求 lim_[x,π/2] (sec x - tan x)
回答 (4)
2013-04-25 4:54 am
✔ 最佳答案
圖片參考：http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540D-5_zpsb616cdb4.png
http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540D-5_zpsb616cdb4.png
2013-04-25 16:03:39 補充：
Correction:
http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540D-5_zpsfd8c9acb.png
參考: My Maths World
2013-04-26 4:41 am
lim(x->π/2) (secx - tanx)
= lim(x->π/2) (1/cosx - sinx/cosx)
= lim(x->π/2) (1 - sinx)/cosx
= lim(x->π/2) -cosx/sinx (L'Hospital's Rule)
= -0/1
= 0
= lim(x->π/2) (1/cosx - sinx/cosx)
= lim(x->π/2) (1 - sinx)/cosx
= lim(x->π/2) -cosx/sinx (L'Hospital's Rule)
= -0/1
= 0
2013-04-25 7:32 am
✡ HYPERCUBE ✡ :step wrong
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2013-04-25 5:02 am
lim_[x,π/2] (sec x - tan x)
lim_[x,π/2] (1/cos x - sin x /cos x)
lim_[x,π/2] ( 1-sin x) / cos x
Let y=π/2-x
lim_[y,0] (1-sin(π/2-y)) / cos (π/2-y)
lim_[y,0] (1-cos y) / sin y
lim_[y,0] (2sin^2 (y/2)) / (2sin(y/2)cos(y/2))
lim_[y,0] sin(y/2) / cos (y/2)
lim_[y,0] tan (y/2)
tan 0
0
lim_[x,π/2] (1/cos x - sin x /cos x)
lim_[x,π/2] ( 1-sin x) / cos x
Let y=π/2-x
lim_[y,0] (1-sin(π/2-y)) / cos (π/2-y)
lim_[y,0] (1-cos y) / sin y
lim_[y,0] (2sin^2 (y/2)) / (2sin(y/2)cos(y/2))
lim_[y,0] sin(y/2) / cos (y/2)
lim_[y,0] tan (y/2)
tan 0
0
收錄日期: 2021-04-13 19:26:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130424000051KK00319