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Prove

$\lim_{n\to\infty} 5^{\frac{1}{n}}=1$

Observe that we have \begin{align*} |5^{\frac{1}{n}}-1|&< \epsilon\\ 5^{\frac{1}{n}}&<\epsilon +1 \\ (\frac{1}{n})\ln{5}&<\ln{(\epsilon +1)}\\ \ln{5}&<n[\ln{(\epsilon +1)}]\\ \frac{\ln{5}}{\ln{(\epsilon +1)}}&<n \\ \end{align*}

Therefore let $\epsilon >0$ be arbitrarily given. Then choose $N> \frac{\ln{5}}{\ln{(\epsilon +1)}}$; then when $n\geq N$ that implies that

$|5^{\frac{1}{n}}-1|< \epsilon$ therefore $\lim5^{1/n}=1$.

However showing the last line i'm having trouble with now that I have my $N$.

HighSchool15
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    What's the connection between all those five expressions in the middle of your answer? – José Carlos Santos Jan 30 '18 at 08:32
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    @JoséCarlosSantos If you can't see it, here is a quick run-down: 1-2 remove absolute value signs because $5^{1/n}>1$ because $\frac1n>0$. 2-3 take logarithm. 3-4 multiply by $n$ and finally 4-5 divide by $\ln(\epsilon + 1)$. – Arthur Jan 30 '18 at 08:35
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    @Arthur What I cant see are the $\implies$ and $\iff$ signs that should be there. – José Carlos Santos Jan 30 '18 at 08:41
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    @JoséCarlosSantos I never knew those were obligatory. I mean, they can certainly be of use, but writing lines of an inequality under one another without such arrows is entirely standard, and I daresay more common than using them, generally. I don't use them unless they are important to a specific point I'm making, for instance, and I have never had people complain like you do here. – Arthur Jan 30 '18 at 08:53
  • I agree with Jose. Usually if you have a string of inequalities like this, the assumed connection between the lines is $\Rightarrow$. However, for the proof to work, we require the reverse implication between each line. This should be more explicit. – Harambe Jan 30 '18 at 09:03
  • Also see https://math.stackexchange.com/questions/1867269/use-delta-epsilon-to-show-that-lim-n-to-infty-a-frac1n-1/. – StubbornAtom Jan 30 '18 at 10:05

2 Answers2

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It seems ok, you have found a value for $N$ such that for each $\epsilon>0$ for $n>N\quad |5^{\frac{1}{n}}-1|< \epsilon$, thus the limit is proved.

user
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I think your proof works.

Below is another proof.

Let $1+h_n=5^{\frac{1}{n}}$, then $h_n>0$ and $(1+h_n)^n=5$.

So, $\displaystyle 5>1+\binom{n}{1}h_n$ for $n>1$.

$\displaystyle h_n<\frac{4}{n}$.

For $\epsilon>0$, take $N\in\mathbb{N}$ such that $\displaystyle N>\frac{4}{\epsilon}$. Then for $n>N$,

\begin{align*} \left|5^{\frac{1}{n}}-1\right|&=h_n\\ &<\frac{4}{n}\\ &<\frac{4}{N}\\ &<\epsilon \end{align*}

CY Aries
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