己知 lim_[x,0] (sinx)/x =1
證明 lim_[x,0] (sin[kx])/x = k
lim_[x,0] (sin[kx])/x
2013-04-25 4:27 am
回答 (2)
2013-04-25 4:40 am
✔ 最佳答案
Let kx = uwhen x tends to 0, u also tends to 0.
So lim (sin kx)/x tends to 0 = lim (sin u)/(u/k) with u tends to 0
= lim k (sin u)/u = k lim (sin u)/u = k x 1 = k.
2013-04-25 5:09 am
l=im_[x,0] sin(kx)/x
=lim_[x,0] k[sin(kx)/(kx)]
丨Let kx=u
丨∵x→0
丨∴u→0
=im[u,0] k [sin u / u]
=k lim[u,0] [sin u / u]
=k(1)
=k
=lim_[x,0] k[sin(kx)/(kx)]
丨Let kx=u
丨∵x→0
丨∴u→0
=im[u,0] k [sin u / u]
=k lim[u,0] [sin u / u]
=k(1)
=k
收錄日期: 2021-04-13 19:27:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130424000051KK00317