Limit

Dec 2016
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161
Earth
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Determine:


\(\displaystyle \displaystyle\lim_{n \to \infty} \sqrt[n]{n^\sqrt{n} \ }\)
 
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SlipEternal

MHF Helper
Nov 2010
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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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SlipEternal

MHF Helper
Nov 2010
3,728
1,571
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.
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
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.