posted by Robert on .
lim (1+x)^1/x. Give an exact answer.
The limits as x approaches zero is (1 plus x) to the 1 divided by x.
The log of each term is
(1/x) ln (1 + x) = ln (1+x)/x
Using L'Hopital's rule for the limit of
f(x)/g(x), the limit if the log is
lim f'/g' = lim [1/(1+x)]= 1
and the antilog of that is e. 2.71828...