~(1?x), 于是limx?1arctan(1?x)1?x?lim??1 x?1?(1?x)1?e1?x所以x?1时,arctan(1?x)与1?e1?x是同阶无穷小量非等价. 3. 当x?0?时,下列函数中哪些是比x的高阶无穷小量?哪些是与x的同阶无穷小量?哪些是比x的低阶无穷小量? (1)x?tan2x; (2)1?cosx; (3)cos(1?x); (4)sinx.
2?解:(1)因为 limx?0?x?tan2x2tan2x?lim(1?)?3 , ?x?0x2x故当x?0?时,(x?tan2x)是x的同阶无穷小量.
1?cosxx22?lim?0 , (2)因为 lim??x?0x?0xx故当x?0?时,1?cosx?o(x).
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(3)因为 limx?0cos?2(1?x)x?sin(x)?2?? , ?lim2x?0??x22?故当x?0?时,cos(1?x)是x的同阶无穷小量.
2?(4)因为 limsinxx1?lim?lim??? , ??x?0?x?0x?0xxx故当x?0?时,sinx是比x更低的无穷小量. 4. 证明当x?0时,tanx?sinx?o(x).
1?1)tanx?sinx1cosx证:lim?lim?lim(?1)?0, x?0x?0x?0xxcosxtanx?sinxtanxsinx?lim(?)?1?1?0 或limx?0x?0xxxsinx(证毕.
5. 利用等价无穷小量代换,求下列极限
x2tan2x(1)lim (2)lim; x?0sin5xx?02xsin()33arctan2x1?2x?1lim?4x(3)lim; (4). x?0arcsin3xx?0e?1解:(1)limx?0tan2x2x2?lim?. x?0sin5x5x5x2?lim?9. (2)limx?0x?0xxsin2()()233x2
(3)limx?03arctan2x2x2?lim?. x?0arcsin3x3x3
2x1?2x?11(4)lim?4x?lim3??.
x?0x?0?4xe?16练习2.5
f(x)是否存在?反之,1. 如果函数f(x)在点x?a处连续,问极限limx?a 17
f(x)存在,f(x)在点x?a处是否一定连续?为什么? 如果limx?a答:根据函数f(x)在一点连续的定义,函数f(x)在x?a处连续,
f(x)一定存在.反之不一定成立,f(x)存在,则lim即若lim不能说f(x)在x?ax?ax?a处一定连续,因为limf(x)存在,并不一定f(x)在x?a处有定义,
x?af(x)也不一定等于f(a),所以不一定连续. 即使有定义limx?a2. 用定义证明下列函数是连续函数. (1)y?3x2?1 (2)y?cosx. 证明:(1)设x0是(??,??)内任意一点,则
y0?3x0?1
2又limy?lim(3x2?1)?3x02?1
x?x0x?x0x?x0limy?y0?3x0?1,所以函数y?3x2?1在点x0处连续.
2又因为x0是(??,??)内任意一点,由任意性知函数y?3x2?1 在(??,??)内连续,故y?3x2?1是(??,??)内的连续函数. (2)设x0是(??,??)内任意一点,给自变量x在x0处取得改变量?x,则相应函数的增量为?y?cos(x0??x)?cosx0??2sin(x0?lim?y?lim[?2sin(x0?而?x?0?x?0?x?x)sin]?0 22?x?x)sin, 22所以函数y?cosx在点x0处连续.
又由于x0是定义域(??,??)内的任意一点,所以y?cosx 在(??,??)内是连续函数.
?2x,3. 函数f(x)???3?x,f(x)的图形.
0?x?1,在闭区间?0,2?上是否连续?画出
1?x?2解:当0?x?1时,f(x)?2x 初等函数连续.
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当1?x?2时, f(x)?3?x 初等函数连续. 在x?1处,f(1)?2,limf(x)?lim2x?2,
x?1?x?1?x?1?limf(x)?lim(3?x)?2,limf(x)?limf(x)?f(1) ???x?1x?1x?1所以函数f(x)在闭区间[0,2]上连续.
?ex,x?04. a取何值,函数f(x)??,在点x?0处连续.
