第一章 一元函数的极限

第一章 一元函数的极限

§1.1 利用定义及迫敛性定理求极限

设R表示实数集合,R*表示扩张的实数集,即R*?R????,???. 例1 若liman?a?R*.证明limn???a1?a2???ann?0?a?R* (算术平均值收敛公式).

?a?n???证明 (1)设a?R,由liman?a,??n???,?N1?0,当n?N1时, an?2.

因此

a1?a2???ann?a

?(a1?a)?(a2?a)???(an?a)n

?a1?a?a2?a???aN?a1nAnAn?aN1?1?a???an?an

??n?N1n??2

???2,

An?其中A?a1?a?a2?a???aN?a.又存在N2?0,当n?N2时,

1?2.因此当

n?max{N1,N2}时,

a1?a2???ann?0?a??2??2??.

（2）设liman???,则?Mn???,?N1?0,当n?N1时,an?3M.

因此

?a1?a2???ann

aN1a1?a2???aNn1??1?aN1?2???annAn1?An?n?N1n?3M,

?0其中A?a1?a2???aN.由于时,

An?M2?0,

n?N1n?1(n???),所以存在N2,当n?N2,

n?N1n?12.因此

a1?a2???ann?12?3M?12M?M.

(3) 当liman???时,证明是类似的.(或令bn??an转化为(2)).

n???注 例1的逆命题是不成立的.反例an???1?n(n?1,2,?),容易看出lim但是极限liman不存在.

n???a1?a2???ann?0,

n???例2 设{an}为单调递增数列, ?n?a1?a2???ann.证明若lim?n???n?a,则liman?a．

n???证明 由{an}为单调递增数列,当m?n时有am?an.固定n,则有

??a1?a2???anm?an?1?an?2???ammm?Am?m?nman,

其中A?a1?a2???an.令m???,则a?lim?m?an.

m???又由于?n?a1?a2???ann?nann?an

所以?n?an?a.令n???,由迫敛性定理得liman?a．

n???注 当{an}为单调递减数列时,上述结论也成立.

例3 设数列{an}收敛,且an?0(n?1,2,?),证明limnn???a1a2?an?liman.(几何平均值收敛公式).

n???证明 设liman?a,则由极限的不等式性质得a?0.

n???(1)若a?0,则limlnan?lna,

n???由例1,因此limlimn???1n(lna1?lna2???lnan)?lna1.

lnann???a1a2?an?limenn????lna1?lna2???lnan??e?a

(2)若a?0,则limlnan???.因此

n???limn???1nn(lna1?lna2???lnan)???1,

?limn???a1a2?an?limenn????lna1?lna2???lnan?0.

注 可以证明当a???时结论也成立.

例4 设an?0(n?1,2,?),证明:若liman?1an存在,则limn???nn???an也存在且limn???nan?liman?1an.

n???证明 令b1?a1,b2?由例3得, limn???a2a1,…,bn?anan?1,….

nb1b2?bn?limbnn???.

所以limn???nan?limanan?1?limn???an?1an.

n???例5 证明limnn!nn???n?e.

n证明1 设an?由例4得 limn,则

nn!an?1ann!??n?1?n?1?n?1?!an?e

?n!nn1????1??n???e(n???).

n???n?limn???n证明2 利用司特林(Stirling)公式n!~

例6 设an?a,bn?b(n???).令cn证明 cn?ab?1n1n1n?1n?n?2?n???e?n得 limnn!n???n?limn???e?2?n?2n1?e

?1n?a1bn?

?a2bn?1???anb1?.证明 limcn?abn???.

?a1bn?a2bn?1???anb1?nab[?a1bn?ab)?(a2bn?1?ab)???(anb1?ab?]

[?a1bn?a1b?a1b?ab)?(a2bn?1?a2b?a2b?ab)???(anb1?anb?anb?ab?] [a1?bn?b)?a2(bn?1?b)???an(b1?b?]?bn[?a1?a)?(a2?a)???(an?a?]??.

由于数列{an}收敛,故是有界的.设an?M(n?1,2,?),则

cn?ab?Mbn?b?bn?1?b???b1?bn?ba1?a???an?an.

