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Math Help - Limit

  1. #1
    Junior Member mathfun's Avatar
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    Limit

    Let f,g be two functions such that {x^2}f\left( x \right) + {x^2}g\left( x \right) - {f^2}\left( x \right){g^2}\left( x \right) = 0 for every x > 0. If f,g have positive values for every x > 2012, find the limits \mathop {\lim }\limits_{x \to \infty } f\left( x \right),\mathop {\lim }\limits_{x \to \infty } g\left( x \right).

    I firstly divided with {x^2}{f^2}\left( x \right){g^2}\left( x \right), so that i could use this Squeeze theorem - Wikipedia, the free encyclopedia but with no result. Can you help?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Limit

    Quote Originally Posted by mathfun View Post
    Let f,g be two functions such that {x^2}f\left( x \right) + {x^2}g\left( x \right) - {f^2}\left( x \right){g^2}\left( x \right) = 0 for every x > 0. If f,g have positive values for every x > 2012, find the limits \mathop {\lim }\limits_{x \to \infty } f\left( x \right),\mathop {\lim }\limits_{x \to \infty } g\left( x \right).
    We can start our investigation in the following way:

    (i) If we choose f(x)=g(x) , you'll easily find that f(x)=g(x)=\sqrt[3]{2x^2} satisfies the given hypothesis. In this case, we have \lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}g(x)=+\infty .

    (ii) If we choose

    f(x)=\begin{Bmatrix} \sqrt[3]{x^2} & \mbox{ if }& x\in (0,2]\\\dfrac{x^2+x\sqrt{x^2-4}}{2} & \mbox{if}& x>2\end{matrix}

    g(x)=\begin{Bmatrix} \sqrt[3]{x^2} & \mbox{ if }& x\in (0,2]\\ 1 & \mbox{if}& x>2\end{matrix}

    then, f(x) and g(x) also satisfies the given conditions. In this case, \lim_{x\to +\infty}f(x)=+\infty,\lim_{x\to +\infty}g(x)=1 .

    Taking into account these results my questions are:

    (a) Have you quoted the exact formulation of the problem ?.
    (b) Are there additional hypothesis for f and g ?
    (c) In what terms we have to find those limits ?
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  3. #3
    Junior Member mathfun's Avatar
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    Re: Limit

    Hmm, now that i check more, it must be to find \[\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{f\left( x \right)}},\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{g\left( x \right)}}\]
    Sorry for the inconvinience.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Limit

    Quote Originally Posted by mathfun View Post
    Hmm, now that i check more, it must be to find \[\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{f\left( x \right)}},\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{g\left( x \right)}}\]
    Sorry for the inconvinience.
    Things do not change. Sometimes \lim_{x\to +\infty}1/g(x)=0 and sometimes \lim_{x\to +\infty}1/g(x)=1. I insist: Could you please transcribe the exact formulation of the problem ?. Alternatively: Could you please provide the source ?
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  5. #5
    Junior Member mathfun's Avatar
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    Re: Limit

    Its a book of mine and i have translated it exactly i think. It just might be wrong.

    Thanks for help
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