## tunahan Group Title Why $$lim_{n \to \infty}(1-\frac{1}{n})^{n} = e^{-1}$$ ? OR How can i find this rule, whats it name ? 2 years ago 2 years ago

1. tunahan Group Title

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

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 :)

3. tunahan Group Title

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

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

5. tunahan Group Title

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

6. tunahan Group Title

what is this number name $e^{-1}$ ?

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

8. tunahan Group Title

ok..

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

10. tunahan Group Title

ok thank you very much, its clearer

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

12. tunahan Group Title

hmm...

13. tunahan Group Title

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