?a?x,x?0解:设f(x)在点x?0处连续,则 limf(x)?limf(x)?f(0),
x?0?x?0?而limf(x)?lim(a?x)?a?f(0)?a,
x?0?x?0?x?0xlimf(x)?lime?1?f(0)?a, 所以a?1. ??x?05. 求下列函数的间断点,并指出其类型: (1)y?x1y?; (2); 2sinx(x?2)1?cosx; 2xx?11(5)y?; (6)y?sinxsin.
tanxx(3)y?xcos; (4)y?1x解(1)xlim??21??? 2(x?2)所以x??2是f(x)的第二类无穷间断点. (2)因为x?0处函数无定义,但 limx?0x?1 sinx所以x?0是函数的第一类可去间断点; 而当x?n? (n为不为0的整数)时,xlim?n?x??, sinx故x?n?(n??1,?2?)为f(x)的第二类无穷间断点.
xcos?0,x?0时函数无定义,所以x?0是函数的 (3)limx?01x第一类可去间断点.
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x2x2?1, (4)因为x?0时函数无定义,但 lim1?cos?limx?0x?0x2x22所以x?0是函数的第一类可去间断点. (5)因为xlim?n?x?1??,所以x?n? (n?Z)是函数的第二类 tanx无穷间断点;
当x??2?n?(n?Z)时函数无定义,
但limx??n?2?x?11?lim(x?1)?0 tanxx???n?tanx2故 x??2?n? (n?Z)是函数的第一类可去间断点.
sinxsin?0(无穷小量与有(6)当x?0时函数无定义,但 limx?01x界变量之积仍是无穷小量),
所以x?0是函数的第一类可去间断点. 6. 求下列极限:
e(1)limx?0sinxx2?4). ; (2)limarctan(2x?23x?6xln(1?x). sinx1x?3x(1?)x; (6)lim(); (5)xlim???x??xx?6?sinx)cscx; (4)lim(3)lim(1x?0x?0111?tanxx3); (8)lim(cos?sin)x; (7)lim(x??x?01?sinxxx1解:(1)limex?0sinx?ex?0limsinx?e0?1
x2?4x2?4)?arctanlim(2) (2)limarctan(x?2x?23x?6x3x2?6x?arctanlim[x?2(x?2)(x?2)x?22]?arctanlim?arctan x?23x3x(x?2)3cscx?sinx)(3)lim(1x?0
?lim(1?sinx)x?0?1?(?1)sinx?e?1
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1ln(1?x)x?lim[?ln(1?x)x]?1?lne?1 (4)解法1:limx?0x?0sinxsinx解法2: limx?0ln(1?x)x?lim?1(x?0,ln(1?x)~x) x?0sinxsinxx1(1?)(5)解法1:xlim???x?limex???1xln(1?)x?e?1?xx???xlim?ex???lim?1x?e0?1
(x???,?111?0,ln(1?)~?) xxx解法2:xlim(1?)x?lim(1????x???1x1?)xx?(?1)?(1?1)xx?e?1?e?1
u(x)?lim(?)?0,limv(x)?limx??? 解法3:因为xlim???x???x???x???(?)?x?lim(?而xlim???x???1x1x)?0,所以 原式?e0?1
1x1解法4:lim(1?)x???xx???x1?x?x?lim[(1?)]x x????x1?x)?e?0 ?x?limu(x)?lim(1?x???x???limv(x)?limx???x1??lim?0,所以 原式?e0?1 x????xxx?6?9?69x?3x9x9()=lim(1?)=lim(1?(6)解法1:lim)x??x?6x??x??x?6x?6
9(1?)=[limx??x?6x?699]lim(1?x??969)=e. x?6x3x33?3(1?)lim(1?)e3x?3xx??9xx()=lim???e解法2:lim x?6x??x?6x??6x6??(?6)e(1?)lim(1?)6xx??x(解法3:limx??x?3x9x)=lim(1?) x??x?6x?6u(x)?lim(因为xlim???x???9)?0,limv(x)?limx??? x???x???x?6 21
