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Limit as x tends to infinity

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

    calculate the limit as x tends to infinity of:

    [tex]\sqrt[3]{x}[/tex] ((x+1)[tex]^{(2/3)}[/tex]-(x-1)[tex]^{(2/3)}[/tex])

    2. Relevant equations

    3. The attempt at a solution

    using the identity: a-b=(a^2-b^2)/(a+b) ; and dividing top and bottom by x,
    = lim [tex]\frac{x^{(1/3)}[(1+1/x)^{(4/3)}-(1-1/x)^{(4/3)}}{(1+1/x)^{(2/3)}+(1-1/x)^{(2/3)}}[/tex]
    = infinity
    is that right? it seems very wrong.
  2. jcsd
  3. Nov 29, 2008 #2


    Staff: Mentor

    The limit is 4/3.
    Just to make it simpler to write, let u = (x + 1)^(1/3) and v = (x - 1)^(1/3). Then your expression is
    [tex]x^{1/3}(u^2 - v^2)[/tex]
    [tex]= x^{1/3}(u - v)(u + v)[/tex]
    Now multiply by 1 in the form of [tex]\frac{u^2 + uv + v^2}{u^2 + uv + v^2}[/tex]
    This will give you [tex]x^{1/3}\frac{(u^3 - v^3)(u + v)}{(u^2 + uv + v^2)}[/tex]
    Now undo the substitution and take the limit.
  4. Nov 29, 2008 #3
    thank you
  5. Nov 29, 2008 #4
    so, that means we have to use the binomial expansion inorder to divide by x^1/3 ??
  6. Nov 29, 2008 #5


    Staff: Mentor

    No, not at all. You'll have terms with x + 1 and x - 1 to the 1/3 and 2/3 powers. For each of these terms factor as x(1 + 1/x). Depending on the power the original terms are raised to, you'll pull out a factor of x^(1/3) or x^(2/3). In the end, you'll have x to the same power in the numerator as in the denominator, so they cancel.
  7. Nov 30, 2008 #6
    ok, but then i get:
    after cancellin the x^(2/3) from top and bottom, the last term in the denominator cancels out as x tends to infinity so isnt the limit 4/2 instead of 4/3 beacuse i get 1+1 at the denominator ??
  8. Nov 30, 2008 #7


    Staff: Mentor

    My denominator looked like this:
    [tex]x^{2/3}[(1 + 1/x)^{2/3} + (1 + 1/x)^{1/3}(1 - 1/x)^{1/3} + (1 - 1/x)^{2/3}][/tex]

    The first factor of x^(2/3) cancels with the same factor in the numerator. The part in square brackets approaches 3 as x gets large, so I don't understand what you're saying about the last term cancelling.

    I'm certain that the limit is 4/3, both from the work I did and verifying the limit with Excel.
  9. Nov 30, 2008 #8
    oh right, i see. ur absolutely correct and thank you. :)
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