# Prove: $\lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1$ [duplicate]

x is a real number between 0 and 1

$\lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1$

SO far i know you need to do this to prove it:

let b>1, prove that $\root n \of b \rightarrow 1$ as $n \rightarrow\infty$

I was given this hint: let $a_n = \root n \of b - 1$ , prove that $a_n \rightarrow 0$

Thanks!

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## marked as duplicate by Aryabhata, ncmathsadist, leo, hardmath, Hanul JeonNov 30 '13 at 5:00

Why is this giving you a problem? It is not a undefined form .... if you directly use $n\to \infty$ it becomes $x^0$ and that is 1. – ABC Apr 5 '13 at 3:36

The proof you are looking for will follow from the binomial theorem - or, more precisely, the derivative Bernoulli inequality, which says that

$$(1+h)^n \geq 1 + nh, \qquad h>0$$

Set $\sqrt[n]{x} = (\frac{1}{1+h})$ for $h > 0$. If you flip both sides of the Bernoulli inequality, you get

$$\left(\frac{1}{1+h}\right)^n \leq \frac{1}{1+nh}$$

So, substituting for $x$, $$x \leq \frac{1}{1+nh} = \frac{1}{1+n(\frac{1}{\sqrt[n]{x}}-1)}$$ $$\therefore 1 + n(\frac{1}{\sqrt[n]{x}}-1) \leq \frac{1}{x}$$ $$\implies \frac{n}{\sqrt[n]{x}} \leq \frac{1}{x} + n-1$$ $$\implies {\sqrt[n]{x}} \geq \frac{n}{n - 1 + \frac{1}{x}} = \frac{1}{1 + \frac{k}{n}}$$ where in the last step, we set $k$ s.t. $\frac{1}{x} = 1 + k$.

From this, we see that $1 \geq \sqrt[n]{x} \geq \frac{1}{1 + \frac{k}{n}}$, where $k$ is fixed. It follows that, as $n$ gets really big,

$$\lim_{n \to \infty} \sqrt[n]{x} = 1.$$

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I'll admit I'm not entirely sure if this will be of too much help, but it's worth a try.

Recall that $1+x+x^2+\dots+x^{n-1}=\frac{1-x^n}{1-x}=\frac{x^n-1}{x-1}$.

Let $x=\sqrt[n]b$ so that $x^n=b$ and thus we have

$$1+\sqrt[n]b+\sqrt[n]{b^2}+...+\sqrt[n]{b^{n-1}}=\frac{b-1}{\sqrt[n]b-1}$$

Rearrange to yield:

$$\sqrt[n]b-1=\frac{b-1}{1+\sqrt[n]b+\sqrt[n]{b^2}+...+\sqrt[n]{b^{n-1}}}$$

Now, can you show $\sqrt[n]b-1\to0$ as $n\to\infty$?

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Let $x$ be a fixed real number with $0<x\leq 1$. The sequence $\{x^{1/n}\}_{n=1}^\infty$ is increasing, bounded above by $1$, so by the monotonic sequence theorem, it has a limit $L$, with $0<x\leq L\leq 1$. Say $$\lim_{n\to \infty} x^{1/n}=L$$ and take logarithms of both sides. Since the logarithm is continuous on $(0,\infty)$, we can bring the logarithm inside the limit: $$\log(\lim_{n\to \infty} x^{1/n}) = \lim_{n\to \infty} \log(x^{1/n}) = \lim_{n\to \infty} \frac{1}{n} \log x = 0.$$ Hence, $\log L=0$, and so $L=1$.

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