而xlim???9?x?9 所以 原式?e9 x?699limxln(1?)lim?xx?3x9xx??x??x?6x?6()=lim(1?)?e?e?e9 解法4:limx??x?6x??x?6tanx?sinxx31?tanxx3(7)解法1:lim() )?lim(1?x?0x?01?sinx1?sinxu(x)?lim因为 limx?0x?0tanx?sinx?limx?01?sinx1v(x)?lim3?? limx?0x?0xsinx(1?1)sinx(1?cosx)cosxlim?0, x?01?sinx(1?sinx)cosx111?1)1tanx?sinxsinx(1?cosx)cosxlim而lim== limx?0x?0x3x?0x3(1?sinx)cosxx3(1?sinx)1?sinxsinx(xx2x?2sinx?1112?lim?3?lim?32? x?01?sinxx?01?sinxxcosxxcosx221?tanxx32 故 lim()=e x?01?sinx1?tanxx3tanx?sinxtanx?sinxx3(1?sinx)解法2:lim(, )=lim[(1?)]x?01?sinxx?01?sinx1?sinxtanx?sinxtanx?sinxtan?0有limu(x)?lim(1?)x?sinx?e?0 因为 limx?0x?0x?01?sinx1?sinx1111?sinxtanx?sinxlimv(x)?limx?0tanx?sinx1? x?0x3(1?sinx)2111?tanxx3v(x) 故由结论有 lim()?limu(x)?e2. x?01?sinxx?011112(8)解法1:lim(cos?sin)x?lim[(cos?sin)2]2?lim(1?sin)2 x??x??x??xxxxxxx2x222x2x1111(cos?sin)x?lim[1?(cos?sin?1)]x 解法2:limx??x??xxxx11(cos?sin?1)=0, limx??, 而 limx??x??xx
22
xxsin?0,lim?? lim?sin?lim??1,原式?e1?e 而limx??x??x??x??2x1111?sin?1)?lim(?x)?(1?cos)?limxsin x??x??xxxx??x1()211 ?lim(?x)?x?limx??lim??1?1,原式?e1?e x??x??2xx??2xlimx?(cosxln(cos?sin)limxln(cos?sin)11xxxx(cos?sin)x=lime解法3:lim=, ex??x??x??xx111111ln(cos?sin)11xx(0)xln(cos?sin)=lim而 lim x??1xxx??0x111?1(?sin?cos)(2)1111xxxcos?sin?sin?cosxxxx=1, =lim=limx??x???111cos?sinx2xx11?sin)x?e1=e. 故lim(cosx??xx7. 证明方程x5?3x?1在1与2之间至少存在一个实根. 证:设f(x)?x5?3x?1,则f(x)在?1,2?上连续,
且f(1)?1?3?1??3?0,f(2)=25?3?2?1?25?0,即f(1)?f(2)<0 由零值定理知至少存在一点??(1,2),使f(?)?0, 即方程x5?3x?1在1与2之间至少存在一个实根.
8. 设f(x)在闭区间?a,b?上连续,且f(a)?a,f(b)?b,证明在区间(a,b)内至少存在一点?,使f(?)??.
证:设g(x)?f(x)?x,
由于f(x)在?a,b?上连续,所以g(x)在?a,b?上连续, 又g(a)?f(a)?a?0 ,g(b)?f(b)?b?0, g(a)g(b)?0, 故至少存在一点??(a,b),使g(?)?0, 即至少存在一点??(a,b),使f(?)??.
证毕
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练习2.6
1.设某企业从利润中提出20000元存入银行,准备若干年后建造一栋职工宿舍楼,假设造价要400000元,银行年利率为8%。问需要存多少年才能达到建房所需的款项?
解:设需要存n年才能达到建房所需的款项,则
400000?20000(1?8%)nlg20?nlg1.08?n?38.09520?1.08n
1.3010?n?0.0334
答:需要约存39年才能达到建房所需的款项.
2.假设某公司从盈余中提出45120元存入银行,希望六年能得到80000元上对某设备更新.试问需向银行要求多高的利率?
解:设需向银行要求年利率为r,则
80000?45120(1?r)61.773?(1?r)6
lg(1?r)?0.0414,r?0.1?10%答:需要向银行要求10%的利率.
3.设某企业决定用200000元进行投资,希望今后八年内每年末能得到相等金额的款项发放奖金,若投资报酬率为10%,求每年末可得到多少金额的款项?
