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Homework Help: Aggravating limit

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x\to\,-\infty}x + \sqrt{x^2 + 2x}[/tex]



    2. Relevant equations



    3. The attempt at a solution
    I tried conjugating by multiplying both the numerator and denominator by [tex]x + \sqrt{x^2 + 2x}[/tex], which gave me [tex]\frac{2x}{x - \sqrt{x^2 + 2x}}[/tex], but from there I have no idea on what to do.
     
    Last edited: Oct 13, 2008
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  3. Oct 13, 2008 #2

    Dick

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    sqrt(x^2+2x)=sqrt((x^2)*(1+2/x))=|x|*sqrt(1+2/x), yes?
     
  4. Oct 13, 2008 #3
    from there, I have limited out and made the algebra equal to [tex]\frac{2x}{x - \left|x\right|}[/tex]. From there, what should I do?
     
  5. Oct 13, 2008 #4

    Dick

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    Duh. I missed a sign mistake in your numerator. Can you correct it? Now you have to get rid of the absolute value. How can you do that?
     
  6. Oct 13, 2008 #5
    Ohhh, I double checked, and it was -2x. And, since its approaching negative infinity, the absolute value becomes --x, or +x. and, with this in mind, the equation becomes [tex]\frac{-2x}{2x}[/tex], or -1. And when a limit acts on a constant, the limit is the same as the constant. so, the limit equals -1.
     
  7. Oct 13, 2008 #6

    Dick

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    The limit is -1 alright, but the function isn't a constant. You already took most of the limit when you replaced sqrt(1+2/x) with 1. And |x|=-x if x<0. So x-|x|=x-(-x) is 2x. Your conclusions are right, but some of your words to describe the reasons are off.
     
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