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Math Help - the limit as x approaches infinity

  1. #1
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    Question the limit as x approaches infinity

    I know the answer to this problem, but I do not know how to get to it.

    lim (sqrt(9x^2+x)-3x) = (1/6)
    x-> infinity
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  2. #2
    o_O
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    Multiply top and bottom by its conjugate: \sqrt{9x^2 + x} {\color{red}+} 3x:

    \lim_{x \to \infty} \left(\sqrt{9x^2 + x} - 3x\right) \cdot \frac{{\color{red}\sqrt{9x^2 + x} + 3x} }{{\color{red}\sqrt{9x^2 + x} + 3x}}

    The numerator is an application of the difference of squares formula: (a-b)(a+b) = a^2 - b^2

    So: = \lim_{x \to \infty} \frac{9x^2 + x \ - \ 9x^2}{\sqrt{9x^2 + x} + 3x}

    etc. etc.
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  3. #3
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    \frac{\sqrt{9x^2 + x} - 3x}{1} \cdot \frac{\sqrt{9x^2 + x} + 3x}{\sqrt{9x^2 + x} + 3x}

    \frac{(9x^2 + x) - 9x^2}{\sqrt{9x^2 + x} + 3x}<br />

    \frac{x}{\sqrt{9x^2 + x} + 3x}

    divide every term by x ...

    \frac{1}{\sqrt{9 + \frac{1}{x}} + 3}

    now let x \to \infty
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  4. #4
    Moo
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    Hello,

    Another approach is to use asymptotic equivalences (but I don't know your level, so maybe you can't use it...)

    \sqrt{9x^2+x}-3x=3x \left(\sqrt{1+\frac{1}{9x}}-1\right)

    Since x \to \infty, we have \frac{1}{9x} \to 0

    \sqrt{1+\frac{1}{9x}}=\left(1+\frac{1}{9x}\right)^  {1/2} \sim_{\substack{x \uparrow \infty}} 1+\frac 12 \times \frac{1}{9x}+ \mathcal{O}\left(\frac{1}{x^2}\right)

    So \lim_{x \to \infty} \sqrt{9x^2+x}-3x=\lim_{x \to \infty} 3x\left(1+\frac 12 \times \frac{1}{9x}+ \mathcal{O}\left(\frac{1}{x^2}\right)-1\right)

    =\lim_{x \to \infty} 3x \left(\frac 12 \times \frac{1}{9x}+\mathcal{O} \left(\frac{1}{x^2}\right)\right)=\lim_{x \to \infty} \frac 16+\mathcal{O} \left(\frac 1x\right)=\boxed{\frac 16}
    Last edited by Moo; September 15th 2008 at 01:17 PM. Reason: ...
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