The first thing I was trying to do is to substitute $a$ with something like $1 \over b$ or $1 + \frac{1}{b}$, but I got no results. Then I tried to use Squeeze theorem here, but I also got confused because, for example, $(a + 1)^{1/n}$ is not a decreasing function.
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What does $(a+1)^{1/n}$ have to do with it? – Thomas Andrews Dec 10 '16 at 21:04
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Do you know the mean value theorem? – Thomas Andrews Dec 10 '16 at 21:06
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Yes, I know about it, but it holds for $(a,b)$, not for $(a, b]$, doesn't it? – Welez Dec 10 '16 at 21:10
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2@Purifyingflames,look here http://math.stackexchange.com/questions/1867269/use-delta-epsilon-to-show-that-lim-n-to-infty-a-frac1n-1/1867299#1867299 – haqnatural Dec 10 '16 at 21:12
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You need to specify the the limit. Does $n \to 1$? Does $n \to 0$? Does $n \to \infty$? – Fly by Night Dec 10 '16 at 21:21
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Without Loss of Generality, we assume $0<a<1$. Let $x_n=(1/a)^{1/n}-1$ so that $0\le x_n$. Then, we see using Bernoulli's Inequality that
$$a=\frac{1}{(1+x_n)^n}\le \frac{1}{1+nx_n}$$
whereupon isolating $x_n$ reveals
$$0 \le x_n \le \frac{1-a}{a\,n}\tag 1$$
Applying the squeeze theorem to $(1)$ yields the coveted limit
$$\lim_{n\to \infty}a^{1/n}=\lim_{n\to \infty}\frac{1}{1+x_n}=1$$
Note for $a>1$, simply analyze the limit of $(1/a)^{1/n}$ using the same way forward.
Mark Viola
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Let $b_n = a^{1/n}$.
Assume $a\neq 1$ ($a=1$ is easy.)
Use that $$\frac{1-b_n}{1-a} =\frac{1-b_n}{1-b_n^n}= \frac{1}{1+b_n+b_n^2+\cdots + b_n^{n-1}}$$ Use this to get a good lower and upper bound on $1-b_n$ for use with the squeeze theorem.
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The Mean Value Theorem version is that $\frac{b_n^n-1}{b_n-1} = nc^{n-1}$ for some $c$ in $[b_n,1]$, and again using this to get upper and lower bounds on $1-b_n$.
Thomas Andrews
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