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How can I prove that $$\lim_{n \rightarrow \infty} \sqrt[n]x=1$$ for all $x>0$

I know I've got to use the monotone convergence theorem somehow. It's easy to show that $\sqrt[n]x$ is bounded, but having trouble showing that it is strictly increasing for $0<a<1$ and strictly decreasing for $a>1$. Also how can prove the infimum and supremum is 1 for the two cases?

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    $\begingroup$ Are you allowed to use logarithms? $\endgroup$ – Henry Oct 5 '15 at 7:10
  • $\begingroup$ See here and here and here. (Probably you can find more similar questions.) $\endgroup$ – Martin Sleziak Oct 5 '15 at 9:46
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Note that for $a\gt 0$, $$a^b = e^{b\ln a}.$$ So $$\lim_{n\to\infty}x^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln x}.$$ Since the exponential is continuous, we have $$\lim_{n\to\infty}e^{\frac{1}{n}\ln x} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln x}.$$

Can you conclude now?

OR Use a sledgehammer,

We use the following theorem:

If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$

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First assume $x\ge 1$. Write $x_n = \sqrt[n]{x} - 1$. Notice that $x_n\ge0$, because $\sqrt[n]x\ge1$ if $x\ge1$. Using Bernoulli inequality we get $$ x = (1+x_n)^n \ge 1 + n\cdot x_n $$ for all $n$. This yields $0\le x_n\le \frac{x-1}n$ for all $n$. This shows $\lim_{n\to\infty} x_n = 0$ or $\lim_{n\to\infty} \sqrt[n]x = 1$.

If $0<x<1$, then we have $\lim_{n\to\infty}\sqrt[n]{1/x} = 1$ by the above and hence $$\lim_{n\to\infty}\sqrt[n]{x} = \frac1{\lim_{n\to\infty}\sqrt[n]{1/x}} = \frac11 = 1. $$

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Use the binomial theorem:

Case I. $x\geqslant1.$ For each $n,$ let $x_n:=\sqrt[n]{x}-1.$ Then we have $x=(1+x_n)^n\geqslant nx_n$ and hence $0\leqslant x_n\leqslant\dfrac{x}{n}$ and so $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}(\sqrt[n]{x}-1)=0$ which means that $\lim\limits_{n\to\infty}\sqrt[n]{x}=1.$

Case II. $0<x<1.$ Let $a:=\dfrac{1-x}{x}.$ Then $1+a>1$ and you can apply case I.

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