显然有xn?0,又由xn?2得xn?2,从而xn?1?xn?0即xn?1?xn,
即数列?xn?是单调递增的.
由极限的单调有界准则知,数列?xn?有极限. 设limn??xn?a,则a?2a,于是a2?2a,a?2,a?0(不合题意,舍去),?limn??xn?2.
(2) 因为x1?1?0,且xn?1?1?xn1?x, n所以0?xn?2, 即数列有界
又 x?xn??xn?1?n?1?xn???1?1?x???1???xn?xn?1n??1?xn?1?(1?xn)(1?xn?1) 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 x11?1,x2?1?2?32,x2?x1?0 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limxan??n?a, 则a?1?1?a， 解得 a?1?52,a?1?52(不合题意，舍去). 所以 limn??x1?5n?2. 27. 用函数极限定义证明：
(1)sinx3x2?1x2?4xlim???x?0; (2)limx??x2?4?3; (3)xlim??2x?2??4; 1?4x2
(4)limx??12x?1?2; (5)limx?0xsin1?0.2x证：(1)???0,要使
sinxsinxx?0?x?1x??, 11
只须x?1?,取X?1?，则当x?X时，必有
sinx?0??, xsinx?0.
x???x(2)???0,要使
故lim23x?1x2?4?3?13x2?4?13|x|2??, 只须x?13?,取X?13?，则当x?X时，必有
3x2?1x2?4?3??, 3x2故lim?1x??x2?4?3. (3) ???0,要使
x2?4x?2?(?4)?x?2??, 只要取???，则
0?x?2??时，必有x2当?4x?2?(?4)??,
故x2?4xlim??2x?2??4. (4) ???0,要使
1?4x22x?1?2?2x?1?2x?12??,
只须x?12??2，取???2，则
0?11?4x2当x?2??时，必有2x?1?2??
1?4x2故limx??1?2. 22x?1 12
(5) ???0,要使
11xsin?0?xsin?x??,
xx只要取???，则
当0?x?0??时，必有xsin1x?0??, 故lim1x?0xsinx?0. 28. 求下列极限：
(1)limx2?3x2?xx?3x2?1; (2)limx?1x4?3x2?1;x2(3)lim?1x3?xx??2x2?x?1; (4)limx??x4?3x2?1;(5)limx2?1(n?1)(n?x??2x?1; (6)lim2)(n?3)n??5n3;(7)若lim??x2?1??1?ax?b???12，求a和b. x???2x解：(1)limx2?3limx?3?x2?3?9?33x?3x2?1?lim??. x?3?x2?1?9?15(2)limx2?xlim(x?1x2?x)12?1x?1x4?3x2?1?lim(x?1x4?3x2?1)?14?3?12?1??2.x2?11?1(3)xlimx2??2x2?x?1?limx???1.2?112x?x211?(4)limx3?xx?x3lim?1?1? 3?x??x4?3x2?1?limx??1?3?x???xx??0.x?1?31?2x4limx????1?x2?x4??2x?1(5)?lim2x?1x2lim?21?x????x?x2??x??x2?1?limx???01?1?1?x2lim?x????1?x2??由无穷大与无穷小的关系知， limx2?1x??2x?1??. 13
(6)lim(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?5x2?1(1?a)x2?(a?b)x?(1?b)?ax?b?(7)因为 x?1x?1?x2?1?1由已知lim??ax?b??知，分式的分子与分母的次数相同，且x项的系数之
x???x?1?2比为
1，于是 21?a?0 且
?(a?b)1? 12解得 a?1,b??3. 229. 通过恒等变形求下列极限：
(1)lim1?2?3???(n?1)1??1; (2)lim;1?????2n?n??n??n?22?x?1(3)limx?2x?1x?6x?8; (4)lim;22x?4x?1x?5x?43222
(5)limxx???3?x3?2?x3?2?; (6)limx?0x21?1?x2;x?351?cot3x(7)lim; (8)lim;π2?cotx?cot3xx?5x?5x?4(1?x)(1?3x)?(1?nx)(9)lim(1?x)(1?x)?(1?x) (x?1); (10)lim;n?1x??x?1(1?x)22nx2?x?13??1(11)lim?; (12)lim;?23?x?1?1?xx?1(x?1)1?x?loga(1?x)ax?1(13)lim; (14)lim;x?0x?0xx3sinx(15)lim(1?2x)sinx; (16)limln.x?0x?0x解：(1)lim1?2?3???(n?1)n(n?1)1?1?1?lim?lim?1???. n??n??n??2?n22n2n?2 14
1?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(5)limxx???32n?1?x2x?2?x?2??xlim???334x3x?2?x?233?lim41?22?1?x3x3x????2.x2(1?1?x2)(6)lim?lim??lim(1?1?x2)??2.2x?0x?0?x1?1?x2x?0?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?.36325351?cot3x1?cot3x(8)lim?lim3π2?cotx?cot3xπx?x?(1?cotx)?(1?cotx)44(1?cotx)(1?cotx?cot2x)?lim
π(1?cotx)(1?1?cotx?cot2x)x?41?cotx?cot2x3?lim?.π2?cotx?cot2x4x?4(9)lim(1?x)(1?x2)?(1?x2) (x?1)x??n ?lim(1?x)(1?x)(1?x)?(1?x)x??1?xn?122n
1?x21 ?lim?.x??1?x1?x 15