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How can I prove by definition (with $\epsilon$ and $N$) that, for $c>0$,

$$\lim_{n \to \infty}\ \sqrt[n]{c} = 1 ?$$

Thanks.

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    go buy a copy of Rudin, Principles of Mathematical Analysis. you wont regret it!2011-03-15

2 Answers 2

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Assume first $c>1$ and write $c^{1/n}=1+h_n$ with $h_n>0$. Taking $n$-th powers and using the binomial theorem get $ c=(1+h_n)^n=1+nh_n+\dots>1+nh_n\implies 0 Then $ 0 You can use this inequality to apply the standard $\epsilon$-$N$ argument.

If $0, use $c^{1/n}=1/(1/c)^{1/n}$.

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    I am sure you know that if $a_n\to a\ne0$ then $1/a_n\to1/a$.2013-04-03
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$\lim_{n\to\infty}\exp\left(\frac{\ln(c)}{n}\right)=\exp\left(\ln(c)\lim_{n\to\infty}\frac1{n}\right)=\dots$

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    And thanks for the downvote @Nir. :) Julián gave you the machinery required for $\epsilon-\delta$, so you might want to accept that one.2011-03-15