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Math Help - how to find the bounds of this siquence..

  1. #1
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    how to find the bounds of this siquence..

    (n^2+2)^0.5 - (n)^0.5

    i thought of doing a limit where n->infinity

    but here i get undefined form and even if i whould get some finite limite
    it will only be one bound

    and i cant do limit n->-infinity because its a sequence must be positive??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    (n^2+2)^0.5 - (n)^0.5

    i thought of doing a limit where n->infinity

    but here i get undefined form and even if i whould get some finite limite
    it will only be one bound

    and i cant do limit n->-infinity because its a sequence must be positive??
    We have the sequence a_n=\sqrt{n^2+2}-\sqrt{n}. To find the lower bound define f(n)=\sqrt{n^2+2}-\sqrt{n}. A little work shows that f'\left(\approx.886388\right)=0 and the the second derivative is positive at that value, therefore it is an relative minimum. So now checking 0 the left-endpoint of the domain we find that f(.886388)\approx.72755 is a lower bound. Now we can show that a_n is unbounded above, the reason being that for every \varepsilon>0 you can find a number N such that N\leqslant{n}\implies\varepsilon<a_n. Also you can note that as n gets arbitrarily large that a_n\sim\sqrt{n^2}-\sqrt{n}=n-\sqrt{n} which clearly is unbounded above.
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  3. #3
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    but its sequence "n" must be positive and whole
    you cant input n=0.8
    ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    but its sequence "n" must be positive and whole
    you cant input n=0.8
    ?
    I know that, but think about that if x is a local min/ max then the values around it are closer (if continous) to it. So test the biggest integer after the number I gave you and the smallest before.
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