计算需要用到拉格朗日中值,和泰勒式
lim [(1+x)^(1/x)-e ]/x
=lim(x→0) [e^[ln(1+x)/x]-e]/x
=lim(x→0) (e^ξ)[ln(1+x)/x-1]/x {用到e^ξ[ln(1+x)/x-1]=e^[ln(1+x)/x]-e,其中f(x)=e^x}
=lim(x→0) (e^ξ){[x-(1/2)x^2+o(x^2)]/x-1}/x
=lim(x→0) (e^ξ)[o(x^2)/x-x/2]/x [ln(1+x)/x1]
=lim(x→0) -e/2