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习题1.1 1. 用定义(“ε-n0”语言)证明: (1)11
lim
=+n
n 证:对0>∀ε, ε
1
=∃N,当[]Nn>时:
ε<=−+nnn111。故由定义得11
lim=+n
n (2)0sinlim
=n
n
证:对0>∀ε, ε
1
=
∃N,当[]Nn>时:
ε<<−nnn10sin。故由定义得0sinlim=n
n
(3)2
1
121lim2
2=++nn 证:对0>∀ε, ε
1
=∃N,当[]Nn>时:ε<<−++2
221
21121nnn。故由定义得2
1
121lim2
2=++nn (4)01
11
lim
=++n 证:对0>∀ε, 2
1
ε=
∃N,当[]Nn>时:
ε<< 本文来自文库分享网www.wkfxw.com
−++n
n101
11。故由定义得