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Homework Help: Limits question

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data

    L = lim x-> infinity ([tex]\sqrt{3x^{2}+4x}[/tex] - [tex]\sqrt{3x^{2}-2x}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I first multiplied the whole equation by [tex]\sqrt{1}[/tex] to get rid of the square root.
    Thus I got (3x^2 + 4x - (3x^2 - 2x)). With this, I divided the terms by x to get ultimately 4 + 2 = 6, but I don't think this answer is right because I didn't really substitute x anywhere. It just turned out to be some random algebra at the end.

  2. jcsd
  3. Jul 29, 2008 #2
    Your first step is incorrect.
    You cant just square everything to get rid of the square roots.
    (a + b)^2 does not equal to a^2 + b^2.
  4. Jul 29, 2008 #3


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    Homework Helper

    To simplify something like sqrt(A)-sqrt(B) multiply by (sqrt(A)+sqrt(B))/(sqrt(A)+sqrt(B)) i.e. '1' to get (A-B)/(sqrt(A)+sqrt(B)). You seem to be losing the denominator.
  5. Jul 29, 2008 #4
    But doing that doesn't really help me because now I have square roots on the denominator.
  6. Jul 29, 2008 #5
    Trying some factoring out, I got

    6x/(x sqrt(3 + 1/4x) + x sqrt(3 + 1/2x))

    and got stuck
  7. Jul 29, 2008 #6
    Ah ha, I think I got it now.

    Using l'hopital's rule I think? dividing by x, and therefore getting 6/2sqrt(3) as x heads to infinity in 1/4x and 1/2x and thus becoming 0 there.

    Thanks to Dick and others who helped.

  8. Jul 29, 2008 #7
    Yup thats the correct answer. Or in simpler terms its sqrt(3). I dont think l'hopitals rule should be used here.
  9. Jul 29, 2008 #8


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    It's certainly not neccesary to use L'Hopital's rule.

    JFonseka, do exactly what you were told to do initially: multiply [itex]\sqrt{3x^2+ 4x}- \sqrt{3x^2- 2x}[/itex] by
    [tex]\frac{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}[/tex]
    to get
    [tex]\frac{3x^2+ 4x- 3x^2+ 2x}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}[/tex]
    [tex]\frac{6x}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}[/tex]

    Yes, you now have square roots in the denominator but they are added so the "[itex]\infty- \infty[/itex]" problem that you had initially doesn't occur here. Since
    Do the "standard" method when dealing with limits at infinity: divide both numerator and denominator by the highest power of x- here just x itself. Of course, the "x" you divide by in the denominator become "x2" in the square root.
    Doing that you have:
    [tex]\frac{6}{\sqrt{3+ 4/x}+ \sqrt{3- 2/x}}[/tex]

    Now, as x goes to infinity, those fractions with x in the denominator go to 0. The limit is:
    [tex]\frac{6}{2\sqrt{3}}= \frac{3}{\sqrt{3}}= \sqrt{3}[/tex]
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