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How can I prove by definition(with epsilon and N) that the nth root of c (c>0), when n tends to infinity, is 1?

Thanks.

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

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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

If $0

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    You don't need the binomial theorem, just Bernoulli's inequality for integer exponents2011-03-15
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    I have edited my answer, but I think you need some kind of inequality before using epsilons.2011-03-15
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    Julian Thank you very much!2011-03-15
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    @kahen: Can you please explain me more about what to do when c<1? thanks2011-03-15
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    Can someone please post further elaboration when c<1. I've tried putting what Julian suggested in the inequality. I also tried saying that $c=(1+d)^n$ and using Bernoulli inequality2013-04-03
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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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    Jerry, your last two answers have been flagged by the software as low quality because you aren't using very many words. It would help if you added a little more explanation to them.2011-03-15
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    I would think any further elaboration would fall in the spoonfeeding realm, which I am told the majority of the users here frown upon.2011-03-15
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    The OP asked for a prove by definition. As I read it the answer should involve some $\delta$ and $\epsilon$.2011-03-15
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    Well, when I answered, there was no indication he wanted an $\epsilon-\delta$ proof. (He edited after I posted.) :P I'll delete this later.2011-03-15
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    @Jerry: who told you that? My impression is that people here generally prefer more exposition to less.2011-03-15
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    I won't reveal the identity of the guilty party, sorry. I was told by my acquaintance that people who give elaborate answers to homework get downvoted here, and since this seemed to be homework... anyway, I've my downvote, and I don't really care for reputation anyway (otherwise I'd register).2011-03-15
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    @jerry: I wrote "by defenition" before your answer and added the epsilon and N after, anyway I thank you for the answer, I belive you used here Continuity, right?2011-03-15
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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