从洛必达法则谈起 下载本文

XXXX数学与计算科学学院2011届毕业论文

推论3.4.1 设f(x)是一个正的连续函数,常数p?1满足limx?0f(x)?a?0,令 xpxn?1?xn?f(xn),n?1,2,???,如果{xn}为正数列且收敛于0,有

limnxnp?1?n??1.

a(p?1)证明方法与结论3.4.1相同.

结论3.4.2 设a?0,p?1,取x0为正数使得0?ax0p?1?1,令xn?1?xn?axnp,那么有

n[1?a(p?1)nxnp?1]p lim (5) ?n??lnn2(p?1)p证明:由递推公式xn?1?xn?axn,可得到

1xnp??111?(1?axnp?1)p?1 (6) ?p?1?p?1p?1p?1xnxn(1?axn)1n[1?a(p?1)nxnp?1]n1n由 ??lnna(p?1)nxnp?1lnna(p?1)nxnp?1lnnn?11?(1?axkp?1)p?1111n ? [p?1??p?1]?p?1p?1lnna(p?1)x0lnnk?1xk(1?axk) ?n111??

lnna(p?1)xnp?1lnna(p?1)1?(1?axkp?1)p?1?a(p?1)xkp?1(1?axkp?1)p?1 ? p?1p?1p?1x(1?ax)k?1kkn?1由施笃兹定理得

n[1?a(p?1)nxnp?1]n[1?a(p?1)nxnp?1] lim?limn??n??lnnlnna(p?1)nxnp?1 ?lim[n??11??

a(p?1)ln(1?n?1)1?(1?axnp?1)p?1?a(p?1)xnp?1(1?axnp?1)p?1 ]

xnp?1(1?axnp?1)p?11?(1?axnp?1)p?1?a(p?1)xnp?1(1?axnp?1)p?11n? ?lim[] p?1p?1p?1n??a(p?1)xn(1?axn)第13页 共18页

XXXX数学与计算科学学院2011届毕业论文

由于 (1?axn)(p?1)(p?2)2p?2(axnp?1)2?o(xn),于是

2(p?1)(p?2)1?(1?axnp?1)p?1?a(p?1)xnp?1(1?axnp?1)p?1??(axnp?1)2?

2p?1p?1?1?(p?1)axnp?1?p?12p?2 a(p?1)xn(p?1)axnp?1?o(xn) (7)

将(7)式代入(6)式得

n[1?a(p?1)nxnp?1]1?(p?1)(p?2)2p?1lim?lim[anxn?a(p?1)2anxnp?1] n??n??a(p?1)lnn2 ?1?(p?2)p. [?p?1]?p?122(p?1)n?? 例3.4.1 设xn?1?xn(1?xn),n?0,1,2,???,0?x0?1,由结论3.4.1,limnxn?1.再由结论3.4.2,可得到

lim 例3.4.2 设0?x0?n(1?nxn)?1

n??lnnn???2,令xn?sinxn?1,n?1,2,???,求证 limnsinxn?1. 3证明:显然{xn}收敛于0,设f(x)?sinx?x,则

limx?0f(x)sinx?xcosx?11?lim?lim??, x?0x?0x3x33x262n??于是由推论3.4.1,我们得到limnxn?3.证毕.

4 托普里兹定理及其应用

4.1 托普里兹定理的证明

在数列极限得计算中,经常会出现“

?”型的极限问题,本文前面介绍了用洛必达法?n则和施笃兹定理来解决这类问题.下面介绍解决这类问题的另一重要工具——托普里兹定理.

托普里兹定理:若(1)pnk?0;n?1,2,???,k?1,2,???,n.(2)

n?pk?1nk(3)?1;

limpnk?0;(4)limtn?l.则lim?pnktk?l.

n??n??n??k?1证明:由(4)知,???0,?n?N,?n?N,有

tn?l??2.

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XXXX数学与计算科学学院2011届毕业论文

由(4)又知,?M?0,?n?N,有

tn?l?M

由(3)知,?N1?N,N1?N,?n?N1,有

pnk??2MN,k?1,2,???,N(N固定)

由(1)、(2)知,?n?N1,有

?pk?1Nk?1nnnkkt?l??pnktk?l

k?1n ??pnktk?l??pk?1nkkk?N?1?npnktk?l??2MN(NM)??2k?N?1?npnk??2??2??

即 limn?? t?l

4.2 托普里兹定理在证明某些经典定理上的应用

例4.2.1(柯西定理)如果limxn?a,则limn??x1?x2?????xn?a

n??nn1证明:令pnk?,k?1,2,???,n,则显然满足条件:(1)pnk?0,(2)?pnk?1

nk?1(3)limpnk?0,又知(4)limxn?a,由托普里兹定理:

n??n??lim?pnktk?limn??k?1nx1?x2?????xn?a.

n??nn??例4.2.2(施笃兹定理)若(1)yn?yn?1,n?1,2,???,(2)limyn???(3)

limn??xn?xn?1x?l,则limn?l

n??yyn?yn?1n证明:令tn?xn?xn?1y?yk?1,pnk?k,k?1,2,???,n,x0?y0?0,则满足条件:

yn?yn?1ynyk?yk?1?0; yn(1)由yn?yn?1,yn??知pnk?nn(2)

?pnk??k?1k?1yk?yk?1?1; yn第15页 共18页

XXXX数学与计算科学学院2011届毕业论文

(3)limpnk?limn??n??yk?yk?1?0(yn??) yn(4)limtn?limn??n??xn?xn?1?l

yn?yn?1由托普里兹定理

lim?pnktk?lim?n??k?1n??k?1nnyk?yk?1xk?xk?1 ?ynyk?yk?1 ?limn???k?1nxk?xk?1x?limn?l n??yynn由此可见,柯西定理和施笃兹定理是托普里兹定理的特殊情形,能用柯西定理或施笃兹定理解决的问题都能用托普里兹定理来处理,当然托普里兹定理还可以解决一些不便用柯西定理或施笃兹定理解决的计算问题.

4.3 托普里兹定理在解决一些数列极限上的应用 例4.3.1 设{pn}是正数数列,且limn??pn?0,而limtn?b,求

n??p1?p2?????pnlimt1pn?t2pn?1?????tnp1.

n??p1?p2?????pn解:令pnk?npn?k?1,k?1,2,???,n,则由题意知,满足条件:(1)pnk?0;

p1?p2?????pnn(2)

?pnk??k?1k?1pn?k?1p?pn?1?????p1?n?1;

p1?p2?????pnp1?p2?????pn(3)pnk?pn?k?1pn?k?1; ??0(假设)

p1?p2?????pnp1?p2?????pn?k?1(4)limtn?b;

n?? 由托普里兹定理: limn???pnktk?lim?k?1n??k?1n??nnpn?k?1tk

p1?p2?????pn ?limpnt1?pn?1t2?????p1tn?b

p1?p2?????pn例4.3.2 已知两个数列{an}、{bn}满足条件:(1)bn?0且b1?b2?????bn????发散,

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