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 $n<\infty$?

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.
