Question

Why \(lim_{n \to \infty}(1-\frac{1}{n})^{n} = e^{-1} \) ?
OR How can i find this rule, whats it name ?

Answer 1

I thought it should be \[\lim{n \to \infty}(1-0)^{n}\]

Answer 2

hiii,
binomial expansion of \[(1+x)^{n}\]
substitute for n=1/x
you'll get binomial expansion as : \[(1+x)^{1/x}\]
= 1+ x*1/x + (1-x)/2! + (1-x)(1-2x)/3!+............
now apply limit x tending to 0
you'll get : 1+1+1/2!+ 1/3!+.......... = e
similarly for \[(1-x)^{x}\]
we get e-1
:)

Answer 3

can we find this rule in internet, is it possible to send a link for it ?

Answer 4

yeah you can google that out !! I answerd that on what I recently studied :)

Answer 5

ok i think your answer is good enough, but to understand better i need to read once more i think from other source..

Answer 6

what is this number name \[e^{-1}\] ?

Answer 7

it's actually exponential to the power -1... we write it as " e " it's value approximately equals to 2.7

Answer 8

ok..

Answer 9

e-1 = 1/e
is it more clear now..?

Answer 10

ok thank you very much, its clearer

Answer 11

we can expand e^x as 1+ x/1! + x^2/2! + x^3/3!+......

Answer 12

hmm...

Answer 13

ok thank you meera now its much clearer, thanks for your efforts

Answer 14

:)