Question

lim x趋于0+ x^[(1/(ln(e^x-1))] 用罗比达

Answer 1

直接不好求，先取对数
lim(x→0+) ln ( x^[(1/(ln(e^x-1))] )
=lim(x→0+) lnx/ln(e^x-1) ∞/∞ 罗比达
=lim(x→0+) (e^x-1)/(xe^x) 0/0 继续
=lim(x→0+) e^x/(e^x+x*e^x)
=lim(x→0+) 1/(1+x)
=1
则
lim(x→0+) x^[(1/(ln(e^x-1))]

=e^1
=e

Answer 2

先取自然对数，化成e^【lnx/ln(e^x-1)】，
现在再用罗比达求【lnx/ln(e^x-1)】
上下求导，得【(e^x-1)/xe^x】，继续上下求导得到1/【（x+1）】，x趋于0+时1/【（x+1）】=1
那么x^[(1/(ln(e^x-1))] =e^1=e

Answer 3