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Math Help - limit of a sequence

  1. #1
    iva
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    limit of a sequence

    The answer to this is 1/3 but I don't know why. I'm obviously using the wrong technique but i don't know which one I am supposed to use:

    Evaluate lim n-> infinity an

     \sqrt{n^2 + n} - \sqrt{n^2 - 1}

    i divide each term by n^2 to get:

     \sqrt{1 + n/1} - \sqrt{1-1/n^2} = \sqrt{1 - 0} - \sqrt{1 - 0} = 0

    I know its wrong but i don't know why

    Thank you
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    Re: limit of a sequence

    Quote Originally Posted by iva View Post
    The answer to this is 1/3 but I don't know why. I'm obviously using the wrong technique but i don't know which one I am supposed to use:

    Evaluate lim n-> infinity an

     \sqrt{n^2 + n} - \sqrt{n^2 - 1}
     \sqrt{n^2 + n} - \sqrt{n^2 - 1}=\frac{n+1}{\sqrt{n^2 + n} + \sqrt{n^2 - 1}}
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  3. #3
    iva
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    Re: limit of a sequence

    Wow thanks i just figured out that identity ( if thats what its called) that i had seen before but have never used. I get that part now.

    But now if i divide through by n^2 i get

    (0-0)/1+1 = 0

    which is still wrong..if the answer is 1/3 then should i be dividing through by n rather than n^2? Well i just tried that but then i get

    1/(\sqrt{n+1} + \sqrt{n-1}) = 0

    This doesn't make sense, as i need a 3 in the denominator so i guess one of these terms, i guess the first , should evaluate to 2, and the second to 1. Sorry now i am just taking stabs in the dark
    Last edited by iva; August 28th 2011 at 10:22 AM. Reason: correction
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    MHF Contributor alexmahone's Avatar
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    Re: limit of a sequence

    Quote Originally Posted by iva View Post
    Wow thanks i just figured out that identity ( if thats what its called) that i had seen before but have never used. I get that part now.

    But now if i divide through by n^2 i get

    (0-0)/1+1 = 0

    which is still wrong..if the answer is 1/3 then should i be dividing through by n rather than n^2? Well i just tried that but then i get

    1/(\sqrt{n+1} + \sqrt{n-1}) = 0

    This doesn't make sense, as i need a 3 in the denominator so i guess one of these terms, i guess the first , should evaluate to 2, and the second to 1. Sorry now i am just taking stabs in the dark
    \frac{n+1}{\sqrt{n^2 + n} + \sqrt{n^2 - 1}}=\frac{1+1/n}{\sqrt{1+1/n}+\sqrt{1-1/n^2}}
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  5. #5
    iva
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    Re: limit of a sequence

    you divided the numerator by n but the denominator by n^2, i thought we have to divide by the same thing. OK thanks for that equation, so that gives us 1/(1+1) = 1/2 right?
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    MHF Contributor alexmahone's Avatar
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    Re: limit of a sequence

    Quote Originally Posted by iva View Post
    you divided the numerator by n but the denominator by n^2, i thought we have to divide by the same thing. OK thanks for that equation, so that gives us 1/(1+1) = 1/2 right?
    Actually, I divided the denominator by \sqrt{n^2}=n.

    Yes, the limit is 1/2.
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    Member anonimnystefy's Avatar
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    Re: limit of a sequence

    hi iva

    no he divided both by n,but when n goes under the root sign it turns into n^2.
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  8. #8
    iva
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    Re: limit of a sequence

    Thanks, i have so much to learn
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  9. #9
    iva
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    Re: limit of a sequence

    And you guys found an error in my textbook! their answer was 1/3
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