We have to prove that lim x-->0 [(a^x - 1)/x] = ln a

First, if we substitute x = 0, we get the indeterminate form 0/0. This allows the use of l"Hopital's rule and we can substitute the numerator and the denominator by their derivatives

lim x-->0 [(a^x) * ln a ]

substitute x = 0, we get ln a

Next we can write : a^x - 1 = h.

=> a^x = 1 + h

=> x = log(a) ( 1 + h )

=>x = [ln( 1 + h )]/( ln a)

As a^x - 1 = h , x--> 0 => h --> 0

So we have lim h-->0 [h /(ln(1 + h)/ln a)]

=> lim h-->0 [ln a/(ln(1 + h)/h)]

=> lim h-->0 [ln a/(ln(1 + h)^(1/h))]

=> ln a * [ 1/ lim h-->0 [(ln(1 + h)^(1/h))]

=> ln a * ln e

=> ln a * 1

=> ln a

This proves lim x-->0 [(a^x - 1)/x] = ln a