→ yoyooyooo:第二題的arccos(0)可以等於3π/2嗎140.113.188.253 04/15 11:58
→ yoyooyooo:可以的話此極限是不是就不存在了?140.113.188.253 04/15 11:59
→ suhorng:arccos端看你怎麼定義 通常會定義在[0,π] 61.217.35.77 04/15 12:25
推 yoyooyooo:謝謝你!140.113.188.253 04/15 13:09
※ 引述《yoyooyooo (yoyoyoo)》之銘言：
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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
--
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※ 編輯: suhorng 來自: 61.217.35.77 (04/15 11:39)