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anonymous

  • 5 years ago

limit of ( n/(n-4) )^n as n goes to infinity??

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  1. anonymous
    • 5 years ago
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    positive infinitiy

  2. anonymous
    • 5 years ago
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    or wait, nvm

  3. anonymous
    • 5 years ago
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    1

  4. anonymous
    • 5 years ago
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    yea..the calculator says e^4...have no idea how to get there

  5. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty} [\frac{n}{n-4}]^n\] ?

  6. anonymous
    • 5 years ago
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    yes thats the prob

  7. anonymous
    • 5 years ago
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    are you sure that your answer is e^4?

  8. anonymous
    • 5 years ago
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    because I got 1 too lol

  9. anonymous
    • 5 years ago
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    ok yea thats wat i was thinking too..thank u!

  10. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty} e^{n \ln \frac{n}{n-4}}\] first you find the limit of the fraction : \[\lim_{n \rightarrow \infty} \frac{n}{n-4} = \lim_{n \rightarrow \infty} \frac{n}{n} = 1\] so now plug this value in the first equation I wrote: \[\lim_{n \rightarrow \infty} e^0 = 1\] np :)

  11. anonymous
    • 5 years ago
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    I get \[e^4\]

  12. anonymous
    • 5 years ago
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    how? .-.

  13. anonymous
    • 5 years ago
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    hmm, I prolly missed out a point that I can't seem to find ._.

  14. nowhereman
    • 5 years ago
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    No, that is not how it works. This way there wouldn't be any constant e. The problem with your approach is, that when the ln goes to zero, at the same time n goes to infinity and so you can't say that their product goes to zero! But you can do the following substitution: \[\lim_{n→∞}{\left(\frac{n}{n-4}\right)^{n}} = \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{m+4}} = \lim_{m→∞}{\left(1+\frac{4}{m}\right)^{m}} \cdot \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{4}} = e^4 \cdot 1

  15. anonymous
    • 5 years ago
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    Set the expression equal to y and take the log of both sides. You'll end up with\[\log y = n \log \frac{n}{n-4}=n \log \frac{1}{1-4/n}=-n \log (1-4/n)=-\frac{\log (1-4/n)}{1/n}\]That's now indeterminate, so you can apply L'Hopital's and take the limit. Then undo the log.

  16. nowhereman
    • 5 years ago
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    \[\lim_{n→∞}{\left(\frac{n}{n-4}\right)^{n}} = \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{m+4}} = \lim_{m→∞}{\left(1+\frac{4}{m}\right)^{m}} \cdot \lim_{m→∞}{\left(\frac{m+4}{m}\right)^{4}}\]\[ = e^4 \cdot 1\]

  17. anonymous
    • 5 years ago
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    when the degree is same on the top and bottom, dont you jus look at the coeffecients of the highest degree which is 1/1= 1

  18. anonymous
    • 5 years ago
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    You should get what nowhereman got.

  19. anonymous
    • 5 years ago
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    /I got.

  20. nowhereman
    • 5 years ago
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    misha: true, but you can only pull the limit inside, if the outer part does not depend on the variable (n here)

  21. anonymous
    • 5 years ago
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    then I was wrong, misha follow both of their ways ^_^

  22. nowhereman
    • 5 years ago
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    You should probably take a look at http://en.wikipedia.org/wiki/Exponential_function or something similar.

  23. anonymous
    • 5 years ago
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    ok thanks for clarifying!

  24. anonymous
    • 5 years ago
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    I gave you a medal sstarica

  25. anonymous
    • 5 years ago
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    lol, thank you :) but I didn't work for it

  26. anonymous
    • 5 years ago
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    for being a good sport :)

  27. anonymous
    • 5 years ago
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    ^_^ thank you

  28. anonymous
    • 5 years ago
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    You can have one too, nowhereman...

  29. anonymous
    • 5 years ago
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    lol I gave both of you a medal

  30. anonymous
    • 5 years ago
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    loki

  31. anonymous
    • 5 years ago
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    I prolly am just seeing it, but just to make sure, is there something wrong b/w you and nowhereman?

  32. anonymous
    • 5 years ago
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    Not that I am aware of. Why?

  33. anonymous
    • 5 years ago
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    I've never talked to him.

  34. anonymous
    • 5 years ago
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    mm, lol nvm, you guys just act weird around each other?

  35. anonymous
    • 5 years ago
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    ?

  36. anonymous
    • 5 years ago
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    lol, I'm prolly just seeing it, it's nothing ^_^

  37. anonymous
    • 5 years ago
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    he doesn't 'chat'

  38. anonymous
    • 5 years ago
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    noticed :)

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