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Math Help - Limit as X approaches negative infinity

  1. #1
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    Limit as X approaches negative infinity

    i need to find the limit as x -> negative infinity for the function

    (sqrt(x^2-2x))/(2x+7)
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  2. #2
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    The limit is -1/2. \sqrt{x^2 - 2x} : \sqrt{x^2 - 2x + 1} = \sqrt{(x-1)^2} = |x-1|, and \frac{|x-1|}{2x + 7} : \frac{|x|}{2x}. Here I am using ":" to mean "approximately equal."
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  3. #3
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    You have already asked this before here:
    http://www.mathhelpforum.com/math-he...ing-limit.html
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  4. #4
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    yes but i needed the first answer and didn't know how to ask for it differently
    Last edited by Manizzle; October 2nd 2008 at 10:48 AM.
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  5. #5
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    He wasn't wrong, Both methods are correct.
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  6. #6
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    Actually, I did make a mistake and I'm sorry about that. Regardless, you could've still pointed it out in the original thread.

    The problem is when I multiplied \sqrt{x^2 -2x} by \frac{1}{x}. I'll leave it to you to see what went wrong.

    An alternative way to do this is to take x^2 and x common factor:
    \frac{\sqrt{x^2-2x}}{2x+7}

    = \frac{\sqrt{x^2(1-\frac{2}{x})}}{x(2+\frac{7}{x})}

    = \frac{|x|\sqrt{1-\frac{2}{x}}}{x(2+\frac{7}{x})}

    Since limit goes to negative infinity, absolute value of x is turned into -x. And I'm sure you can follow from there.
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