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Math Help - cannot find limit of a sequence because of a non factoring fraction

  1. #1
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    cannot find limit of a sequence because of a non factoring fraction

    Hello,

    I am trying to find the limit of this sequence:

    \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0

    but I cannot factorize the fraction at all.

    Any idea is very much appreciated!

    Thank you all!
    Last edited by Melsi; July 31st 2011 at 01:19 AM. Reason: removed a small part that was of no use
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  2. #2
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    Re: cannot find limit of a sequence because of a non factoring fraction

    Quote Originally Posted by Melsi View Post
    I am trying to find the limit of this sequence:
    \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0
    You should know this theorem:
    If 0<a<b then \left( {\sqrt[n]{{a^n  + b^n }}} \right) \to b.
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  3. #3
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    Re: cannot find limit of a sequence because of a non factoring fraction

    Hello,

    no it is the very first time I see this theorem, I will check it out tomorrow (02:00 at midnight now) and come back with a full solution!

    Very kind of you!
    Thank you very much!
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  4. #4
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    Re: cannot find limit of a sequence because of a non factoring fraction

    Here is the proof of that theorem.
    b = \sqrt[n]{{b^n }} \leqslant{\color{blue} \sqrt[n]{{a^n  + b^n }}} \leqslant \sqrt[n]{{b^n  + b^n }} = b\sqrt[n]{2} \to b
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  5. #5
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    Solved!

    The proof... that's a good one, cool! Now it can be solved..

    the initial formula of the general term could be written like this:

    \sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}=\frac{(k^{n}+l  ^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}

    and the limit like this:

    \lim_{n  \to \infty}{a_{n}=\lim_{n\to \infty }\frac{(k^{n}+l^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}=\frac{\lim_{n\to \infty }(k^{n}+l^{n})^{1/n}}{\lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}}

    if we apply the theorem for the numerator we have:



    l\leq \lim_{n\to \infty }(k^{n}+l^{n})^{1/n}\leq \lim_{n\to \infty }(l+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{n}+l^{n})^{1/n} = l

    and for the denominator it goes like this:

    l^{2}\leq \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}\leq \lim_{n\to \infty }(l^{2}+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n} = l^{2}

    Finally having found those two limits we can write:

    \lim_{n  \to \infty}a_{n}=\frac{l}{l^{2}}=\frac{1}{l}

    and all these supposing that a\leq b and a,b> 0 this means that if b\leq a it will go the other way around, the limit will be
    \frac{1}{k}

    Thank you once again, for your valuable help!
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