題目如下：
lim(n→π/2)(cosn/(π/2-n)
希望給我祥細答案。
AMATH 唔識做！
2012-11-24 4:07 am
回答 (2)
2012-11-25 4:40 pm
✔ 最佳答案
Let x = (pi/2) - n2012-11-25 08:47:10 補充：
so, when n->pi/2, x->0, and n = (pi/2) - x, therefore
lim(n->pi/2) (cos n) / [(pi/2) - n]
= lim(x->0) {cos[(pi/2) - x]} / x
= lim(x->0) (sin x) / x
= 1 ... [by definition]
2012-11-24 4:30 am
lim(n→π/2)(cos n)/(π/2 - n)
= lim(n→π/2)(-sin n)/(-1)
= sin π/2
= 1
= lim(n→π/2)(-sin n)/(-1)
= sin π/2
= 1
收錄日期: 2021-04-16 15:16:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121123000015KK05251