How can I prove that $\lim_{n \rightarrow \infty} \sqrt[n]x=1$ How can I prove that $$\lim_{n \rightarrow \infty} \sqrt[n]x=1$$
for all $x>0$
I know I've got to use the monotone convergence theorem somehow. It's easy to show that $\sqrt[n]x$ is bounded, but having trouble showing that it is strictly increasing for $0<a<1$ and strictly decreasing for $a>1$. Also how can prove the infimum and supremum is 1 for the two cases?
 A: Note that for $a\gt 0$, 
$$a^b = e^{b\ln a}.$$
So
$$\lim_{n\to\infty}x^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln x}.$$
Since the exponential is continuous, we have
$$\lim_{n\to\infty}e^{\frac{1}{n}\ln x} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln x}.$$
Can you conclude now?
OR Use a sledgehammer,
We use the following theorem:
If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$
A: First assume $x\ge 1$. Write $x_n = \sqrt[n]{x} - 1$. Notice that $x_n\ge0$, because $\sqrt[n]x\ge1$ if $x\ge1$. Using  Bernoulli inequality we get
$$
x = (1+x_n)^n \ge 1 + n\cdot x_n
$$
for all $n$. This yields $0\le x_n\le \frac{x-1}n$ for all $n$. This shows $\lim_{n\to\infty} x_n = 0$ or $\lim_{n\to\infty} \sqrt[n]x = 1$.
If $0<x<1$, then we have $\lim_{n\to\infty}\sqrt[n]{1/x} = 1$ by the above and hence
$$\lim_{n\to\infty}\sqrt[n]{x} = \frac1{\lim_{n\to\infty}\sqrt[n]{1/x}} = \frac11 = 1.
$$
A: Use the binomial theorem:
Case I. $x\geqslant1.$ For each $n,$ let $x_n:=\sqrt[n]{x}-1.$ Then we have $x=(1+x_n)^n\geqslant nx_n$ and hence $0\leqslant x_n\leqslant\dfrac{x}{n}$ and so $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}(\sqrt[n]{x}-1)=0$ which means that $\lim\limits_{n\to\infty}\sqrt[n]{x}=1.$
Case II. $0<x<1.$ Let $a:=\dfrac{1-x}{x}.$ Then $1+a>1$ and you can apply case I.
A: There is a theorem that states that
If $x_n>0$ and $\lim_\limits{n \to \infty } \frac{x_{n+1}}{x_n} = L$ then $\lim_\limits{n \to \infty} \sqrt[n]{x_n} 
 = L$. 
For $x_n = x$ we get $\lim_\limits{n \to \infty} \sqrt[n]{x} = \lim_\limits{n \to \infty} \frac{x}{x} = 1.$
The theorem comes from Stolz-Cesaro and here is a proof:
$\lim_\limits{n \to \infty } \frac{x_{n+1}}{x_n} =e^{\lim_\limits{n \to \infty} \ln(\frac{x_{n+1}}{x_n})} = e^{\lim_\limits{n \to \infty} \ln(x_{n+1})-\ln(x_n)} = e^{\lim_\limits{n \to \infty} \frac{\ln(x_{n+1})-\ln(x_n)}{n+1-n} } = e^{\lim_\limits{n \to \infty} \frac{\ln x_n}{x}} = e^{\lim_\limits{n \to \infty} \ln \sqrt[n]{x_n} } = \lim_\limits{n \to \infty} \sqrt[n]{x_n} $.
But this is of course overly complicated. An alternative is $\lim_\limits{n \to \infty} \sqrt[n]{x} = x^{\lim_\limits{n \to \infty} \frac{1}{n} } = x^0 = 1$ 
