Posted by
**Robert** on
.

lim (1+x)^1/x. Give an exact answer.

x->0

This reads:

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...