利用例1得limcn?ab.

n???

例7 设liman?a.证明limn???a1?2a2???nan2n?a2.

?a?n???证明 由liman?a,??n????0,?N1?0,当n?N1时, an?2.

a2n所以

a1?2a2???nan2n?a2?(a1?a)?2(a2?a)???n(an?a)n2?

?(a1?a)?2(a2?a)???N1(aN?a)1n2??N1?1?(aN1?1?a)???n(an?a)n2?a2n

?An2?1n2?aan?n?1??A????2??222n22nna2n,其中A?a1?a?2a2?a???N1aN?a.

1又存在N2?0,当n?N2时,

An2???2.故当n?max{N1,N2}时,

a1?2a2???nan2n?a2??2??2??.

例8 证明limn???nn?1.

证明 令?n?所以0??n?

nn?1,则n??1??2n?1n?n?1?n?n?12n?n?1??2n???12n?n?1??n2n.

.(n?2)由迫敛性定理得, ?n?0(n???).所以limn?1.

n???例9 求极限limn???1?n2?33???n1?nnn3.

3???nn解 以下不等式是显然的:1?

2?n?nn由例8与迫敛性定理得所求极限为1

例10 设a0,a1是两个定数,且当n证明 由a2?a1?a3?a2?a4?a3?a1?a22a0?a123?2时an?an?1?an?22.证明liman?n???a0?2a13.

a0?a12??,

,

a0?a122,

………

an?an?1???1?n?1a0?a12n?1,

.

相加得an?a1?a0?a1?111?n1??2?????1?n?2??2222??11?12?a0?a13所以lim(a?a)?a0?a1?n1n???.

2这推出liman?n???a0?2a13.

例11 设x1?0,xn?1?3?1?xn3?xn?(n?1,2,?),求极限limxn.

n???分析 若{xn}极限存在且为a,则aa?3?3?1?a?3?a.由此解得a??3.再由xn?0知a?0.故

.

3?1?x1?3?x13?3x1?33?3?x13x1解 由x2?3??3?

?(3?3)x1?3?3?x1?3?3?3?33?x13?3?x1?3?

3得x2?3?3?33?x13?x1?3?3x1?.

同理有x3?3?33?x2x2??3?3??3????3??n?12x1?3.

一般情况有xn??3?3??3????3??x1?3.

所以limxn?3.

n???

例12 设a1?0,an?1?11?an(n?1,2,?),求极限liman.

n???分析 若{an}极限存在且为a,则a?a??1?2??1?25511?a.由此解得a??1?25.再由an?0知a?0.故

.

,我们有

11?a解 令aan?a?11?an?1??an?1?a?1?an?1??1?a??11?aan?1?a?aan?1?a.

由上述递推关系可得an?a?an?1a1?a(n?2,3,?),由于a?1,故得limann?????1?25.

例13 设K是正数,x0?0,对任意自然数n,令xn?1?K?xn?1?2?xn?1???.证明limxn??n????K.

证明 x1?K?1?K??x0????2?x0??1K?1??x0?22??x1?x1?K?1K???x0?2xx0?0??K?2,

同理x1?K?2x0?x0?K?2.两式相除得

nK?x0????x?K?0K??K??2.

n由归纳法得

xn?xn?K?x0????x?K?0K??K??2.由于

x0?x0?KK?x0??1,得到lim?n?????x0?K??K??2?0.

所以limn???xn?xn?KK?0,这证明了limxn?n???K.

§1.2 stolz定理及其应用

定理1 设{an}是趋于零的数列, {bn}严格递减趋于零,则当liman?1?anbn?1?bn存在或为??、

n?????时，有limn???anbn?limn???an?1?anbn?1?bn.

证明 设limn???an?1?anbn?1?bn?l.

(1) 若l是有限实数,则???0,?N?0,当n?N时,有l???an?1?anbn?1?bn?l??.

