（5）lim?1?2x?=lim(1?2x)1xx?0x?01.22x?e2

?e?2（6）

?3?x?lim??x??1?x??x=

2lim(1?)x??1?x1?x.(?2)?12

10. 利用极限存在准则证明: ⑴

11?1?limn?2?2???2??1n??n??n?2?n?n???

2n211?1?2?n?2?2???2n?n?n?n??n??n?2?n???2?n??

n2n2又lim2?1,lim2?1n??n?n?n??n??

的极限存在,

故原式＝1 ⑵ 数列2,并求其极限.

xn?2?xn?1,n?2,3,...解：10.先证单调。2?2,2?2?2,??x2?2?x1?2?2?2?x1,假设xk?xk?1,则xk?1?2?xk?2?xk?1?xk故?xn?单调递增。20.再证有界。x1?2?2,假设xk?1?2,则xk?2?xk?1?2?2?2故?xn?有界。所以limxn?,设limxn?a,由xn?2?xn?1知a?2?an??n??

所以a?2,a??1(舍去)?limxn?2n??

11. 当x?0时, 2x?x与x较高阶的无穷小?

22?x3相比, 哪一个是

x2?x3x2(1?x)lim?lim?0x?02x?x2x?0x(2?x)

2?当x?0时，x2?x3是较高阶的无穷小。12. 当x?1时, 无穷小1?x和1?1?x?是否同阶?2是否等价?

1（1-x2）(1?x)(1?x)lim2?lim?1x?11?xx?12(1?x)1?当x?1时，(1-x2)1-x2所以同阶且等价.

x2secx?1~213. 证明: 当x?0时, 有.

1?1secx?12(1?cosx)1cosxlim?lim?lim.x?0x?0x?0x2x2x2cosx22x4sin22.1?1?limx?0x2cosxx2?当x?0时，secx?12

14. 利用等价无穷小的代换定理, 求极限:

limtanx?sinxx?0sin3x.

12x(x)1tanx?sinxtanx(1?cosx)2lim=lim?lim=x?0x?0x?0sin3xx3x32

15. 讨论其图形.

?x20?x?1f?x????2?x1?x?2 的连续性, 并画出

f(1?0)?limx2?1x?1?f(1?0)?lim(2?x)?1x?1?又f(1)?1,?f(x)在x?1处连续.总之,f(x)在[0,2]上连续.

16. 指出下列函数的间断点属于哪一类.若

是可去间断点,则补充或改变函数的定义使其连续. ⑴

x2?1y?2x?3x?2x?1,x?2

x2?1(x?1)(x?1)lim2?lim??2x?1x?3x?2x?1(x?1)(x?2)?x?1为可去间断点,补充定义:yx?1??2即可.limx?2x?1(x?1)?lim??,2x?2x?3x?2(x?2)2

?x?2为无穷间断点.⑵

yx?1?x?1y???3?xx?1x?1x?1

=0

x?1?x?1?limy?lim(x?1)?0x?1?x?1?limy?lim(3?x)?2x?1为其跳跃间断点.

1?x2nf?x??limxn??1?x2n17. 讨论函数

的连续性, 若有间

断点, 判别其类型。

?1,x?1?1?xf?x??limx??0,x?1n??1?x2n???1,x?1在x?1处,f(?1?0)??1,f(?1?0)?12n?x??1为跳跃间断点.在x?1处,f(1?0)?1,f(1?0)??1?x?1为跳跃间断点.

的连续区间, 并

18. 求函数 求limf?x?,limf?x?.

x?0x??3x3?3x2?x?3f?x??x2?x?6由x2?x?6?0得：x1?2,x2??3?连续区间为（－?，－3）?（－3，2）?（2，＋?）1limf(x)?x?02(x?3)(x2?1)x2?1?8limf(x)?lim?lim?x??3x??3(x?3)(x?2)x??3x?25

19. 求下列极限：