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Math Help - Limits with infinity and radicals. (specific question)

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    Limits with infinity and radicals. (specific question)

    I have found the answer to the following problem both graphically and numerically to be -1/2, but I can't seem to prove that this is the answer algebraically. There must be a mistake with my algebra, but I'm not seeing it. Help?

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    Super Member TheChaz's Avatar
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    Re: Limits with infinity and radicals. (specific question)

    For x < 0 ,  \sqrt{x^2} = -x. There's one mistake...

    I saw a similar thread out on the internets:
    calculus - When can we plug in values in a limit? - Mathematics - Stack Exchange
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Limits with infinity and radicals. (specific question)

    Quote Originally Posted by Jnnash86 View Post
    I have found the answer to the following problem both graphically and numerically to be -1/2, but I can't seem to prove that this is the answer algebraically. There must be a mistake with my algebra, but I'm not seeing it. Help?

    \lim_{x\to-\infty}(\sqrt{x^2+x+1}+x)

    =\lim_{x\to-\infty}\frac{(\sqrt{x^2+x+1}+x)(\sqrt{x^2+x+1}-x)}{(\sqrt{x^2+x+1}-x)}

    =\lim_{x\to-\infty}\frac{x^2+x+1-x^2}{(\sqrt{x^2+x+1}-x)}

    =\lim_{x\to\infty}\frac{-x+1}{(\sqrt{x^2-x+1}+x)}

    =\lim_{x\to\infty}\frac{\frac{-x+1}{x}}{\frac{(\sqrt{x^2-x+1}+x)}{x}}

    =\lim_{x\to\infty}\frac{\frac{-x}{x}+\frac{1}{x}}{\frac{\sqrt{x^2-x+1}}{x}+\frac{x}{x}}

    =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{\frac{x^2-x+1}{x^2}}+1}

    =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+1}

    =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{1+0+0}+1}

    =-\frac{1}{2}
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    Re: Limits with infinity and radicals. (specific question)

    from about your third step ...

    \lim_{x \to -\infty} \frac{x+1}{\sqrt{x^2+x+1} - x}

    divide every term by \sqrt{x^2} = |x|

    \lim_{x \to -\infty} \frac{\frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{\frac{x^2+x+1}{x^2}} - \frac{x}{|x|}}

    \lim_{x \to -\infty} \frac{-1}{\sqrt{1 + \frac{1}{x} +\frac{1}{x^2}} - (-1)} = -\frac{1}{2}
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