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 $n<\infty$?

*

*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.
