Calculating $\lim_{n\to+\infty} x^{1/n}$ $$\lim_{n\to+\infty} x^{1/n}$$
Can I just do the following and consider when $x \geq 1$ and when $x < 1$?
Then for for $x \geq 1$ we have $x^{1/n} - 1 < \epsilon$ and using algebra we get $n > 1/\log_{x}(\epsilon +1)$. But then for $x < 1$ I can't do the same thing.
For $x < 1$ we get a strictly increasing sequence on a bounded interval making it so the sequence converges to the supremum which is $1$? I guess this reasoning would work for the other case anyway...
 A: Obtaining sentence "for $x> 1$ we have $x^{\frac{1}{n}}−1< \epsilon$" is main here, so for it let me suggest following way:
$$x=\left(1+(x^{\frac{1}{n}}−1) \right)^n > n(x^{\frac{1}{n}}−1) \Rightarrow 0<x^{\frac{1}{n}}−1 <\frac{x}{n} $$
For $0<x<1$ we have $\frac{1}{x}>1$ and can go to fraction limit based on already proved case:
$$\lim_{n \to \infty}x^{\frac{1}{n}} = \lim_{n \to \infty}\frac{1}{\left( \frac{1}{x} \right)^{\frac{1}{n}}}=\frac{1}{\lim_\limits{n \to \infty}\left( \frac{1}{x} \right)^{\frac{1}{n}}}=1$$
A: Let $L$ be the given limit. Taking $\log$ on both sides:
$$\log L=\lim_{n\to+\infty} \frac{1}{n}\log x. $$
Since $\log x$ is a finite quantity $\log L$ tends to $0$ as $n$ approaches infinity. Therefore $\log L=0$ i.e. $L=1$.
A: This is a classic Rudin proof. Let $x > 1$ and set $x_n = \sqrt[n]{x}-1 $ so that $x_n\geq 0$ for all $n$. By the Binomial Theorem $$\begin{align*}(x_n+1)^n = \sum_{k=0}^n{n\choose k} x_n^k & \geq {n\choose 0} + {n\choose 1}x_n = 1 + nx_n \\
x = (x_n+1)^n &\geq 1 + nx_n\\
\end{align*}$$
Some rearranging and we find that $0 < x_n \leq\frac{x-1}{n}$. Letting $n\to\infty$ and we get that $x_n\to 0$ which is equivalent to saying that $\sqrt[n]{x}\to 1$. For $x = 1$ the result is trivial. For $0 < x < 1$ we notice that $\frac{1}{x} > 1$ and so by our work above we have that $\sqrt[n]{\frac{1}{x}} \to 1$ as $n\to\infty$. But since $\sqrt[n]{\frac{1}{x}}\to 1$ and none of the terms of the sequence are zero, then the reciprocal converges to $1$. Hence $x > 0$ implies that $\sqrt[n]{x}\to 1$.
