更新1:
一部分太模糊sorry
Maths Limits?
回答 (2)
2016-11-03 3:35 pm
✔ 最佳答案
SolA=lim(x->0+)_(1+2^(1/x)]/[3+2^(1/x)]
Set y=2^(1/x)
x->0+ => y->∞
A=lim(y->∞)_(1+y)/(3+y)=lmi(y->∞)_(1/y+1)/(3/y+1)=1
B=lim(x->0－)_(1+2^(1/x)]/[3+2^(1/x)]
Set z=2^(1/x)
x->0－ => z->0
B=lim(z->0)_(1+z)/(3+z)=1/3
A<>B
lim(x->0)_[1+2^(1/x)]/[3+2^(1/x)]不存在
2016-11-04 2:25 pm
lim(x->0)_[1+2^(1/x)]/[3+2^(1/x)]
=lim(x->0)_[(2^x)/(2^x)][1+2^(1/x)]/[3+2^(1/x)]
=lim(x->0)_[(2^x)+1]/[3(2^x)+1]
=[(2^0)+1]/[3(2^0)+1]
=2/(3+1)
=0.5
=lim(x->0)_[(2^x)/(2^x)][1+2^(1/x)]/[3+2^(1/x)]
=lim(x->0)_[(2^x)+1]/[3(2^x)+1]
=[(2^0)+1]/[3(2^0)+1]
=2/(3+1)
=0.5
收錄日期: 2021-04-18 15:48:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161103071727AAr3vfI