2007113165515976 - 1 lim x 2 = 9 x 3 < x 2 9 < 0.007...

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1 函数极限例题与课外练习 一、典型例题 1 2 3 lim 9 x x = ,问 δ 取何值时,才能由 3 x < ,使 2 9 0.007 x −< 成立 解:因 3 x ,可限制 03 1 x <−< , 3, x 24 x < < ,所以 37 x +< ,因 2 93 3 7 3 xx x x −=− +< − ,显然,当 73 x ε < 3 7 x < 2 9 x ,取 min{ ,1} 7 = , 0.007 = ,取 0.001 7 == ,有 3 x < , 使 2 9 0.007 x 成立 2 证明 2 2 1 12 lim 23 x x = +− 证 因 1 x ,但 1 x ,所以 2 2 ( 1 ) ( 1 ) 2 1 2 1 2 3(2 ) (1 ) 3 2 33 (2 ) x x x x x x x −− + + −= + + + 0 ∀> ,由 1 3( 2) x x < + 有困难,故先限定 1 01 6 x < 57 66 x ⇒<< , 1 x 17 2) 2 x +> ,即 1 7 11 3( 2) 17 2 x x <− < < + 7 min{ , } 62 = ,则当 x < , 2 2 x < 3 据极限定义证明 0 00 lim ( ) lim ( ) lim ( ) fx A fx −+ →→ = ⇔== . 证 必要性:若 0 lim ( ) f xA = ,对 0, 0 >∃> ,使当 0 0 < 时,有
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2 () fx A ε −< . 因而,在 0 0 xx δ <− < 时,有 < 0 lim ( ) f xA + = ,当 0 0 <− < 时,有 0 lim ( ) f = 充分性:若 0 lim ( ) f + = 0 lim ( ) f = ,对于 1 0, 0 >∃> ,当 01 0 时,有 ;也必存在 2 0 > 02 0 < 时,有 ,取 12 min{ , } δδ = ,则当 0 0 < 时,有 ,因 0 lim ( ) f = .
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2007113165515976 - 1 lim x 2 = 9 x 3 &lt; x 2 9 &lt; 0.007...

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