how to prove e^u = limit(x→+∞) (1-u/x)^(-x) ? Please help?
回答 (2)
2017-10-28 2:10 pm
e^u = limit(x→+∞) (1-u/x)^(-x)
e^x=(1+x/n)^n
limit(x→+∞) (1-u/x)^(-x)
=[1+(-u)/x]^x(-1)
=(e^-u)^-1
=e^u
e^x=(1+x/n)^n
limit(x→+∞) (1-u/x)^(-x)
=[1+(-u)/x]^x(-1)
=(e^-u)^-1
=e^u
2017-10-28 9:34 am
Sol
Set y=(1－u/x)^x
lny=xln(1－u/x)
A= lim(x->∞)_(1－u/x)^(－x)
=e^[ lim(x->∞)_(－x)ln(1－u/x)
=e^[ lim(x->∞)_ln(1－u/x)/(－1/x)
=e^[ lim(p->0)_ln(1－up)/(－p) 0/0 type
dln(1－up)/dp=dln(1－up)/d(1－up)*d(1－up)/dp=－u/(1－up)
A= e^[ lim(p->0)_[－u/(1－up)]/(－1)]
=e^u
Set y=(1－u/x)^x
lny=xln(1－u/x)
A= lim(x->∞)_(1－u/x)^(－x)
=e^[ lim(x->∞)_(－x)ln(1－u/x)
=e^[ lim(x->∞)_ln(1－u/x)/(－1/x)
=e^[ lim(p->0)_ln(1－up)/(－p) 0/0 type
dln(1－up)/dp=dln(1－up)/d(1－up)*d(1－up)/dp=－u/(1－up)
A= e^[ lim(p->0)_[－u/(1－up)]/(－1)]
=e^u
收錄日期: 2021-04-24 00:46:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171027135900AAi3agK