由于bn?1?bn?0,所以

?l?l???(bn?1?bn)?an?1?an??l???(bn?1?bn), ???(bn?2?bn?1)?an?2?an?1??l???(bn?2?bn?1),

………

?l???(bn?p?bn?p?1)?an?p?an?p?1??l???(bn?p?bn?p?1),

上述各式相加得 ?l???(bn?p?bn)?an?p?an??l???(bn?p?bn). 在上式中固定n并令p???,由于an?p?0,bn?p?0,得 ?l???bn?an??l???bn.

.所以limn???注意到bn?0,由上式便得

anbn?l??anbn?l.

(2)若l???,则?K?0,?N?0,当n?N时,有

an?1?anbn?1?bn?K.仿照(1)中的证法可得,对任

意自然数p,有

an?p?anbn?p?bn?K,固定n并令p???,得

anbn?K.所以limn???anbn???.

(3)若l???,可用?an代替an转化为(2)的情形.

?定理2 设{an}是任意数列, {bn}严格递增趋于?,则当limn???an?1?anbn?1?bn存在或为??、??时，有limn???anbn?limn???an?1?anbn?1?bn.

证明 设limn???an?1?anbn?1?bn?l.

(1) 若l是有限实数,则??由于bn?1?bn?0,所以??l??0,?N?0,当n?N时,有l??2?an?1?anbn?1?bn?l??2.

??????(bn?1?bn)?an?1?an??l?22????(bn?1?bn)?,

,

???????l??(bn?2?bn?1)?an?2?an?1??l??(bn?2?bn?1)2?2???………

???????l??(bn?p?bn?p?1)?an?p?an?p?1??l??(bn?p?bn?p?1),

2?2???上述各式相加得

???l?2?????(bn?p?bn)?an?p?an??l?2??an?p?anbn?p?bn?l???(bn?p?bn)?.

由此便得 l??2??2.

所以

an?p?anbn?p?bn?l??2.

由恒等式

anbn?l?aN?lbNbn?bN??1??bn??????an?aN???l??b?b?N?n?

得

anbn?l?aN?lbNbn?an?aNbn?bN?l

由于bn???(n???),?N1?0,当n?N1时,有

aN?lbNbn??2.

因此当n?max{N,N1}时,

anbn?l??2??2??.这证明了limn???anbn?l.

(2)若l???,则当n充分大时,有an?1?an?bn?1?bn.由bn???(n???),

bn?1?bnan?1?an可知an???(n???),且数列{an}严格递增.注意到lim?0,

n???由(1)的结论得limn???bnan?0.从而limn???anbn???.

(3)若l???,可用?an代替an转化为(2)的情形.

定理1与定理2统一称为Stolz定理.

例1 利用Stolz定理.证明(§1例7)：设liman?a.证明limn???a1?2a2???nan?2n?a2.

n???证明 令An?a1?2a2???nan, BnAnBnAn?1?AnBn?1?Bn?n2,则{Bn}严格递增趋于??limn???,由定理2,

a2limn????limn????limn???(n?1)an?1(n?1)2n?1?n?1??n22an?1?12n???liman?．

例2 求极限limn???1k?2k???nk?1k,其中k为自然数.

nk解 令ananbn?1k?2???nk, bn?nk?1,由定理2,

(n?1)kk?1limn????limn???an?1?anbn?1?bn?limn????n?1?k?1?limn???(n?1)k?n?k?1?nk???1k?1.

其中倒数第二式中…表示关于n的次数为k 例

?1的一个多项式.

?1k?2k???nk1?3 求极限limn???k?1??n???k?1n????k?1?(1k,其中k为自然数.

解 令ananbn?2k???n)?nkk?1, bn??k?1?nk,由定理2,

klimn????limn???an?1?anbn?1?bn?limn????k?1?(n?1)?nk?1k??n?1?kk?1?k12?1?[?n?1??n]

?limn????k?1?k??k?1?k??2??k?1?knk?1?k?1n???????.

?2的多项式.

