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. ??

例3 设a?0,x1?a,x2?a?a,…, xn?a?a???a.求limxn.

n???解 容易证明{xn}单调递增.现证对任意自然数n,xn?xn?a?1.则xn?1?a?xn?a?1.当n?1时显然成立.归纳假设

a?a?1?a?2a?1?a?xna?1.

由单调有界定理, {xn}有极限.设limxn?x.对xn?1?n???两边取极限得x?a?x.

解得x?

1?1?4a2.由于x?0,故得limxn?n???1?1?4a2.

例4 设x1?4,当n?2时, xn?1xn?12?xn?12.求limxn.

n???解 显然xn?0.由于xn?1?xn?xn?2?2xn2xn2?2?xn2xn2与xn?1xn?1?xn?12?21xn?1?xn?12?2

所以xn?1?xn?0,即{xn}单调递减且有下界.故{xn}极限存在,令limxn?x.

n???由递推关系式得x

?1x?x2.解得x?2,即limxn?n???2.

例5 设x1?0,且对任意自然数n,xn?12?xnxn?3a3x2n?2?其中a?0?a.求limxn.

n???解 由于

xn?1xn?xn?3a3x2n,

?a2xn?1xn?1?xn?3a?3xn?a?xn?a222,与xn?1?a2?xnxn?3a?2?2?a?3xn?a?222(3xn?a)2

?x?x?n2n?3a??a3xn?a(3x2n?2???x?xn2n?3a??a3xn?a?2?a)2n2??

?x?n?a??x32n?2a?3(3xn?a)??x?a?3(3xn?a)22

故xn2?a与xn2?1?a同号.因此当x12?a时有xn2?a(n?1,2,?),此时{xn}递增有上界a;当x12?a时有xn2?a(n?1,2,?),此时{xn}递减有下界a. 所以{xn}收敛,设limxn?x.则xn????xx3x?22?3a?a?.因为x?0,解得x?a,即limxn?n???a.

例6 设xn?1?12?13???1n?lnn,证明{xn}收敛.

11证明 由xn的定义, xn?1?xnn?1??1??????ln(n?1)?lnn??ln(1?).由于??1???n?1n?1nn??????1单

1?调递减趋于e,故??1??n??n?1?e.取对数得?n?1?ln(1?1n)?1,ln(1?1n)?1n?1.所以这证明了

{xn}单调递减.

n?1????又由于??1????n??????单调递增趋于e,可得不等式

1n1n?ln(1?1n).

因此1?12?13????(ln2?ln1)?(ln3?ln2)????ln?n?1??lnn??ln?n?1??lnn.

所以xn?0(n?1,2,?),由单调有界定理,{xn}收敛.设limxn?C,这里C称为Euler常数.

n???可以证明0?C?1(C?0.5775216?).

例7 设x1?1,x2?12,

xn?1?11?xn.求limxn.

n???解 若极限存在,设为A,则A?因xn?0(n?1,2,?),A若xn?A,则xn?1?11?xn11?A,A2?A?1?0,A??1?211?xn5?0.618??????1.618?.

?0.618?.若xn?A,则xn?1?A??11?A?A;

?0.618??11?A.即xn在A的左右来回跳动,而x1?1

知:x1,x3,x5,?,x2n?1,??A,x2,x4,x6,?,x2n,??A (1).

若{xn}收敛于A,则{x2n},{x2n?1}也收敛于A.猜想:是否{x2n}在A左端单调递增到

A,{x2n?1}在A右端单调递减到A.下面来考xn?2?xn察的符号.

xn?2?xn?11?11?xn211?xn?1?xn

?xn

??1?xn?xn2?xn

?(1.618??xn)(0.618??xn)??0,???0,2?xn?若xn?0.618??A,若xn?0.618??A. (2).

式(1),(2)表明{x2n}以A为上界, {x2n?1}以A为下界. 因此二子列收敛.记limx2n??,limx2n?1??.

n???n???在式x2n?1?11?x2n及x2n?11?x2n?1n???中令n???,有??11??,??11??.

所以????A?0.618?.既然limx2n?limx2n?1?A,故limxn?A?0.618?.

n???n???

例8 证明序列2,2?12,2?12?12,?收敛,并求其极限.

解 以序列特征可以看出,相邻两项的关系是xn?1满足方程A?2?1A?2?1xn (1).因此,设{xn}收敛,则极限A2??.又xn?0,所以A2??1?2.令xn?A??n?1?n (2).(2)代入

(1), ?n?1?2?1?1??nn2?? (3).则将满足(1)的序列{xn}?A的问题,转化为满足(3)的序列

12{?n}?0的问题.事实上, ?1?x1?A?1?,即limxn?lim(1?n???2,?1?.由(3)利用数学归纳法,易证

?n?12nn???2??n)?1?2.

