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Thread: Limit

  1. #1
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    Limit

    .


    Determine:


    $\displaystyle \displaystyle\lim_{n \to \infty} \sqrt[n]{n^\sqrt{n} \ }$
    Thanks from topsquark and MarkFL
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  2. #2
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    Re: Limit

    This is basically the same process that Plato used in another thread.

    $$\sqrt[n]{n^{\sqrt{n}}} = n^{1/\sqrt{n}}$$

    Since $f(x) = x^{1/\sqrt{x}}$ is continuous for $x>0$, we can consider the subsequence when $n$ is a perfect square:

    Let $x_n=n^{1/(2n)}-1$. Then $(x_n+1)^{2n} = n$.

    When $n\ge 1$:
    $$\dfrac{n(n-1)}{2}x_n \le n \Longrightarrow x_n \le \dfrac{2}{n-1}$$

    Since $\displaystyle \lim_{n \to \infty} x_n = 0$, we have $\lim_{n \to \infty} \sqrt[n]{n^{\sqrt{n}}} = 1$.
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  3. #3
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    Re: Limit

    Quote Originally Posted by SlipEternal View Post
    This is basically the same process that Plato used in another thread.

    $$\sqrt[n]{n^{\sqrt{n}}} = n^{1/\sqrt{n}}$$

    Since $f(x) = x^{1/\sqrt{x}}$ is continuous for $x>0$, we can consider the subsequence when $n$ is a perfect square:

    Let $x_n=n^{1/(2n)}-1$. Then $(x_n+1)^{2n} = n$.

    When $n\ge 1$:
    $$\dfrac{n(n-1)}{2}x_n \le n \Longrightarrow x_n \le \dfrac{2}{n-1}$$

    Since $\displaystyle \lim_{n \to \infty} x_n = 0$, we have $\lim_{n \to \infty} \sqrt[n]{n^{\sqrt{n}}} = 1$.
    I made a slight mistake. It should have been we consider when $n$ is an even perfect square. There would have been an extra factor of 4 on the RHS, but otherwise, it would proceed the same.
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  4. #4
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    Re: Limit

    How about this for a challenge problem:

    Let $x_0 = \sqrt[n]{n}$ and let

    $$x_k = \sqrt[n]{n^{x_{k-1}}}$$

    Find

    $$\lim_{n \to \infty} \lim_{k \to \infty}x_k$$

    I think this is still going to be 1, but not sure how to prove it.
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