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.

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