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**

We notice that if we'll substitute x by 0, w'ell get an indetermination.

lim (a^x-1)/x = (a^0 - 1)/0 = (1 - 1)/0 = 0/0

We'll apply L'Hospital to determine the limit:

lim (a^x-1)/x = lim (a^x-1)'/x'

lim (a^x-1)'/x' = lim a^x*ln a/1 =a^0*ln a

lim (a^x-1)/x = ln a

Another method to prove that the limit of the function is ln a is substitution method.

a^x-1 = u

a^x = 1 + u

ln a^x = ln(1 + u)

We'll apply power rule:

x*lna = ln(1 + u)

x = ln(1 + u)/ln a

We'll re-write the limit in u:

lim (a^x-1)/x = lim u/[ln(1 + u)/ln a]

lim u/[ln(1 + u)/ln a] = ln a*lim u/[ln(1 + u)]

ln a*lim 1/(1/u)*[ln(1 + u)] = ln a*1/ln lim ln(1+u)^(1/u)

ln a*1/ln lim ln(1+u)^(1/u) = ln a*1/ln e

But ln e =1

**lim (a^x-1)/x = ln a**