n->+inf,inf表示无穷
lim[(1+1/n)^n)/e]^n
=lime^nln[(1+1/n)^n)/e]
下面求limnln[(1+1/n)^n)/e]
limnln[(1+1/n)^n)/e]
=limn[nln(1+1/n)-lne]
又n->+inf,ln(1+1/n)=(1/n)-(1/2)*(1/n^2)+o(1/n^2)
所以=limn[nln(1+1/n)-lne]
=limn[1-(1/2)(1/n)+no(1/n^2)-1]
=lim[(-1/2)+n^2o(1/n^2)]
=-1/2
所以原式=e^(-1/2)