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 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
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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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