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Limit question, evalute for C

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    lim [itex]\frac{\sqrt[3]{1+cx}-1}{x}[/itex]
    x[itex]\rightarrow[/itex]0

    3. The attempt at a solution

    I put 0 in for x, which lead to 0/0 therefore I know I know I need to modify to get a real answer. I have no idea where to start.
     
  2. jcsd
  3. Feb 20, 2013 #2

    SammyS

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    Are you allowed to use L'Hôpital's rule ?

    If not, then rationalize the numerator.

    a3 - b3 = (a - b)(a2+ab+b2)

    So that [itex]\displaystyle \ \ (\sqrt[3]{s}-\sqrt[3]{t})((\sqrt[3]{s})^2+\sqrt[3]{s}\sqrt[3]{t}+(\sqrt[3]{t})^2)=s-t[/itex]
     
  4. Feb 20, 2013 #3
    So it's like multiply by a conjugate for square root rationals, but for a cube root??

    Equivalent to x^2-y^2=(x-y)(x+y) ???
     
  5. Feb 20, 2013 #4

    Ray Vickson

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    Do you know about Taylor series? If so, just take the first few terms of the expansion of ##(1+cx)^{1/3}## about ##x=0##.
     
  6. Feb 20, 2013 #5

    HallsofIvy

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    Yes, but x^3- y^3= (x- y)(x^2+ xy+ y^2).
     
  7. Feb 20, 2013 #6
    or you can use the limit:

    [itex]\displaystyle\lim_{\begin{matrix}f(x)\to 0\\\mbox{when }x\to x_0\end{matrix}}\frac{(1+f(x))^\alpha-1}{f(x)}= \alpha\qquad (\heartsuit)[/itex]

    You have just to multiply and divide by [itex]c\ne 0[/itex] and use [itex](\heartsuit)[/itex]
     
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