Limit x->infinity [2^x - 2^(-x)]/[2^x + 2^(-x)]
點計?
Question about Limit
2006-12-04 6:11 am
回答 (2)
2006-12-04 6:21 am
✔ 最佳答案
lim (x→∞) [2^x - 2^(-x)] / [2^x + 2^(-x)]= lim (x→∞) [1 - 2^(-2x)] / [1 + 2^(-2x)]【分子分母同時乘以 2^(-x)】
= [1 - 0] / [1 + 0]【lim (x→∞) 2^(-2x) = lim (x→∞) 1 / 2^(2x) = 0】
= 1
Hope it helps! ^^
參考: Myself
2006-12-04 6:47 am
=lim x->infinity2^(x)[1-2^(-2x)]/2^(x)[1+2^(-2x)]
=lim x->infinity[1-2^(-2x)]/[1+2^(-2x)]
=1-0/1+0
=1
Remarks:
As x-> infinity
2x->infinity
2^(2x)->infinity
1/2^(2x)->0 which is equivalent to 2^(-2x)->0
=lim x->infinity[1-2^(-2x)]/[1+2^(-2x)]
=1-0/1+0
=1
Remarks:
As x-> infinity
2x->infinity
2^(2x)->infinity
1/2^(2x)->0 which is equivalent to 2^(-2x)->0
收錄日期: 2021-04-13 16:19:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061203000051KK05709