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Help in evaluating limit

  1. Aug 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate: [tex] \lim_{x \rightarrow \infty} ( \frac {x^3}{(3x)^2-4}-\frac{x^2}{3x+2} ) [/tex]

    2. The attempt at a solution

    First i took x^2/3x+2 common, and then substituted y=1/x , y->0.
    simplifying, i get [tex]\frac{-2}{9y} [/tex]...i am unable to eliminate all the y's.
    The answer given at the back of the book is 2/9.
    Plz help me figure this out.
    Last edited: Aug 31, 2007
  2. jcsd
  3. Aug 31, 2007 #2


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    Are you sure that's exactly the question? The limit approaches minus infinity...
  4. Aug 31, 2007 #3
    There is a theorm for this because what you have is an [tex] \infty [/tex]/[tex] \infty [/tex] correct? (L'Hopital)
    Note: If this is a calc one qustion ignore this.
    Last edited: Aug 31, 2007
  5. Aug 31, 2007 #4
    yeah, thats why i am confused

    Say for the 1st term, i divide and multiply by x^3, the Dr. becomes 0 ?
  6. Aug 31, 2007 #5


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    Are you sure the question isn't:

    [tex] \lim_{x \rightarrow \infty} ( \frac {3x^3}{(3x)^2-4}-\frac{x^2}{3x+2} ) [/tex]
  7. Aug 31, 2007 #6


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    i keep getting [tex]\infty[/tex] not 2/9
  8. Sep 1, 2007 #7
    Find a common denominator, then add the fractions together.
  9. Sep 1, 2007 #8
    I checked it again, but the question is exactly as I put it.
  10. Sep 1, 2007 #9


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    (3x)^2 - 4 = 9x^2 - 4 = (3x + 2)(3x - 2)

    use this to simply your function (i.e multply your second term with (3x -2) in both nominator and denominator)

    Then the x^3 term will vanish

    Also, there must be a misprint as "learningphysics" has noticed
    Last edited: Sep 1, 2007
  11. Sep 1, 2007 #10


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    I'm guessing it's a misprint or something that left out the 3... because the limit of the function I just posted is 2/9.
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