How do you evaluate the limit #(4^x-1)/(8^x-1)# as x approaches #0#?

1 Answer
Roy E.
Jan 14, 2017

#2/3#

Explanation:

L'Hôpital's rule, with some preliminary work. Re-write as #(2^(2x)-1)/(2^(3x)-1)#. Then the let #f(x)=(2^x)^2-1#, #g(x)=(2^x)^3-1#,
#f'(x)=2h(x)# and #g'(x)=3h(x)# where #h(x)=d/(dx)(2^x)#. Hence the limit is #(f'(x))/(g'(x))=2/3# because the #h(x)#'s cancel so I don't need to know work out what #h(x)# is (but it might be #(ln2)(e^(x ln 2))# ).

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