§1.4 求函数极限的几种方法

一、利用函数的连续性求极限

定理 (复合函数求极限定理) 设函数f(y)在y?b连续,函数y?g(x)有性质limg(x)?b,则

x?alimf[g(x)]?f[limg(x)]?bx?a.

u(x)v?x?x?a推论 设limu(x)?A?0,limv(x)?B,则limx?a?AB.

x?ax?a证明 由复合函数求极限定理,

limu(x)x?av?x? ?limex?av?x?lnu?x??ex?alimv?x?lnu?x??eBlnA?AB

例1 求极限limx?0ax?1x.(a?0)?0

时y?0.解得xlnaln?1?y?y1解 令ax?1?y,则当x故limx?0?ln?1?y?lna.

ax?1x?limy?0ylnaln?1?y??limy?0?lna.

注 此例中取x 例

?1n?0,得数列极限, limn?na?1??0.(a?0)

n????2 求极限lim?n?????nna?2nb??,?a?0,b?0? ??n解 令由于

a?2nb?1?xnn,则xnnn?a?2nb?1?0(n???).

n???limxn??limn??n????a?2b?n?1??lim?n???2??na?1?nb?1??12?lna?lnb??lnab,

?所以lim?n?????na?2nb????n??lim(1?xn)?lim?1?xnn???n?????n?x1n???nxn?elnab?ab.

1sinx?例3 求极限lim???x?ax?a.?a?k??

sinx?sina1sina?sina?11sinx?解 lim???x?asina??x?a??sinx???lim?1???1??x?a?sina???x?a?sinx?sina?sin??lim??1??x?a?sina???1cosasinax?sina????x?a?.

由于limx?asinx?sinax?a?sinx??cosa,所以lim??x?asina??x?a?esina?ecota.

例4求极限limx?0xcosx.

?12cosx?1x解 注意到limx?0cosxx?1,我们有

1alimx?0xcosx?lim1?cosx?0??x?1????lim?1?cosx?0???x?11??1cosx?1????e.

2练习 求极限 (1)

?cosx?lim??x?0cos2x??x2?; (2) lim?2?3xxx?0??2?3xx22????x.

二、利用微分学方法(L’Hospital法则,Taylor公式)求极限 例5 求极限limx?0?1?x?xx1?e.

?u'?x????v'?x?lnu?x??v?x??u?x???解 由导数公式

ddx1ddxu(x)v?x??u(x)v?x?

得

1(1?x)x?(1?x)xx??1?x?ln?1?x?x2?1?x?

由L’Hospital法则得

limx?0?1?x?xx1?e?elimx?0x??1?x?ln?1?x?x2?1?x??elimx?0?ln?1?x?2x?3x2??e2.

例6 求极限lim?1?2?2?cotx?. x?0?x?解 利用L’Hospital法则与等价无穷小代换得

?1lim?2?cotx?0?x2sin?x??limx?0?2x?xxsin222cosx2x?limx?0sin2x?xx42cos2x(等价无穷小代换)

?limx?0sinx?xcosxxsinxx?sinx?xcosxx3(化简) ?2limx?0cosx?cosx?xsinx3x2(L’Hospital法则)

?23limx?0?23.

1例7求极限lim?a?x?0??x?bx?cx3????x

解 由指数函数的连续性与L’Hospital法则得

1?alim?x?0??x?bx?cx3????xxxx?a?b?cln?3?exp?limx?x?0??3???a?explim??x?0????xlna?baxxlnb?cxx?b?cxlnc???

?lna?lnb?lnc??exp???3??abc.其中expy表示指数函数ey.

例

?x1?xx?8求极限lim???xx???e??x?1??

解 原极限可化简为

?x1?xx?lim????limxx???x???e??x?1??x?1???x?e??1???x??????. (*)

1??e?1??x??x(*)式中分母的极限为e2,因此只要求分子的极限.

x?1???limx?e??1????limx???x???x??????1??e??1??x??1xx?limy?0e??1?y?yy1(利用代换y?1x)

1?lim(1?y)y?0y?1?1??ln1?y??2?y?1?y???y (L’Hospital法则)

因此只要求

?1??1?y?ln?1?y??y?1?y?ln?1?y??y1lim????lim?limln1?y??222y?0y?0y?0y?1?y??y?1?y?y?y

(利用lim?1?y??1)?y?0limy?0ln?1?y??1?12y?12.

因此(*)式中分母的极限为

e2,最后得到

?x1?xx?2lim????xx???e?e??x?1?lim.

例9 设f(x)在?0,???上连续,且limf?x??a,求数列极限

x???n????10f?nx?dx．

解 将数列极限转化为函数极限,然后利用L’Hospital法则

limn????10f?nx?dx?limn????n0f?t?dtn (变量代换t?nx)?limy????y0f?t?dty(将n换成连续变量y)

?limy???f?y?(L’Hospital

法则) ?a

例10 求极限limx?0tanx?xsin3.

x3x解 利用Taylor公式 tanx?x?我们有

x??limx?03?ox??,sin3x?x?o?x?

x3limx?0tanx?xsin33?ox???33x1?limx?03?oxx??333x?x?o?x???13.

o?x????1??x??三、利用定积分求极限

定理2 (1)若f(x)在?a,b?上可积,则

?baf?x?dx?limn???b?ann?i?1b?a?b?a?f?a?i??limn???nn??1n?1?i?0b?a??f?a?i?n??.

