請問lim(1/ln(x+1)-1/x)怎麼算，x趨近於0 (答案是二分之一) 謝謝！?
回答 (1)
2019-05-20 9:41 pm
Sol
A=lim(x->0)_(1/ln(x+1)-1/x)
Set x=e^y-1
x+1=e^y
A=lim(y->0)_[1/y－1/(e^y-1)]
=lim(y->0)_(e^y－1－y)/[y(e^y－1)] 0/0 type
=lim(y->0)_(e^y－1)/[ye^y+(e^y－1)]
=lim(y->0)_(e^y－1)/(ye^y+e^y－1) 0/0 type
=lim(y->0)_e^y/(ye^y+e^y+e^y)
=1/(0+1+1)
=1/2
A=lim(x->0)_(1/ln(x+1)-1/x)
Set x=e^y-1
x+1=e^y
A=lim(y->0)_[1/y－1/(e^y-1)]
=lim(y->0)_(e^y－1－y)/[y(e^y－1)] 0/0 type
=lim(y->0)_(e^y－1)/[ye^y+(e^y－1)]
=lim(y->0)_(e^y－1)/(ye^y+e^y－1) 0/0 type
=lim(y->0)_e^y/(ye^y+e^y+e^y)
=1/(0+1+1)
=1/2
收錄日期: 2021-04-30 22:54:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190520131233AA2jbL2