解:这是一个普通年金现值问题
?PvAn?200000,r?10%,n?8?A?PvAn200000?
1111[1?][1?]r(1?r)810%(1?10%)8200000200000???37481.2611?0.466410(1?)2.14424
答:每年末可得到37481.26元的款项.
习题二
1.单项选择
f(x)存在的( ). (1)函数f(x)在点x0处有定义是极限xlim?x0?A?必要条件 ?B?充分条件 ?C?充要条件 ?D?无关条件
解:因为研究当x?x0时f(x)的极限,只是探讨当x无限趋近x0时,
f(x)的变化趋势,而不涉及x?x0时f(x)的状况,f(x)在x?x0处可以
有定义,也可以无定义,故选?D?.
(2)若f(x0?0),f(x0?0)均存在,则必有( ).
f(x)存在 ?B?limf(x)不存在 ?A?xlim?xx?x00f(x)可能存在也可能不存在 ?D?以上都不对 ?C?xlim?x0解:若f(x0?0),f(x0?0)均存在,但有可能f(x0?0)=f(x0?0) 也有可能f(x0?0)?f(x0?0),故选?C?. (3)数列的极限存在是该数列有界的( ).
?A?必要条件 ?B?充分条件 ?C?充要条件 ?D?无关条件
解:数列的极限存在是该数列有界的充分条件,因为数列-1,1,-1,…有界,但是它不存在极限,所以非必要. 故选?B?. (4)数列1,0,1,0,1…….的极限为( ).
?A? 0 ?B?1 ?C? 发散 ?D? 不能确定
解:根据数列极限定义,由于不存在常数A ,故该数列没有极限或称数列发散,故选?C?.
e=( ). (5)极限limx?0
25
1x
?A? 0 ?B? 1 ?C? ? ?D? 不存在
解:因为 lime??? , lime?limx?0?1x1x1e?1xx?0?x?0??0,
e不存在 故选?D?. 所以limx?01x(6)极限limx??sinx=( ). x?A? 0 ?B? 1 ?C? ? ?D?不存在
解:因为 limx??(7)极限lim1sinxsinx?0 故选?A?. =limx??xxsinx=( ).
x??(x??)ex???A? 1 ?B? –1 ?C? 0 ?D? ?
解:因为 limx??(xsin(8)极限limx?0sinx?sin(x??)=lim??1 故选?B?. x??x??x??(x??)e(x??)e11?sin2x)=( ). 2xx?A? 4 ?B? 1 ?C? 1 ?D? 2
412sin2x11(xsin?sin2x)=limxsin?lim解:lim=0+2=2 x?0x?02xx2xx?02x故选?D?. (9)若limx?03sinkx2?,则k?( ) . 2x3?A? 0 ?B? 3 ?C? 2 ?D? 4
23943sinkx3?kx3?k2?lim??,k?, 解:limx?0x?02x92x23故选?D?.
x(10)极限lim(1?)x=( ).
x?071?A? e ?B? e17?17 ?C? e7 ?D? e?7
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1x?x?(?7)xx?7解:lim(1?)=lim(1?)?e, x?0x?077171故选?B?.
((11)极限limx??x?12x)=( ). x?A? e?2 ?B? e2 ?C?
1 ?D? e e(解:limx??x?12x?1)=lim(1?)?x?(?2)?e?2 x??xx故选?A?.
x1?2x=( ). (12)极限limx?0?A? e2 ?B? e?2 ?C? e ?D? 不存在
x1?2x=lim(1?(?2x))解:limx?0x?01?(?2)?2x?e?2
故选?B?.
(13)若lim(1?kx)?,则k=( ).
x?01x1e?A? 1 ?B?
1x?1
1kkx ?C? 2 ?D? ?2
1?ek?
e解:lim(1?kx)?lim(1?kx)x?0x?0所以k??1,故选?B?.
(14)下列极限中,正确的是( ).
