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Math Help - Limits approaching Infinity

  1. #1
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    Limits approaching Infinity

    I'm having difficulty finding lim (sqrt(9x^2+x)-3x) as x->infinity. I multiplied by the conjugate to get lim(x/(sqrt(9x^2+x)+3x) as x->infinity, but I'm not realy sure how to proceed. The book says that its 1/6 butI have no idea how to get there. Any and all help would be appreciated.
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  2. #2
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    Re: Limits approaching Infinity

    \sqrt{9x^2+x}-3x=3x\left(\sqrt{1+\frac{1}{9x}}-1\right)=3x\left(\left(1+\frac{1}{9x}\right)^{1/2}-1\right)=3x\left(1+\frac{1}{18x}+O\left(\frac{1}{x  ^2}\right)-1\right)=
    \frac{1}{6}+O\left(\frac{1}{x}\right)
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  3. #3
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    Re: Limits approaching Infinity

    Quote Originally Posted by TheDarkNight93 View Post
    I'm having difficulty finding lim (sqrt(9x^2+x)-3x) as x->infinity. I multiplied by the conjugate to get lim(x/(sqrt(9x^2+x)+3x) as x->infinity, but I'm not realy sure how to proceed. The book says that its 1/6 butI have no idea how to get there. Any and all help would be appreciated.
    \displaystyle \begin{align*} \lim_{x\to\infty} \left( \sqrt{9x^2 + x} - 3x \right) &= \lim_{x \to \infty} \frac{\left(\sqrt{9x^2 + x} - 3x\right)\left(\sqrt{9x^2 + x} + 3x\right)}{\sqrt{9x^2 + x} + 3x} \\ &= \lim_{x \to \infty}\frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} + 3x} \\ &= \lim_{x \to \infty} \frac{x}{x\left(\sqrt{9 + \frac{1}{x}} + 3\right)} \textrm{ for }x > 0\textrm{ which we can do because we are making }x \to \infty \\ &= \lim_{x \to \infty}\frac{1}{\sqrt{9 + \frac{1}{x}} + 3} \\ &= \frac{1}{\sqrt{9 + 0} + 3} \\ &= \frac{1}{6} \end{align*}
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    Re: Limits approaching Infinity

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \lim_{x\to\infty} \left( \sqrt{9x^2 + x} - 3x \right) &= \lim_{x \to \infty} \frac{\left(\sqrt{9x^2 + x} - 3x\right)\left(\sqrt{9x^2 + x} + 3x\right)}{\sqrt{9x^2 + x} + 3x} \\ &= \lim_{x \to \infty}\frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} + 3x} \\ &= \lim_{x \to \infty} \frac{x}{x\left(\sqrt{9 + \frac{1}{x}} + 3\right)} \textrm{ for }x > 0\textrm{ which we can do because we are making }x \to \infty \\ &= \lim_{x \to \infty}\frac{1}{\sqrt{9 + \frac{1}{x}} + 3} \\ &= \frac{1}{\sqrt{9 + 0} + 3} \\ &= \frac{1}{6} \end{align*}
    How do you go from x/(sqrt(9x^2+x)+3x) to x/(x(sqrt(9+1/x)+3))? By factoring out the x, wouldn't you be left with x/(x(sqrt9x+1)+3))?
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    Re: Limits approaching Infinity

    \sqrt{9x^2+x}=\sqrt{x^2(9+1/x)}=\sqrt{x^2}\sqrt{9+1/x}=x\sqrt{9+1/x} if everything is positive.
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