先取自然对数得
lim(x→∞)ln{[(1+1/x)^x^2]/e^x }
=lim(x→∞)ln[(1+1/x)^x^2]-lne^x
=lim(x→∞)x^2ln(1+1/x)-x (令x=1/t)
=lim(t→0)ln(1+t)/t^2-1/t
=lim(t→0)[ln(1+t)-t]/t^2 (运用洛必达法则)
=lim(t→0)[1/(1+t)-1]/(2t)
=lim(t→0)[-t/(1+t)]/(2t)
=lim(t→0)-1/[2(1+t)]
=-1/2
所以
lim(x→∞)[(1+1/x)^x^2]/e^x
=lim(x→∞)e^ln{[(1+1/x)^x^2]/e^x }
=e^(-1/2)