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Math Help - Finding the limit

  1. #1
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    Finding the limit

    I am not sure how to prove this by writing it out. I graphed it and the limit of that is 1/3.
    \displaystyle \lim_{ x \to \infty }{ \frac{ x - 3 }{ \sqrt{ 9x^2 - 1 } } }
    Last edited by johnsy123; September 2nd 2012 at 02:42 AM.
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  2. #2
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    Re: Finding the limit

    Quote Originally Posted by johnsy123 View Post
    I am not sure how to prove this by writing it out. I graphed it and the limit of that is 1/3.
    \displaystyle \lim_{ x \to \infty }{ \frac{ x - 3 }{ \sqrt{ 9x^2 - 1 } } }
    \displaystyle \begin{align*} \lim_{x \to \infty}\frac{x - 3}{\sqrt{9x^2 - 1}} &= \lim_{x \to \infty}\frac{\frac{1}{x}\left(x - 3\right)}{\frac{1}{x}\left(\sqrt{9x^2 - 1}\right)} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{\frac{1}{x^2}}\,\sqrt{9x^2 - 1}} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{\frac{9x^2 - 1}{x^2}}} \\ &= \lim_{x \to \infty}\frac{1 - \frac{3}{x}}{\sqrt{9 - \frac{1}{x^2}}} \\ &= \frac{1 - 0}{\sqrt{9 - 0}} \\ &= \frac{1}{\sqrt{9}} \\ &= \frac{1}{3} \end{align*}
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