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Expanding a simple cube root

  1. Oct 12, 2014 #1
    A quiz at the end of Steven Krantz's Calculus Demystified includes the following problem:
    \lim_{x \to \infty} [ \sqrt[3]{x+1}
    \sqrt[3]{x} ]
    I see how one can use the Maclaurin series to get
    \sqrt[3]{x+1} = 1 + \frac {x}{3} - \frac {x^2}{9} + \frac {5 x^3}{81} + . . .
    but trying it with the cube root of x gives me zero plus an endless series of undefined terms.
    Is there a way to expand [itex] \sqrt[3]{x} [/itex] and solve this problem?
  2. jcsd
  3. Oct 12, 2014 #2
    But why do you want expand that in series?
    Isn't it by simple high school algebra :
    for x>0?
  4. Oct 12, 2014 #3
    Oh yeah, you're right zoki85 - thanks!

    I also just realized that I could simply define y = x + 1, plug in y-1 for all the x's in the series above and there's the answer to my other question. :)
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