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Polynomial Infinity Limit

  1. Jan 27, 2010 #1
    Hi, I'm in Engineering Foundation.
    I'm stuck in one limit question.

    Find the limit :_
    Lim (3x + V 9x^2 - x )
    x-> -infinity

    by substitution it gives ( inf - inf )
    I tried to solve it and get -inf

    Can anyone help me please ?
  2. jcsd
  3. Jan 27, 2010 #2


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    Welcome to PF!

    Hi mohamedibr752! Welcome to PF! :smile:

    (have a square-root: √ and an infinity: ∞ :wink:)
    (no need to shout! :redface:)

    Hint: it's the same as limx->-∞ 3x + 3|x|√(1 - 1/9x)) :wink:
    Last edited: Jan 27, 2010
  4. Jan 27, 2010 #3
    Thank you for your help .. I get it now ..
    but why did you write it 3x(1 + ...
    I think it is 3x(1 - ...
    because the x is approaching -∞ so as to remove absolute 3x we have to put - to make sure that the 3x is positive ??

    The answer I get is -∞.

    Thank you
  5. Jan 27, 2010 #4


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    Hi mohamedibr752! :smile:
    I goofed! :redface:

    I did correct it before you answered, but I think you missed the change. :wink:
    No, try again … expand the square-root and do a bit of cancelling. :smile:
  6. Jan 27, 2010 #5


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    Wait, what? No the answer isn't [itex]-\infty[/itex].

    Look again at [tex]3x+3|x|\sqrt{1-\frac{1}{9x}}[/tex] as the limit approaches [itex]-\infty[/itex]. It is definitely not what you stated.
  7. Jan 27, 2010 #6
    Multiply by [tex]\frac{3x -\sqrt{9x^2-x}}{3x-\sqrt{9x^2-x}}[/tex] ....... and then divide by x,

    then take the limit
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