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anonymous

  • 5 years ago

lim (1 - 1/x)^(-x) as x->+infinty

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  1. anonymous
    • 5 years ago
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    = 0

  2. anonymous
    • 5 years ago
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    can you please explain to me how you got that

  3. anonymous
    • 5 years ago
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    I can.

  4. anonymous
    • 5 years ago
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    if you let x = infinity. Then what is 1/x really equal to?

  5. anonymous
    • 5 years ago
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    It's wrong though.

  6. anonymous
    • 5 years ago
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    My solution? Please correct me then sir.

  7. anonymous
    • 5 years ago
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    I think this is a special identity representing e, but let me check.

  8. anonymous
    • 5 years ago
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    If my Cal I recall is correct, if the exponent was positive, it is the well-recognized lim that goes to e. The fact that exponent is is negative, I am leaning to agree with you it goes to 0. How did you reach your conclusion.

  9. anonymous
    • 5 years ago
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    Well, You have something that is always smaller than one being raised to big negative numbers. It will certainly not tend toward 0 because the exponentiation will grow faster than the -1/x will shrink. However I think it does converge on e rather than go on to infinity.

  10. anonymous
    • 5 years ago
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    It's 1-1, how do you get -1? And it's to the -x, meaning the equation is actually 1/(1-1/x) which is certainly zero.

  11. anonymous
    • 5 years ago
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    No. It is \[(1-\frac{1}{x})^{-x}\] I agree that the fraction will tend toward 0, but the whole expression tends toward \(1^\infty\)

  12. anonymous
    • 5 years ago
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    Actually I contend that inside the bracket goes to 1 because \[(1 \div \infty)\] tends to zero.

  13. anonymous
    • 5 years ago
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    Oh my bad didn't see 1-1/x, even then the resultant inside is still zero. 0^infinity = 0.

  14. anonymous
    • 5 years ago
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    it's not \(1-\infty\) it is 1- 0

  15. anonymous
    • 5 years ago
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    God I need to stop failing and go to bed lol. I'm sorry polpak you're right.

  16. anonymous
    • 5 years ago
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    But that is leaves us with \(\frac{1}{1^\infty}\) which is undetermined form and needs special handling with l'hopital.

  17. anonymous
    • 5 years ago
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    Have to take the log to bring down the x out of the exponent, then take the derivative, etc. It's kind of a bear. It looks like it might be e though.

  18. anonymous
    • 5 years ago
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    We need Lokisan for this one. We need someone who has read everything concerning the number or the limit e.

  19. anonymous
    • 5 years ago
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    Heh. Well I checked with Alpha, and it's definitely e. Glad I don't have to do this for my homework though.

  20. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=lim+as+x+goes+to+infinity+%281+-+1%2Fx%29^%28-x%29

  21. anonymous
    • 5 years ago
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    Good job.

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