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tunahan

  • 3 years ago

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

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  1. tunahan
    • 3 years ago
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    I thought it should be \[\lim{n \to \infty}(1-0)^{n}\]

  2. meera_yadav
    • 3 years ago
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    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
    • 3 years ago
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    can we find this rule in internet, is it possible to send a link for it ?

  4. meera_yadav
    • 3 years ago
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    yeah you can google that out !! I answerd that on what I recently studied :)

  5. tunahan
    • 3 years ago
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    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
    • 3 years ago
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    what is this number name \[e^{-1}\] ?

  7. meera_yadav
    • 3 years ago
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    it's actually exponential to the power -1... we write it as " e " it's value approximately equals to 2.7

  8. tunahan
    • 3 years ago
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    ok..

  9. meera_yadav
    • 3 years ago
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    e-1 = 1/e is it more clear now..?

  10. tunahan
    • 3 years ago
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    ok thank you very much, its clearer

  11. meera_yadav
    • 3 years ago
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    we can expand e^x as 1+ x/1! + x^2/2! + x^3/3!+......

  12. tunahan
    • 3 years ago
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    hmm...

  13. tunahan
    • 3 years ago
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    ok thank you meera now its much clearer, thanks for your efforts

  14. meera_yadav
    • 3 years ago
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    :)

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