Lim x--->0+ [ ln(e^x-1)/ln(x) ]

How to simplify
Lim x--->0+ [ ln(e^x-1)/ln(x) ]

Use L'Hopital's Rule.

lim(x→0+) ln(e^x - 1)/ln x; this is of the form ∞/∞
= lim(x→0+) [e^x/(e^x - 1)] / (1/x), by L'Hopital's Rule
= lim(x→0+) xe^x/(e^x - 1)
= lim(x→0+) x/(1 - e^(-x)), multiplying num. and denom. by e^(-x) [now of the form 0/0]
= lim(x→0+) 1/e^(-x), by L'Hopital's Rule
= 1/1
= 1.

I hope this helps!