Re: [多變] 兩題多變數函數的極限

看板trans_math作者 ( )時間12年前 (2012/04/15 11:30), 編輯推噓1(103)
留言4則, 2人參與, 最新討論串2/4 (看更多)
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: 如圖 : 兩題解到一半都卡卡的 1. tanθ lim ------ = 1 θ→0 θ tan(x - y) So lim ------------ = 1 as x - y can be made arbitrary small if (x,y)→(2,2) x - y (x,y) is close enough to (2,2) (*) Another possible answer tan(x - y) lim ------------ doesn't exist since the function is not defined (x,y)→(2,2) x - y in any neighborhood of (2,2) along the line x=y. It depends on the definition of the limit. 2. x^3 - xy^2 r^3 cosθcos(2θ) By writing ----------- in polar form we get ------------------- x^2 + y^2 r^2 which is r cosθcos(2θ) when (x,y)≠(0,0) and its absolute value is less than r. So the limit is -1 π cos 0 = --- 2 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.217.35.77 ※ 編輯: suhorng 來自: 61.217.35.77 (04/15 11:39)

04/15 11:58, , 1F
第二題的arccos(0)可以等於3π/2嗎
04/15 11:58, 1F

04/15 11:59, , 2F
可以的話此極限是不是就不存在了?
04/15 11:59, 2F

04/15 12:25, , 3F
arccos端看你怎麼定義 通常會定義在[0,π]
04/15 12:25, 3F

04/15 13:09, , 4F
謝謝你!
04/15 13:09, 4F
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