[發問] 微積分證明

看板Chen_C_H作者slavedaeric (慘兮兮啊茫茫然)時間17年前 (2006/10/06 17:56)推噓7(7推 0噓 8→)

留言15則, 7人參與討論串1/1

我們的微積分用的是非常怪異的Calculus for biology and medicine 我看了以後覺得寫得好爛好像重要的東西全部都變成optional 後面還出現一堆矩陣和集合真是神奇的東西從開學到現在才剛剛教完A本那些基本的極限證明可是啊我看書上的証法和上課教的好像不太一樣耶寫上課教的會不會被扣分啊 <課本寫的> Use the formal definition of limits to show that lim√x=2 x→4 Sol: We assume throughout that x>0.We need to show that for every ε>0 there corresponds a numberδ>0 so that ∣√x-2∣<ε whenever 0<∣x-4∣<δ. Now ∣√x-2∣<ε is equivalent to -ε<√x-2<ε 2-ε<√x<2+ε To square all sides,we need to make sure that 2-ε≧0 and 2+ε≧0 Since ε>0 by assumption,2-ε≧0 requires us to assume 0<ε≦2.With 0<ε≦2, squaring yields (2-ε)^2<x<(2+ε)^2 4-4ε+ε^2<x<4+4ε+ε^2 Since 4-4ε+ε^2 > 4-4ε-ε^2,we have 4-(4ε+ε^2)< x< 4+(4ε+ε^2) or ∣x-4∣<4ε+ε^2 This suggests that if we set δ=4ε+ε^2 when 0<ε≦2,then 0<∣x-4∣<δ implies ∣√x-2∣<ε.This is indeed the case,namely if 0<∣x-4∣<4ε+ε^2, then -4ε-ε^2 <x-4< 4ε+ε^2, or 4-4ε-ε^2 <x-4< 4+4ε+ε^2. This shows that (2-ε)^2<x<(2+ε)^2.Taking square roots(and observing that x>0) 2-ε<√x<2+ε or -ε<√x-2<ε,which is the same as ∣√x-2∣<ε. When ε>2,we can choose δ=12(or any other smaller value).The value δ=12 comes from pluggingε=2 into δ=4ε+ε^2.With δ=12,if 0<∣x-4∣<12,then -12<x-4<12 or -8<x<16 .Since x>0, we have 0<x<16,or after taking square roots, 0<√x<4,which is the same as -2<√x-2<2 or ∣√x-2∣<2<ε. Orz,可以只寫上課教的那個嗎??? 怎麼覺得他廢話一堆啊 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.129.62.77 ※ 編輯: slavedaeric 來自: 140.129.62.77 (10/06 18:33)

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