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anonymous

  • 5 years ago

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

  2. anonymous
    • 5 years ago
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    why dont u try taking log..

  3. anonymous
    • 5 years ago
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    i dont know how to do it

  4. anonymous
    • 5 years ago
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    if we simply put n=infinity result comes out to be \[1^{0}\]

  5. anonymous
    • 5 years ago
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    is it an indeterminate form?

  6. anonymous
    • 5 years ago
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    take log on both sides ln L = 1/n x sigma ln (1 + r^2/n^2)

  7. anonymous
    • 5 years ago
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    take r/n = x, 1/n= dx

  8. anonymous
    • 5 years ago
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    hence ln L = ln(1 + x^2)dx from 0 to 1

  9. anonymous
    • 5 years ago
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    now use integral by parts to integrate this

  10. anonymous
    • 5 years ago
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    is it an indeterminate form? 1^0

  11. anonymous
    • 5 years ago
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    i told u..did u get it??????????????????????????/

  12. anonymous
    • 5 years ago
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    deeprony? r u getting it?

  13. anonymous
    • 5 years ago
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    i didnt understand what you said

  14. anonymous
    • 5 years ago
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    see ill tell u the general method whenever u hav \[\sum_{0}^{\infty}f(r/n) \times 1/n\]

  15. anonymous
    • 5 years ago
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    yes.. what do i do?

  16. anonymous
    • 5 years ago
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    then put r/n = x and 1/n = dx

  17. anonymous
    • 5 years ago
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    so u hav \[\int\limits_{0}^{1}f(x)dx\]

  18. anonymous
    • 5 years ago
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    samjhe??

  19. anonymous
    • 5 years ago
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    yea but why did u set the upper limit to 1??

  20. anonymous
    • 5 years ago
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    when r=n then r/n is 1 isnt it/?

  21. anonymous
    • 5 years ago
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    oh yes.. i get it now...

  22. anonymous
    • 5 years ago
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    \[given=\lim_{n \rightarrow \infty} e^{\frac{\ln (1+\frac{1}{n^2})+...+\ln(1+\frac{n^2}{n^2})}{n}}\] apply l hospitals rule.

  23. anonymous
    • 5 years ago
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    please i am having a doubt if we directly put n=infinity then the limit is coming out to be 2^0 is this an inderminate form???

  24. anonymous
    • 5 years ago
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    its coming infinity^0

  25. anonymous
    • 5 years ago
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    y so saubhik n^2 is in the denominator as n tends to infinity 1/n^2 , 2/n^2 ,... becomes 0

  26. anonymous
    • 5 years ago
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    see the last term its 1+n^2/n^2. as n -> infty then thus term tends to infty. understood?

  27. anonymous
    • 5 years ago
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    cancel the n^2 last term comes out to be 2

  28. anonymous
    • 5 years ago
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    wow wow man!!! Mankind has not been able to understand till today what is infty/infty and yet u claim its 1!

  29. anonymous
    • 5 years ago
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    ok my bad i should have been more careful

  30. anonymous
    • 5 years ago
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    hey him r u der?

  31. anonymous
    • 5 years ago
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    get it?

  32. anonymous
    • 5 years ago
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    yea.. it worked out..=)

  33. anonymous
    • 5 years ago
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    chek dis one out http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd66882d95c8b0b951d5ec4

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