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Thread: limits

  1. May 14th 2008, 12:09 PM #1
    Legendsn3verdie
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    limits

    using a conjucate.. can someone show me how they would solve this? thanks.



    lim x-> infinite [2x - squareroot(4x^2+1)]
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  2. May 14th 2008, 12:13 PM #2
    Mathstud28
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    Quote Originally Posted by Legendsn3verdie View Post
    using a conjucate.. can someone show me how they would solve this? thanks.



    lim x-> infinite 2x - squareroot(4x^2+1)
    Multiplying by the conjugate we get

    \lim_{x\to\infty}(2x-\sqrt{4x^2+1})\cdot\frac{2x+\sqrt{4x^2+1}}{2x+\sqr t{4x^2+1}}

    Using the fact that (f(x)-g(x))(f(x)+g(x))=f(x)^2-g(x)^2 we get

    \lim_{x\to\infty}\frac{4x^2-(4x^2+1)}{2x+\sqrt{4x^2+12}}=\lim_{x\to\infty}\fra c{-1}{2x+\sqrt{4x^2+1}}=0
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