limx[(1+1/x)^x-e]
=lim[(1+1/x)^x-e]/(1/x)
令x=1/t,則原式化為
lim[(1+t)^(1/t)-e]/t
=lim{e^[(1/t)ln(1+t)]-e}/t
=lim{e^[ln(1+t)/t]-e}/t
=elim{e^[ln(1+t)/t-1]-1}/t (*)
=elim{[ln(1+t)/t]-1}/t
=elim[ln(1+t)-t]/t²(洛必達(dá)法則)
=elim[1/(1+t)-1]/(2t)
=elim(-t)/[2t(1+t)]
=-e/2
(*)用的是等價(jià)無(wú)窮小代換.
e^x-1～x(x→0),令x=ln(1+t)/t-1,因?yàn)?
lim[ln(1+t)/t-1]
=lim[ln(1+t)-t]/t(洛必達(dá)法則)
=lim[1/(1+t)-1]/1
=0
故e^[ln(1+t)/t-1]-1～ln(1+t)/t-1(t→0,即ln(1+t)/t-1→0)