Prove that limit of the function (a^x-1)/x=lna,x->0,using two methods.  

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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...

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giorgiana1976 | Student

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

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