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Math Help - Determing a limit

  1. #1
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    Determing a limit

    The question states: Find the limit of the sequence
    { n^{2/3}[(n+1)^{1/3}-n^{1/3}]}

    I'm assuming they are asking as n approaches infinity, would anyone be able to offer tips on where to begin to solve?
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  2. #2
    MHF Contributor red_dog's Avatar
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    \lim_{n\to\infty}\sqrt[3]{n^2}\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)=

    =\lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n^2+n}+\sqrt[3]{n^2}}=

    \lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{n^2}\left(\sqrt[3]{\left(1+\frac{1}{n}\right)^2}+\sqrt[3]{1+\frac{1}{n}}+1\right)}=

    =\lim_{n\to\infty}\frac{1}{\sqrt[3]{\left(1+\frac{1}{n}\right)^2}+\sqrt[3]{1+\frac{1}{n}}+1}=\frac{1}{3}
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  3. #3
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    red dog used the fact that a^3- b^3= (a- b)(a^2+ ab+ b^2) with a= (n+1)^{1/3} and b= n^{1/3}.
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  4. #4
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    anoher approach could be by putting t=\frac1n and then put t+1=u^3.
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  5. #5
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    Thanks guys!
    one step I don't seem to fully understand is going from:
    \lim_{n\to\infty}\sqrt[3]{n^2}\left(\sqrt[3]{n+1}-\sqrt[3]{n}\right)
    to
    \lim_{n\to\infty}\frac{\sqrt[3]{n^2}}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n^2+n}+\sqrt[3]{n^2}}
    why does part of the factored expression all move into the denominator?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by xxlvh View Post
    The question states: Find the limit of the sequence
    { n^{2/3}[(n+1)^{1/3}-n^{1/3}]}

    I'm assuming they are asking as n approaches infinity, would anyone be able to offer tips on where to begin to solve?
    n^{2/3}[(n+1)^{1/3}-n^{1/3}]=n[(1+1/n)^{1/3}-1]=n\left[\left(1+\frac{1}{3n}+O(n^{-2})\right) -1\right]  =1/3+O(n^{-1})
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  7. #7
    Math Engineering Student
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    \underset{n\to \infty }{\mathop{\lim }}\,\left( \sqrt[3]{n^{3}+n^{2}}-n \right), as i said before, put t=\frac1n and the limit becomes \underset{t\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{t+1}-1}{t}, and then put u=\sqrt[3]{t+1} and the limit is \underset{u\to 1}{\mathop{\lim }}\,\frac{u-1}{u^{3}-1}=\underset{u\to 1}{\mathop{\lim }}\,\frac{1}{u^{2}+u+1}=\frac{1}{3}.
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