其中倒数第二式分子与分母中的…均表示关于n的次数为k注 例3中当k不是自然数时,只要k1n?0(该条件保证

limn????k?1?nk???),利用定理2,并令

x?,我们有limn???anbn?limn???an?1?anbn?1?bn

?limn????k?1?(n?1)k?nk?1k??n?1?kk?1?k?1?[?n?1??n]?limn???1??k?1??n??k1???n?n?1??n??kk

?k??1??1???1??n????k??1???k?1?x???limx?0k1x?1xk?1?x?k?k?1??1?x??1???limx?0kx?1?x??1??1?x?k?k?1?x?1?x??x?k?.

12再利用求函数极限的罗必塔法则,可以求出最后一式的极限为.

例4设limn(An?An?1)?0.试证:极限limn???A1?A2???Annn???存在时,limAn?limn???A1?A2???Ann.

n???证明 因An?An?而极限limn???A1?A2???Ann?A1?A2???Ann,

A1?A2???Ann存在,故只需证明第一项趋于零.

令a1?A1,a2?A2?A1,…,an?An?An?1,则由条件limn(An?An?1)?0知limnan?0,

n???n???且An?(An?An?1)?(An?1?An?2)?A1?A2???An?于是lim??An??

n?????(A2?A1)?A1?an?an?1???a1.

?n?a1??a1?a2?????a1?a2???an????lim??a1?a2???an???n???n???limn???

a2?2a3????n?1?ann(应用定理2)

?limn???(n?1)ann??n?1??limn???n?1n?n?an?0.

例5 设0?x1?1,xn?1?xn?1?xn?(n?1,2,?).证明limnxn?1.

n???证明 由条件

xn?1xn?1?xn.用数学归纳法容易证明对所有自然数

n有0?xn?1,即

存在,

0?1?xn?1.所以数列{xn}是严格单调递减有下界的.由单调有界定理,极限limxnn???设极限值为a.在xn?1?xn?1?xn?中令n???得a由于{1xn}严格单调递增趋于???a?1?a?,由此得a?0.

,根据定理2, limnxn?limn???n1xn

n????limn????n1?1??n?1xn?limn???xnxn?1xn?xn?1?limn???xnxn?1x2n?lim(1?xn)?1.

n???xn?1§1.3 利用压缩影像原理和单调有界定理求极限

压缩影像原理 设f(x)可导且f'?x??r?1,r是常数.给定x0,令xn?f(xn?1)(n?1,2,?).证明序列{xn}收敛.

证明 由拉格朗日中值定理，得

xn?1?xn?f(xn)?f(xn?1)?f'???(xn?xn?1)?rxn?xn?1?r2xn?1?xn?2???rnx1?x0

其中?介于xn,xn?1之间.故对任意自然数p有

xn?p?xn?xn?p?xn?p?1?xn?p?1?xn?p?2???xn?1?xn?rn?p?1

x1?x0?r1?rpn?p?2x1?x0???rrnnx1?x0?rnx1?x01?r???rr?1).

?p?1??rnx1?x0?1?r?x1?x0?1?r?0（n???,0?由柯西收敛准则{xn}收敛.

注 (1)利用压缩影像原理必须保证{xn}是否保持在f'?x??r?1成立的范围之内. (2) f(x)称为压缩映射(因为0?例1 设x1?0,xn?1?解 令f?x??又

f'?x??3?1?x?3?xr?1)．

3?1?xn3?xn?(n?1,2,?),求极限limxn.

n???(x23?0),则xn?1?f(xn)(n?1,2,?). ),故f(x)称为压缩映射.

3?1?xn3?xn6(3?x)2?(x?0由压缩影像原理,{xn}收敛.再对递推公式xn?1?

?,两边取极限即可.

例2设K是正数,x0?0,对任意自然数n,令xn?证明 令f?x??(x?K1?K?xn?1?2?xn?1???.证明limxn??n????f'?x??K.

1?K??x??2?x?(x?12K),则xn?f(xn?1)(n?1,2,?).又

1?K??1?2? 2?x?),从而有0?f'?x??.故f(x)称为压缩映射.由压缩影像原理, {xn}收敛.再对递

推公式xn?1?K?xn?1?2?xn?1???,两边取极限即可. ??

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