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anonymous

  • 5 years ago

Question on limits

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  1. anonymous
    • 5 years ago
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    ask

  2. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty} ((n!)^{1/n})/n\]

  3. anonymous
    • 5 years ago
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    fun!

  4. anonymous
    • 5 years ago
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    omg

  5. anonymous
    • 5 years ago
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    Pretty sure the factorial will dominate, but let's find out =)

  6. anonymous
    • 5 years ago
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    givng up

  7. anonymous
    • 5 years ago
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    how do i go about it polpak?

  8. anonymous
    • 5 years ago
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    Well, on top you have \(\infty^0\)

  9. anonymous
    • 5 years ago
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    Which by itself is an indeterminate form

  10. anonymous
    • 5 years ago
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    agreed

  11. apples
    • 5 years ago
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    The limit does exist and is finite.

  12. anonymous
    • 5 years ago
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    yes apples thats correct

  13. anonymous
    • 5 years ago
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    is anyone working on it?

  14. apples
    • 5 years ago
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    I'd suggest using the Squeeze Theorem.

  15. anonymous
    • 5 years ago
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    what is that? would you explain?

  16. apples
    • 5 years ago
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    The squeeze theorem states that, for any function f(x), if you can find g(x) and h(x) such that g(x) <= f(x) <= h(x) and the limit as x goes to a of g(x) is the same as the limit for h(x), then that is also the limit for f(x).

  17. apples
    • 5 years ago
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    Another way to compute the limit would be to use the following definition of e: \[e = \lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}}\]

  18. anonymous
    • 5 years ago
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    i didnt quite get that theorm.. by using the definition of e it just means that the question i asked yields 1/e as an answer right?

  19. apples
    • 5 years ago
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    Yes.

  20. apples
    • 5 years ago
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    The theorem states that if you can find functions that are upper and lower bounds of your function, and they have the same limit at a certain point, then your function has the same limit at that point.

  21. apples
    • 5 years ago
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    For example, a lower bound for this problem would be \[\left(1 - \frac{1}{n}\right)^n\] which has a limit of 1/e as n goes to infinity. So if you could find an upper bound of the problem that also went to 1/e, you could state that the answer to the problem is 1/e by the squeeze theorem.

  22. anonymous
    • 5 years ago
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    ok i understand now. but how is a the lower limit and the upper limit of a function found?

  23. apples
    • 5 years ago
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    You aren't finding "upper and lower" limits, you're finding limits of the upper and lower bounding functions. Let's consider a simple example, f(x) = 0: \[ \lim_{x \to \infty} f(x) \] An upper bound on this could be something like 1/x, since it is clear that 1/x > 0 for all x > 0. A lower bound could be -1/x, for similar reasons. Since we know that \[ \lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{-1}{x} = 0\] and \[ \frac{-1}{x} <= f(x) <= \frac{1}{x} \] for all x > 0, then \[\lim_{x \to \infty} f(x) = 0\]

  24. anonymous
    • 5 years ago
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    I think it's easy to find the limit using the definition, that apples just mentioned above, for e.

  25. anonymous
    • 5 years ago
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    The limit we're looking for is just the reciprocal of it. So: \[\lim_{n \rightarrow \infty}{n. !^{1 \over n} \over n}=\lim_{n \rightarrow \infty}{1 \over {n \over n.!^{1 \over n}}}={1 \over e}\]

  26. anonymous
    • 5 years ago
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    that was a good explanation apples. thank you.

  27. anonymous
    • 5 years ago
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    yea i got it anwar..=)

  28. anonymous
    • 5 years ago
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    You already found the answer.. Never mind then :)

  29. apples
    • 5 years ago
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    No prob

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