1()x??1 ?B? ?A?xlim?0?2alim()x?0(b?a?0) x???b?C? lime?? ?D? limx?01xx?0x?1 x解:因为lim()x?1??1;
x?0?12由b?a?0有0?aa);0?1 得lim(x?x???bb
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e不存在,而不是lime??; 又由lime????lime?0,得limx?0x?0x?0?x?0?1x1x1x1x由f(x)?x?1x?0不存在; 有f(0?0)??1?f(0?0)?1,lim??x?0xx??1x?0x故选?B?. (15)设f(x)???x,x?0,则limf(x)?( ). x?0?1,x?0?A? 0 ?B? 1 ?C? ? ?D? 不存在
f(x)?limx?0 解:由题设知 limx?0x?0故选?A?.
f(x)??,limg(x)??,则( ). (16)若limx?ax?a?f(x)?g(x)??0 ?f(x)?g(x)??? ?B?lim?A?limx?ax?a?C?limx?a1?f(x)?g(x)?可能存在可能不存在 ?0 ?D?limx?af(x)?g(x)解:取 f(x)?11,g(x)??, x?1x?111f(x)?lim??,lim(?)?? 则 limx?1x?1x?1x?1x?1lim?f(x)?g(x)??lim0?0??, lim?f(x)?g(x)??limx?1x?1x?1x?12???0 x?1 limx?11???0,?A??B??C?都不成立,故选?D?.
f(x)?g(x)(17)下列变量在给定的变化过程中为无穷小量的是( ).
?A?ex(x?0) ?B?1?sinx(x?0)
x?C?lnx(x?0?) ?D??x2(x??)
ex?1,lim(1?解:limx?0x?0sinx)?0,limlnx???,lim(?x2)??? ?x??x?0x根据无穷小量的定义,故选?B?. (18)当x?0时,sinx与1?cosx相比是(
28
).
?A?高阶无穷小 ?B?低阶无穷小 ?C?同阶无穷小 ?D?等阶无穷小
解:由题意知 limsinx?0x?0,lim(1x?0?cosx)?0, 因为 limsinxx?01?cosx?limx2x?0x2?limx?0x??, 2所以当x?0时,sinx是比1?cosx低阶的无穷小量,故选?B?. 19)下列变量在给定变化过程中为无穷大量的是( ).
?A?x2?1xx3?1(x???) ?B?e(x?0?)
?C?(x?1)x(x?1)sin1x3?1(x?1) ?D?x?1(x??) 解:因为 xlimx2???x3?1?xlim1???1???;
x?1x4?1x1(x?1)xx1xlim?0?e?xlim?0?1?0;limexx?1x3?1?limx?1x2?x?1?3; sin1lim(1x?1x??x?1)sinx?1?limx??1?1; 故选?A?. x?120)设f(x)为多项式函数,且limf(x)?2x3x??x2?1,limf(x)x?0x?3,则f(x)?(?A?x2?3x ?B?2x3?x2 ?C?2x3?x2?3x ?D?2x3?x2?3x?3
解:由已知f(x)为多项式函数,且limf(x)?2x3x??x2?1, 可知f(x)?2x3是与x2同次幂的多项式且二次项系数为1,
令f(x)?2x3?x2?ax?b其中a,b待定
29
).
((
又因为limx?0x?0f(x)0?3所以此极限为()型, x0x?0即有limf(x)?lim(2x3?x2?ax?b)?0,得b?0
2x3?x2?ax?0于是lim?lim(2x2?x?a)?3,得a?3 x?0x?0xf(x)?2x3?x2?3x 故选?C?.
(21)函数f(x)?x2在x?1处的改变量?y?( ).
?A?2x?x ?B?2?x?(?x)2 ?C?2x?x?(?x)2 ?D?2?x
解:由题设得
?y?f(1??x)?f(1)?(1??x)2?1?1?2?1??x?(?x)2?1?2?1??x?(?x)?2?x?(?x)22
故选?B?.
(22)函数f(x)在点x0处有定义是f(x)在该点连续的( ).
?A?必要条件 ?B?充分条件 ?C?充要条件 ?D?无关条件
解:根据函数f(x)在点x0处连续的定义知在点x0处有定义是在该点连续的必要条件. 故选?A?. (23)函数y?x?3的间断点有( ).
(x?1)(x?2)?A? 1个 ?B? 2个 ?C? 3个 ?D? 0个
解:因为函数的定义域为D?[3,??) 所以函数的间断点为0个,故选?D?. 2.求下列极限.
