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Homework Help: Rationalizing cubed expressions

  1. Nov 19, 2007 #1
    I know how to rationalize most equations when trying to find limits. However, this problem seem give me trouble.

    (((1+x)^3)-1)/x as x approaches 0

    I tried the method of multiplying by the conjugate but it doesn't seem to get me anywhere. Mainly its the cubed (1+x) that troubles me.
  2. jcsd
  3. Nov 19, 2007 #2
    So you don't want to just expand it traditionally and simplify? You want some sort of easy trick (like multiplying by the conjugate and whatnot)?
  4. Nov 19, 2007 #3
    Could you elaborate and show me how I could expand it traditionally and solve. Maybe I'm just over tired but i can't see how that would work.
  5. Nov 19, 2007 #4


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    I don't see anything to rationalize. Just expand (1+x)^3 subtract 1 and divide by x.
  6. Nov 19, 2007 #5
    Alright, sure. Let's do it in steps.

    First, do:

    [tex] (1+x)^2 = (1+x)(1+x)[/tex]

    This should be quickly determined to be:

    [tex] 1+2x+x^2 [/tex]

    So, now note:

    [tex] (1+x)^3 = (1+x)(1+x)(1+x) = (1+2x+x^2)(1+x) [/tex]

    Then distribute:

    [tex] (1+2x+x^2)(1+x) = (1)(1+x) + (2x)(1+x) + (x^2)(1+x) = 1+x+2x+2x^2+x^2+x^3 [/tex]

    Then simplify:

    [tex] 1+x+2x+2x^2+x^2+x^3 = 1+3x+3x^2+x^3 [/tex]

    Now, remember your original expression and substitute:

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

    So, then, we have:

    [tex] \lim_{x\rightarrow 0} \left(\frac{(1+x)^3-1}{x}\right) = \lim_{x\rightarrow 0} \left(x^2+3x+3\right) [/tex]

    You can take it from here.
    Last edited: Nov 19, 2007
  7. Nov 20, 2007 #6
    thank you for clarifying that. Maybe i just needed a quick refresher to show me how again.
  8. Nov 21, 2007 #7
    Wait, so are you good? Or are you still confused?
  9. Nov 21, 2007 #8
    yes im good thank you
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