Re: e
- From: bill <b92057@xxxxxxxxx>
- Date: Tue, 30 Oct 2007 14:46:32 -0700
On Oct 29, 10:27 am, Andersen <andersen_...@xxxxxxxxxxx> wrote:
Is there a simple proof of the identity lim n->oo (1+x/n)^n = e^x ?
If 1+x/n)^n = e^x, then
LN((1+x/n)^n) = LN(e^x). then,
n * (LN((1+x/n) = x * LN(e) = x, and
LN((1+x/n) = x/n
Let x/n = y. The series expansion for LN(1+y) is
LN(1+y) = y - (y^2)/2 + ... )
Now it is necessary to prove that
Lim(y - (y^2)/2 + ...) = y for y -> 0
Factor out a "y" to get
Lim {y * (1 - y/2 + ...) = y * Lim(1 - y/2 ... ).
If Lim(1 - y/2 ... ) = 1, we are all done. If not,
then what?
Bill J
.
- References:
- e
- From: Andersen
- e
