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Compute $\displaystyle\lim_{n\rightarrow\infty} n^{\frac{1}{n}}$

If there exists the sequence $x_n = n^{\frac{1}{n}} - 1$ and we have established the fact: $x_{n}^2 \leq \frac{2}{n}$. Compute $\displaystyle\lim_{n\rightarrow\infty}n^{\frac{1}{n}}$

Truly frustrated at this point because I know it is just a small step in my observations that will allow me to draw the final conclusion and not accomplishing it has me very upset. What am I not seeing that would help me along the path ?

user0102
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D.C. the III
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3 Answers3

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HINT

\begin{align*} n^{\frac{1}{n}} = \exp\left(\frac{\ln(n)}{n}\right) \end{align*}

user0102
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  • WHat does your hint imply? I understand the squeeze theorem form of arriving at the conclusion. But what is your way of doing it? – D.C. the III Oct 06 '18 at 21:14
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    Apply the L'Hôpital rule to the argument in order to obtain $0$. Since the exponential function is continuous, the answer is $\exp(0) = 1$. – user0102 Oct 06 '18 at 21:20
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If you've already shown that $x_n^2 \leq 2/n$, this implies that $-\sqrt{\frac{2}{n}} \leq x_n \leq \sqrt{\frac{2}{n}}$.

Now just apply the squeeze theorem, as both the left and right terms clearly go to $0$ as $n \to \infty$.

A. Thomas Yerger
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I propose a folk-solution based on AM-GM inequality. We have $$ 1\leqslant\sqrt[n]{n}=\sqrt[n]{\sqrt{n}\cdot\sqrt{n}\cdot 1\cdot\ldots\cdot 1}\leqslant\frac{2\sqrt{n}+n-2}{n} $$

The right hand side tends obviously to $1$, so the Squeeze Theorem concludes the proof that $\lim\limits_{n\to\infty}\sqrt[n]{n}=1.$

szw1710
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