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x2?1(x2?1)(5?x?2)[?(1?x)(5?x?2)]??8. (1)lim=lim=limx?1x?1x?11?x5?x?21(2sinx?1)(sinx?1)sinx?12?12sinx?sinx?1(2)lim=lim=lim=?3. ?(2sinx?1)(sinx?1)?sinx?1?2sin2x?3sinx?11x?x?x??166621a2a(n?1)alim[(x?)?(x?)???(x?)] (3)n??nnnn2
1?2???(n?1)a]
n11?2???(n?1)lim[(1?)x?a] =n??nn21n?1lim[(1?)x?a] =n??n2na =x?.
2lim[(n?1)x? =n??1n(4)xlim???1111??1?x?xx?x?xxx3x=lim=lim?1. x???x???11x?11?1?xx2x?1203x?23020330330(2x?1)20(3x?2)30lim()()=1?()=(). (5)解1:lim=x??2x?1x??22x?12(2x?1)50解2:lim(2x?1)(3x?2)=lim50x??x??(2x?1)2030(2x)20(1?1202)(3x)30(1?)302x3x?(3)30
12(2x)50(1?)502xlim(6)n??2n?1?32n?3nn?122()n?3=lim3?3. n??2n()?131?0.
n?1?nlim(n?1?n)=lim(7)n??n??xnxn2lim2sinn=lim(8)n?x?x. ??2n??x2nsin 31
x1?cosx22=2. (9)lim?=lim?limxx?0?xx?0x?0x222?22sinaxsinbxsinax?sinbxlima?limb?a?b. (10)lim=x?0x?0x?0xaxbx2sinx2?sin(11)xlim???x2sin1x=lim2x2?1x???x2sin2x1?1x12x2=limx???11. ?21121?2x2xnsin1xn?2n?122?222?2?1lim()=lim(1?)n?1=lim(1?)2=lim(1?)2lim(1?)?e2. (12)n??n??n??n??n??nnnnnnx2?1x22(2)=lim(1?2)(13)解法1:limx??x?1x??x?1x2?1?2?12
2=lim(1?2)x??x?1x2?1?22lim(1?x??2)?e2. 2x?11x21x2)(1?)22ex2?1x2xx??1?e2 解法2:lim(2)?lim?limx??x?1x??12x??12e(1?2)x(1?2)?x?(?1)xx(1?(cosx)(14)limx?011?cosx?(cosx?1)]=lim[1x?011?cosx?(cosx?1)]=lim[1x?01?(?1)cosx?1?e?1.
(15)limx?1x?1lnx11limlim?==.(x?1时,lnx与x?1等价) x?1x?1x2?1x?1x2?12(16)解法1:limx?02?1?cosx1?x?1x22=limx?02?(1?cosx)x2(2?1?cosx)2
=limx?02sin22x(2?1?cosx)2x?02=limx?0sin2x()2(2?1?cosx)22?(1?cosx)x2?122.
解法2:lim2?1?cosx1?x?1?limx?0x2(2?1?cosx)21?cosx
?limx?0x(2?1?cosx)232
2
?limx?0x2(2?1?cosx)2x22?limx?011 ?. 222?1?cosxx2?ax?b?5,求a,b. 3.设limx?11?x0x2?ax?b?5,所以此极限为()型, 因为limx?101?x于是lim(x2?ax?b)?0,即1?a?b?0 a??1?b
x?1x2?ax?bx2?(?1?b)x?b(x?1)(x?b)lim?lim?lim?lim(b?x)?b?1?5 x?1x?1x?1x?11?x1?x1?x所以b?6,a??7.
x2?1(?ax?b)?3,求常数a,b. 4.已知limx??x?1x2?1(1?a)x2?(a?b)x?1?b?ax?b)?lim?3 解:因为lim(x??x?1x??x?1而x??, 所以 1?a?0,?(a?b)?3,从而a?1,b??4 1x2?1(1?a)x2?(a?b)x?1?b?ax?b)?lim?3 或lim(x??x?1x??x?11?a?0从而a?1
?(1?b)?1?1x1?bx??(1?b)?3, b??4.
