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Limits Problem

  1. Feb 8, 2009 #1
    Evaluate if the following limit exists:

    [tex]lim_{x\rightarrow1}\frac{\sqrt[3]{x}-1}{x-1}[/tex]


    My work:

    [tex]lim_{x\rightarrow1}\frac{\sqrt[3]{x^{2}}-1}{(x-1)(\sqrt[3]{x}+1)}}[/tex]

    I did the conjugate, but I'm still left with a radical in the numerator, and I can't seem to factor any further. Can someone help me out?
     
  2. jcsd
  3. Feb 8, 2009 #2

    Mark44

    Staff: Mentor

    You have the right idea, but haven't quite carried it off. The radical term in your numerator is x^(2/3) != x^(1/3) in the original limit expression.

    x - 1 can be thought of as the difference of cubes, as [itex](\sqrt[3]{x})^3 - 1^3[/itex].

    A difference of cubes can be factored like so:
    [tex](a^3 - b^3) = (a - b)(a^2 + ab + b^2)[/tex].
     
  4. Feb 8, 2009 #3
    Thanks, that helped for that question. But how about a question like the following(we have not yet been taught the difference of fourth root):

    [tex]lim_{x\rightarrow0}\frac{\sqrt[4]{1+x}-1}{x}[/tex]

    Could I do a difference of squares and then the conjugate?
     
  5. Feb 8, 2009 #4

    Mark44

    Staff: Mentor

    How about if you multiplied by 1 (which is always legal)?

    The 1 I am thinking of looks like
    [tex]\frac{(u + 1)(u^2 + 1)}{(u + 1)(u^2 + 1)}[/tex]

    where [tex]u = \sqrt[4]{1 + x}[/tex].

    The idea is that you have something that looks like u - 1 in the numerator, and I want to turn it into u^4 - 1. The other factors to make this happen are (u + 1) and (u^2 + 1).

    No guarantees that this will work, but it might make it so that some things cancel so that you don't get zero in both the numerator and denominator.
     
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