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Lim of An=(n+1)^(1/3) - n^(1/3)

  1. Feb 10, 2012 #1
    Hi all!
    Been trying to look for some examples with no luck.. all I found is
    related to square roots, not cube roots..
    Anyway I'm trying to solve: [tex]\lim_{n\to\infty}\sqrt[3]{n+1} - \sqrt[3]{n}[/tex]

    The limit is obviously 0.. But how do I simplify this expression to show it?
    Or should I use the ratio\root test? Even though I couldn't..
    Thanks a bunch
     
    Last edited: Feb 10, 2012
  2. jcsd
  3. Feb 10, 2012 #2

    kai_sikorski

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    [tex] (n+x)^{1/3} = n^{1/3}+\frac{x}{3 n^{2/3}}-\frac{x^2}{9 n^{5/3}}+\frac{5 x^3}{81 n^{8/3}} + ... [/tex]
    [tex] (n+1)^{1/3} - (n)^{1/3} = \frac{1}{3 n^{2/3}} + O(n^{-5/3}) [/tex]
    [tex] \lim_{n \to \infty} \left((n+1)^{1/3} - (n)^{1/3}\right) = 0 [/tex]
     
  4. Feb 10, 2012 #3
    This is the binomal theorem right? Is this the only way to solve this lim?
     
  5. Feb 10, 2012 #4

    kai_sikorski

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    I just expanded using a Taylor series. There's probably other ways.
     
  6. Feb 10, 2012 #5
    Yep, the trick is using [tex] a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]

    Thanks for your help anyway.
     
  7. Feb 10, 2012 #6

    kai_sikorski

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    Not sure how that helps, but if you found some way it does great!
     
  8. Feb 10, 2012 #7

    kai_sikorski

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    Oh okay I see. You know it's positive and then you bound it above using what you wrote.
     
  9. Feb 12, 2012 #8
    [itex]Let \ a \ = \ \ \sqrt[3]{n + 1} \ \ and \ \ b \ \ = \ \sqrt[3]{n}.[/itex]


    [itex]Then \ \ \dfrac{a - b}{1} \cdot \dfrac{a^2 + ab + b^2}{a^2 + ab + b^2} \ =[/itex]


    [itex]\dfrac{a^3 - b^3}{a^2 + ab + b^3} [/itex]



    What does the numerator become?


    As n --> oo, what happens to the denominator?
     
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