Prove: $ \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 $ 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!
 A: 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, 
$$\lim_{n \to \infty} \sqrt[n]{x} = 1.$$  
A: 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 $n\to\infty$? 
A: 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 $L=1$.
