How can one prove that $\lim_{n \to \infty}x^{1/n}=1$ for every $x>0$? Two questions:
1) Let $x \in \mathbb{R}$ and $y \in \mathbb{R}$ and $x>0$ and $y>0$. Let $n \in \mathbb{N}$. We know that
For every real $x>0$ and every integer $n>0$ there is one and only one real $y>0$ such that $y^n=x$. The number $y$ is written $x^{1/n}$. (Rudin, 1976, p.10 with addition of $y>0$)
Is it correct to affirm that
$$\lim_{n \to \infty}x^{1/n}=1$$ 
for any $x$?
2) When I write $n \to \infty$ I can affirm that $n<\infty$?
 A: *

*The simplest way to show this fact is using the Intermediate Value Theorem: If $f:[a,b]\to \mathbb R$ is continuous, and $f(a)<\xi<f(b)$, then there exists a $c\in(a,b)$, such that $f(c)=\xi$. Further, if $f$ is strictly increasing, then the $c$ is unique.


So, here $f(x)=x^n$ is continuous and strictly increasing in $[0,1+x_0]$, and 
$$
f(0)=0<x_0<f(1+x_0)=(1+x_0)^n
$$
and hence there exists a unique $y_0\in (0,1+x_0)$, such that 
$$
f(y_0)=y_0^n=x_0.
$$
Now, if $x^{1/n}=1+y_n$, then 
$$
x=(1+y_n)^n\ge 1+n y_n\ge n y_n>0,
$$
and thus
$$
0<y_n\le\frac{x}{n}\to 0,
$$
and hence
$$
x^{1/n}=1+y_n\to 1.
$$
While $n\to\infty$, the value of $n$ is positive, eventually larger than any positive numbers, but still a number and not infinite.
A: 1) Bernoulli's Inequality says that for all $n\ge1$ and $u\ge-1$,
$$
1+nu\le(1+u)^n\tag{1}
$$
Let $x=1+nu$ and take $n^\text{th}$ roots:
$$
x^{1/n}\le1+\frac{x-1}{n}\tag{2}
$$
Taking reciprocals of $(1)$:
$$
\frac1{1+nu}\ge\left(1-\frac{u}{1+u}\right)^{\large n}\tag{3}
$$
Let $x=\frac1{1+nu}$ and take $n^\text{th}$ roots:
$$
x^{1/n}\ge1-\frac{1/x-1}{1/x-1+n}\tag{4}
$$
Using $(2)$ and $(4)$ and the Squeeze Theorem gives
$$
\lim_{n\to\infty}x^{1/n}=1\tag{5}
$$
2) since $\infty$ is not a real number, $n$ will never be $\infty$. When we write $\lim\limits_{n\to\infty}\,$, we simply mean that $n$ gets arbitrarily big, not that it ever becomes infinite.
A: One good way to obtain the result is divide the proof. First try to show that for every $M>0$ and every $\varepsilon>0$ there is a $n$ such that $M^{1/n}\le 1 + \varepsilon$ (Hint: use the fact that if $x>1$, then  $x^n$ diverges). After this you may treat the cases $x\ge 1$ and $x<1$ separately.
With respect to the last point of the question number 1) you can also define the $n$-th root as follows, let $n \ge 1$, $x>0$ and $E= \{z\in \mathbb{R}^{\ge0}:z^{n}\le x\}$. Then $x^{1/n}: = \sup E$. 
We need to show that this definition make sense. 
Existence: First of all we have to show that least upper bound property can be apply. Clearly the set is non-empty because $0 \in E$. So only we need to show that the set is bounded above. We divide the proof in two cases, $x\le1$ and $x>1$. For the first case, we claim that $1$ is an upper bound. If were not the case, there is an element $y\in E$ such that $y>1$, then $y^n>1$, $y^n > x$ and hence $y^n \notin E$, a contradiction. Now suppose that we are in the case  $x>1$,  we claim that $x$ itself is an upper bound. If were not the case, there is some element $y\in E$ such that $y>x$ so $y^n>x^n$, and since $x>1$ we have $x^n\ge x$. So  $y^n>x$,  $y \notin E$, a contradiction. Thus in both cases $E$ has an upper bound. 
Now we have to  show that this definition has the desired property. In other words,  if $y =x^{1/n}$, then $y^n=x$
Suppose to the contrary that $y^n<x$.  Choose $\varepsilon<\min \bigg(y, \frac{x-y^n}{2^ny^{n-1}}\bigg)$. Then  we have 
$$(y+\varepsilon)^n<y^n+2^n\varepsilon y^{n-1}<x$$
Then $y+\varepsilon \in E$ which contradicts that $y$ is an upper bound. A very  similar argument can be used to show that $y^n>x$ also yields a contradiction (in this case, choosing $\varepsilon>0$ small enough such that $(y-\varepsilon)^n>x$, contradicting that $y$ is the least upper bound). Thus the only alternative is $y^n=x$ as desired. 
Notice that the uniqueness follows immediately from the definition of least upper bound.
A: The fact that $\lim_{n\to \infty} x^{1/n}=1$ can be shown in two easy steps:


*

*Show that the sequence $\{x^{1/n}\}$ is either strictly increasing and bounded above by $1$, if $0<x\leq 1$, or strictly decreasing and bounded below by $1$, if $x>1$. Then use the monotone convergence theorem to prove that the sequence has a limit.

*Let the limit be $L$. Then, the limit of the sequence $\{\log(x^{1/n})\}$ must be $\log L$, because $\log $ is continuous on $(0,\infty)$. Since
$$ \log(x^{1/n})=\frac{\log x}{n} \to 0,$$
as $n\to \infty$, it follows that $\log L=0$, and therefore $L=1$.