(2) 若f(x)在?0,1?内单调,且积分?f?x?dx存在(可以是非正常积分),则

0?10f?x?dx?limn???1n?ni?1?i?f???n?(当0是瑕点时)?limn?????1n?1?ni?0?i?f???n? (当1是瑕点时).

(3) 若f(x)在?0,???内单调递减,且积分?0f?x?dx存在,则

???0f?x?dx?limh?f?nh?h?0n?0???.

证明 (1)由定积分定义直接得.

(2)当f(x)在?0,1?上可积时结论显然成立.

设?f?x?dx是非正常积分,不妨设0是瑕点并设f(x)在?0,1?内单调递减,显然有

0i?11?nin?i?f???f?x?dx?n?n?1i?nni?1nf?x?dx ?i?1,2,?,n?1?.

对i求和得 ?11f?x?dx?1nf?1??1n?i?1n?i?f????n??10f?x?dx.

令n???,得 ?01f?x?dx?limn???1n?ni?1?i?f???n?.注意到?11f?x?dx??1n0f?x?dx

（3）由于f(x)在?0,???内单调递减,对任意正数h有

??n?1?hnhf?x?dx?hf?nh???k?1?h??nhn?1?hf?x?dxk.?n?1,2,?,k?

对n求和得 ?hf?x?dx?h?f?nhn?1??????kh0f?x?dx.

令k???得 ?h??f?x?dx?h?f?nhn?1??????0f?x?dx.

再令h?0得 ?0

例11 求极限limn?????f?x?dx?limh?f?nh?h?0n?0?.

1k?2k???nk?1k,其中k是大于-1的实数.

n解 利用定理2得

limn???1k?2k???nk?1kn?limn???1n?ni?1?i????n?k??10xdx?k1k?1.

例12 设xnn??n?1??n?2???n?n?nnk,求极限limxn.

n???解 因为lnxn?2ln2?11n?i?1?i????n?,所以limlnxn?limn???1nnn????i?1i??ln?1???n???10ln?1?x?dx?2ln2?1,故

n???limxn?e?4e.

例13求极限limtann???1n?ln1nn!nn.

1n解 由于n???时, tan故原极限??limn???~.

limn???1n?ln?lnn!?nln1??limn???1n??ln1?lnn???ln2?lnn?????lnn?lnn??

1nn?i?1lnin??0lnxdx??1.

练习 (1) 求极限lim??n???n2?n?12?nn2?22???nn2?n22???.

.

(2) 求极限limn???1?22n?1?2?n?n2?22???n2??n?1???? 例14

? 求极限lim?n?????n?1n?2n?12n???n?1??2n??2.

解 原极限可变为

nlimn????k?1kn?1kn?limn???1n?nnk?1k2k?1??lim???n???nk?1?1nnk?1k2?1n???limn???nk??n?k?1nk.

1由于

1???nk?1??nnk?1k2?n???1k?n?n?k?1knk?1k2?1n2?nk?n1k2?0(n???),

k?1所以由定理2得 原式?limn???1n?nnk?k?1?1dxx?2.

0

例15 证明limt?1???1?t?tn?0n2??24.

分析 对于级数?n?0??tn2?1?t?t?t9?t16??,没有现成的求和公式可用.但是我们知道,当

t?1时级数收敛,当t??1时级数发散.又当t?1???时, ?n?0??tn2???.另一方面,注意到

?0e?x2dx??2,由定理2可得如下证明.

????2证明

?2??????0e?x2dx?limh?e?h?0n?0??nh?2?lim?t?1?lnt?tn?0n2(令t?e?h)

?lim?t?11?t?tn?0n2?lim?t?1lnt1?t???lim?t?11?t?tn?0n2

其中limt?1lnt??1是容易证明的.

1?t

例16 求极限limn???nn2?n!?.

解 令an?nn2?n!?,则limn???an?1an?limn????n?1?n?1?n!?2???n?1?!?2nn1???lim??1??n???n?1n??nn21n?0?1,所以级数

???n?1an收敛.从而由级数收敛的必要条件,liman?0,即limn???n????n!?n?0.

练习 求极限 (1) limn???5n?n!n?2n?,(2) limn???2n?n!nn,(3) limn???n3n.

?n!

例17 求极限limsinn?????n2?1??. ??0解 limsinn????nn2?1??limsinn???2n2?1??n??n???limn?????1?nsin?n2?1??n??

?limn?????1?sinn2?.

?1??n?2练习 求极限limsinn?????n2?n?.

.

23?45???2n2n?1

例18 求极限limn???1?3????2n?1?2?4????2n?2n?12n解 令xn0?xn??12?34???,

yn?,则xn?yn且xn2?yn?xn?12n?1.于是

12n?1?0(n???),即limxn?0.

n???

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