于是limx???(1?b)x?(1?b)?limx??x?1 5. 证明当x?0时,e?exsinxx3~. 6ex?esinxex(1?esinx?x)?lim证明 : ?lim 33x?0x?0xx66 33
?ex?(sinx?x)?lim(x?0,esinx?x?1~sinx?x) 3x?0x6?(?6)?e0?limsinx?xcosx?1?sinx?(?6)?lim?(?6)?lim?1(洛必达法则) x?0x?0x?0x33x26xxsinx0()00()0?当x?0时,e?ex3. ~6?sinx,?x?0x?6.a,b为何值时,函数f(x)??a,x?0在点x?0处连续.
?1x?0?xsin?b,x?解:因为f(x)在x?0处连续,所以有limf(x)?limf(x)?f(0)
x?0?x?0?而limf(x)?limsinx?1
x?0?x?0?x1limf(x)?lim(xsin?b)?0?b?b f(0)?a x?0?x?0?xx?0x?0于是limf(x)?limf(x)?f(0)即1?b?a ??故当a?b?1时,f(x)在点x?0处连续.
7.求下列函数的连续区间,并求极限. (1)f(x)?132x?3x?2,并求limf(x).
x?0解:因为f(x)?13x2?3x?2?1在点x?1,x?2处无意义 3(x?2)(x?1)D?(??,1)?(1,2)?(2,??)
1),(1,2),(2,??). 所以f(x)的连续区间为:(??,f(x)?lim且 limx?0x?031x?3x?22?132.
f(x). (2)f(x)?x?4?6?x,并求limx?5 34
解:因为 ??x?4?0?x?4,?即4?x?6得D?[4,6]
?6?x?0?x?6所以f(x)的连续区间为[4,6]
limf(x)?lim(x?4?6?x)?2.
x?5x?5(3)f(x)?lnarcsinx,并求limf(x). 1x?2?arcsinx?0?0?x?1解:因为?即0?x?1得D?(0,1] ?x?0??1?x?1?所以函数的连续区间为(0,1]
limf(x)?limlnarcsinx?lnlimarcsinx?ln1x?21x?21x?2?6
f(x) (4)f(x)?lg(2?x),并求xlim??8解:因为f(x)的定义域为2?x?0,即x?2
所以函数的连续区间为(??,2),
f(x)?limlg(2?x)?lg10?1. 且xlim??8x??88.判定下列函数在指定点的连续性,并求函数的连续区间.
?ex,x?0(1)f(x)??在x?0处
?x?1,x?0解:f(x)的定义域为(??,??)
当x?0时,f(x)?ex连续, 当x?0时,f(x)?x?1连续,
而在x?0处,因为f(0)?(x+1)x?0?1 limf(x)?limex?1
x?0?x?0?x?0?limf(x)?lim(x?1)?1,即limf(x)?f(0) ?x?0x?0所以函数f(x)在x?0处连续, 故函数f(x)的连续区间为(??,??).
35
?ln(1?x)?x?(2)f(x)??0?1?x?1?x?x?x?0x?0在x?0处 ?1?x?0解:f(x)的定义域为[?1,??) 当x?0时,f(x)?ln(1?x)连续 x当?1?x?0时,f(x)?1?x?1?x 连续,x而在x?0处,f(0)?0
1ln(1?x)xxln[lim(1?x)]?lne?1 =limf(x)?lim?limln(1?x)?x?0x?0?x?0?x?0?x1x?0limf(x)?lim??x?01?x?1?x2?lim?1 ?x?0x1?x?1?xlimf(x)?1?f(0)?0,所以f(x)在x?0处不连续
x?0??). 函数f(x)的连续区间为 [?1,0),(0,??12?(x?1),(3)f(x)??e??0,x??1 在x??1处
x??1解:f(x)的定义域为(??,??) 当x??1时,f(x)?e?1(x?1)2连续;
在x??1处,f(?1)?0
x??1limf(x)?limex??1?1(x?1)2?lim11x??1?0?f(?1)
2e(x?1)f(x)在x??1处连续,
所以函数f(x)连续区间为(??,??).
