Question

利用ε-δ极限定义证明

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Answer 1

∀ ε > 0 , ∃ δ = ε / (1 + ε) , ∀ x 满足 0 < | x - 1 | < δ , 都有
| x - 1 | < δ
-δ < x - 1 < δ
-ε / (1 + ε) < x - 1 < ε / (1 + ε)
1 / (1 + ε) <散缓 x < (1 + 2ε) / (1 + ε)
(1 + ε) / (1 + 2ε) < 1/x < 1 + ε
1 - ε / (1 + 2ε) < 1/x < 1 + ε
- ε / (1 + 2ε) < 1/x - 1 < ε
- ε < 1/x - 1 < ε
| 1/x - 1 | < ε
证得极丛掘穗限成立

∀ ε > 0 , ∃ δ = ε , ∀ x 满足 0 < | x + 3 | < δ , 都有
| x + 3 | < δ
| x + 3 | < ε
| (x - 3) + 6 | < ε
| (x - 3)(x + 3) / (x + 3) + 6 | < ε
| (x² - 9) / (x + 3) + 6 | < ε
证得极限成立

∀ ε > 0 , ∃ δ = ε / 5 , ∀ x 满足 0 < | x - 1 | < δ , 都有
| x - 1 | < δ
| x - 1 | < ε / 5
| 5x - 5 | < ε
| (5x - 3) - 2 | < ε
证得极限成立

怎么简单的都放后面渗卜.. 晕@@

Answer 2

任取€(伊普斯冷)>0,｜1－或喊1/x｜<€,|(x-1)/x|<€设x>0.|x-1|<x€,令x€=￡>0（得尔塔），于是对任意的€>o,存在￡>0,只要|x-1|<￡,使｜1－1/x｜<€成立弊兄，故lim1/x=1,x→1时。租团袭