?lim[(n?1)k?nk]?0n??.
(2)记a?max{a1,a2,?,am}
n则有
nnan?na1n?a2???am?nm?an 即 a?而 故 即
1nnnn??a?a???a?m?a
1nn??n1n2nm1nlima?a, limm?a?a,nnlimna1n?a2???am?an??
.
nnlimna1n?a2???am?max{a1,a2,?,am}n??(3)?(3)?(1?2?3)?(3?3) 即 3?(1?2?3)?3而 故
n1nnn?1n1nn1nn
lim3?3, lim3n??n??n1nnn?1n?3lim(1?2?3)?3n??.
(4)
?1?1?11?1?nn
1lim1?0, lim(1?)?1n??n而 n??
故
22. 证: (1)?x1?lim1?n??1?1n.
2?2,不妨设xk?2,则
xk?1?2xk?2?2?2.
x故对所有正整数n有xn?2,即数列?n?有上界.
又
xn?1?xn?2xn?xn?xn(2?xn)
xn?0,又由xn?2得xn?2,从而xn?1?xn?0即xn?1?xn,
x即数列?n?是单调递增的.
显然有由极限的单调有界准则知,数列?设n??xn?有极限.
?limx?22a,于是a2?2a,a?2,a?0(不合题意,舍去),n??n.
xxn?1?1?n1?xn, (2) 因为x1?1?0,且
,则a?所以0?xn?2, 即数列有界
limxn?a
x??xxn?xn?1??xn?1?xn??1?n???1?n?1???1?xn??1?xn?1?(1?xn)(1?xn?1) 又
由1?xn?0,1?xn?1?0知xn?1?xn与xn?xn?1同号，
从而可推得xn?1?xn与x2?x1同号， 而
x1?1,x2?1?13?,x2?x1?022
故xn?1?xn?0, 即xn?1?xn
所以数列{xn}单调递增，由单调有界准则知，{xn}的极限存在. 设n??limxn?a, 则
a?1?a1?a，
解得 所以
a?1?51?5,a?22(不合题意，舍去).
limxn?n??1?5.2
22. 证：(1)???0,要使
1sinxsinx????0?xxx,
只须
x?1?,取
X?1?，则当x?X时，必有
sinx?0??x,
sinx?0故x???x. (2)???0,要使 lim13133x2?1?????3222|x|x?4x?4,
只须
x?13?,取
X?13?，则当x?X时，必有
3x2?1?3??2x?4,
3x2?1lim2?3x??x?4故. (3) ???0,要使
x2?4?(?4)?x?2??x?2,
只要取???，则
x2?4?(?4)??当0?x?2??时，必有x?2,
x2?4lim??4x??2x?2故. (4) ???0,要使
11?4x2?2?2x?1?2x???22x?1,
只须
x?1?????22，取2，则
1?4x210?x????2??2当时，必有2x?1 1?4x2lim?212x?1x??故
2.
(5) ???0,要使
xsin只要取???，则 当0?x?0??时，必有
11?0?xsin?x??xx,
xsin1?0??x,
1?0x?0x故 .
?x2?3?9?33x2?3limx?3(1)lim2???2x?3x?1lim?x?1?9?15x?323. 解：.
limxsinx?x12?1x?1(2)lim4????2.x?1x?3x2?1lim(x4?3x2?1)14?3?12?1x?12lim(x2?x)x2?11(3)lim2?lim?.x??2x?x?1x??112??22xx11?11lim??3??3?3x???xx?xx??0.(4)lim4?limxx?2x??x?3x?1x??3131?1?2?4lim?1????x???xxx2x4?21lim?21??2x????2?2x?1?xx??0xx?(5)?lim2?limx??x?1x??11?1?2lim?1???x???xx2?x2?1lim??x??2x?1由无穷大与无穷小的关系知， .
1?1x2
(n?1)(n?2)(n?3)1?1??2??3??lim?1???1???1??n??5n35n???n??n??n?1?1??2??3?1?lim?1???lim?1???lim?1???.5n???n?n???n?n???n?5
x2?1(1?a)x2?(a?b)x?(1?b)?ax?b?x?1(7)因为x?1
?x2?1?1lim??ax?b??x???x?1?2知，分式的分子与分母的次数相同，且x项的系数之由已知(6)lim1比为2，于是
?(a?b)1?1?a?0 且 12
解得
a?1,b??(1)lim32.
25.解：
1?2?3???(n?1)n(n?1)1?1?1?lim?lim?1???.22n??n??n??n2n2?n?2
n?11?1????11??(2)lim?1????n??lim?2??2.n???22?n??1?12x2?2x?1(x?1)2(3)lim?lim?lim(x?1)?0.x?1x?1x?1x?1x?1x2?6x?8(x?2)(x?4)x?22(4)lim2?lim?lim?.x?4x?5x?4x?4(x?1)(x?4)x?4x?13
334x2(5)limx?x3?2?x3?2??lim?lim33x???x???x?2?x?2x???41?22?1?x3x3?2.x2(1?1?x2)2(6)lim?lim??lim(1?1?x)??2.22x?0x?0x?0?x1?1?xx2?3x?35??3x2?35x?325??x?5?x?35(7)lim?limx?5x?5x?5?x?5??x?5??3x2?35x?325?3?lim?lim
(x?5)?x?5?x?5(x?5)32?x?35x?325?x?5x2?35x?325x?53?252?.3632535
1?cot3x1?cot3x(8)lim?lim3π2?cotx?cot3xπx?x?(1?cotx)?(1?cotx)44(1?cotx)(1?cotx?cot2x)?lim2πx?(1?cotx)(1?1?cotx?cotx)41?cotx?cot2x3?lim?.π2?cotx?cot2x4x?4
(9)lim(1?x)(1?x2)?(1?x2) (x?1)x??n(1?x)(1?x)(1?x2)?(1?x2) ?limx??1?x1?x21 ?lim?.x??1?x1?x(1?x)(1?3x)?(1?nx)(10)limx?1(1?x)n?1 ?limx?1n?1n
(1?x)n?1(1?x)n?1(1?x)(1?3x?3x2)(1?4x?4x2?4x3)?(1?nx?nx2???nxn?1)1(1?x)(1?3x?3x2)(1?4x?4x2?4x3)?(1?nx?nx2???nxn?1)11 ??.2?3?4???nn!1?x?x2?3x2?x?23??1(11)lim??lim?lim?23?x?1?1?xx?1(1?x)(1?x?x)x?1(1?x)(1?x?x2)1?x?(x?1)(x?2)?(x?2) ?lim?lim??1.x?1(1?x)(1?x?x2)x?11?x?x2
lim(x?1)2(x?1)2(12)?lim2?x?12?0x?1x?x?1lim(x?x?1)x?1 ?limx?1x2?x?1 ?lim??.x?1(x?1)21loga(1?x)(13)??loga(1?x)xx
而x?0lim(1?x)?e.1x 而u?elimlogau?logae?1lna
?limloga(1?x)1?.x?0xlna
xu?a?1,则x?loga(1?u),当x?0时，u?0. (14)令
ax?1u1lim?lim??lnax?0u?0log(1?u)xloga(1?u)limau?0u所以(利用(13)题的结果).