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answers:
va2906 say: I feel so stupid when I enter this Yahoo category :(
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ted s say: y = e^(n ln (1 + 1/n) ) =====> lim { n ----> ∞ } y = e ^(lim { n ---> ∞} n ln ( 1 + 1/n) )
n ln ( 1 + 1 / n ) = [ ln (1 + 1/n) ] / [ 1 / n ] ≈ 0 / 0 in the limit...
so apply L'Hopital's techniques on this...... f ' / g ' , limit,etc
{ ( - 1 / n² ) / ( 1 + 1/n) } / { - 1 / n² } = 1 / ( 1 + 1 / n) ---> 1 as n ---> ∞ ====>
y ----> e^1 = e
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Markus say: You are right that 1/n ~ 0 when n is large. But no matter how large n is, 1 + 1/n will always be greater than 1. And you are raising this number that is a "little bit" greater than one by also a large power, multiply it by itself many many times. Just imagine the number x = 1.000000000001, which is also almost 1. But if you multiply it by itself many times, it will no longer remain a "number close to 1"
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Philip say: In general, e^x = lim, x--> +oo, of (1 + x/n)^n.
The binomial expansion of (1 + x/n)^n = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! +.......ad.inf..
Putting x = 1 gives e = 1 + 1/1! + 1/2! + 1/3! + 1/4! +......ad.inf. = 1 + 1 +(1/2) +(1/6) +(1/24) +......
until e is approx. = 2.718281828(10 sig. fig.).
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Elizabeth M say: Set y=(1+1/n)^n so
ln(y)=nln(1+1/n) and using the log series, valid for n>=1,
=n(1/n- 1/(2n^2)+1/(3n^3)-1/(4n^4)+.....)
=1- 1/(2n) + 1/(3n^2)-.....
As n->infinity, RHS->1 so ln(y)->1
So y->e.
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Michael say: Well,
let
xn = (1 + 1/n)^n
and
zn = Ln( xn )
then
zn = n * Ln(1 + 1/n)
and we know that :
lim ( x ---> 0) Ln(1 + x)/x = 1
therefore
lim ( n ---> +oo ) Ln(1 + 1/n) /(1/n) = 1
therefore
when n ---> +oo
zn ~ n * 1/n = 1
therefore
lim (n ---> +oo) zn = 1