?sinxx?0(4)f(x)??x??x?0?1
在x?0处
36
?sinx?x,x?0?sinx,x?0?解:因为f(x)????1,x?0 D?(??,??) ?xx?0?sinx?x?0?1,??x,?所以f(x)在x?0时连续, 在x?0处,f(0)?1
x?0limf(x)?lim??x?0sinxsinx?1,limf(x)?lim(?)??1 limf(x)不存在 ??x?0x?0x?0xx所以f(x)在x?0处不连续,函数f(x)连续区间为(??,0),(0,??).
1x?0?2?xsin(5)f(x)?? 在x?0处 x,x?0??0,解:D?(??,??),函数f(x)在x?0时连续,
f(x)?limx2sin?0?f(0), 在x?0处,limx?0x?01x所以f(x)在x?0处连续,其连续区间为(??,??) 解:函数f(x)在x?0时连续,但在x?0处
limf(x)?limx2sinx?0x?01?0?f(0), x所以f(x)在x?0处连续,
(??,??)因而函数在内连续.
9.给f(x)补充定义一个函数值f(0),能使f(x)在x?0处连续?
1(1)f(x)?sinxcos (2)f(x)?ln(1?kx)x
xm解:(1)函数f(x)?sinxcos在点x?0处无定义,
f(x)?limsinxcos?0,所以补充f(0)?0, 但由于limx?0x?0f(x)?f(0),可使f(x)在x?0处连续. 则有limx?01x1x(2)函数f(x)?ln(1?kx)在点x?0处无定义,
37
mxln(1?kx)?limln(1?kx)但由于limx?0x?0mxmkkx?ln[lim(1?kx)]x?01kxkm?lnekm?km
f(x)?f(0),可使f(x)在x?0处连续. 所以补充f(0)?km,则有limx?010.设f(x)为连续函数,x?a与x?b是f(x)?0的两个相邻的根, 证明:若已知(a,b)内存在一点c,使f(c)?0(或?0),则f(x)在(a,b)内处处为正(或负).
证明:(反证法)仅证处处为正(为负同理) 假设在(a,b)内存在一点d,使得f(d)?0 若f(d)?0,则与已知矛盾 若f(d)?0,又f(c)?0
由零值定理知在(c,d)或(d,c)(?(a,b))内存在一点?,
使得f(?)?0,
即?是介于a,b之间的一个根, 这与a,b是相邻两根矛盾,
故假设不成立.所以f(x)在(a,b)内处处为正.
11.设f(x)与g(x)均在[a,b]上连续,且f(a)?g(a),f(b)?g(b),证明:在
(a,b)内至少存在一点c,使得f(c)?g(c).
证明:设h(x)?f(x)?g(x),由于f(x)与g(x)均在[a,b]上连续,
所以h(x)在[a,b]上连续, 又 h(a)?f(a)?g(a)?0h(b)?f(b)?g(b)?0(f(a)?g(a)) (f(b)?g(b))
由零值定理知:在(a,b)内至少存在一点c,
使得h(c)?f(c)?g(c)?0,即在(a,b)内至少存在一点c, 使得f(c)?g(c).
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12*.利用极限存在准则证明: (1)xlim???lnx?0 2x(2)证明数列?xn?,其中a?0,xn?(xn?1?其极限.
12a),n?2,3,?收敛,并求xn?1证明:(1)因为x?1时,lnx?ln[1?(x?1)]?x?1
lnxx?11111????,lim0?lim?0, ,而x???x???xx2x2xx2xlnx?0. 由夹逼定理得xlim???x20?(2)证明:因(xn?1?a)2?0,故xn?12?a?2axn?1,
xn?12?a2axn?11a所以 xn?(xn?1?)???a2xn?12xn?12xn?1(n?2,3,?)
所以数列?xn?有下界a,其中x1未算入数列中,若算入, 则下界可取为min(x0,a),又由于
(xn2?a)?2xn21a xn?1?xn?(xn?)?xn?2xn2xn?(a?xn)(a?xn)?0 (xn?a)
2xn2所以数列?xn?单调递减,?xn?单调递减有下界必有界, 由单调有界数列必收敛得数列?xn?收敛.
limxn?A,由xn?a?0,所以A?0, 设n??对等式xn?(xn?1?121aa)两边取极限得A?(A?),
2Axn?1解方程得A?a(负值舍去).
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