prove: $\lim_{n \rightarrow \infty} x^{1/n} = 1$ I have to prove that $\lim_{n \rightarrow \infty}$ $x^{1/n} = 1$ for $x > 0$.
I splitted it up in 3 cases:
$x = 1:$
$1^{1/n} = 1$ $\forall$ $n$, so $\lim_{n \rightarrow \infty}$ $x^{1/n} = 1$ if $x =1$
$x > 1:$
I already proved that $(x^{1/n})_{n = 1}^{\infty}$ is decreasing if $x > 1$, so I only have to prove that it's infimum is 1. How do I prove that? 
$x < 1:$
I don't know how the proof works in this case.
Could you please help me completing the second and third case?
Thanks in advance!
 A: I don't think whether this would be a standard proof,
but when $n\to\infty,1/n\to0$ so $x^{1/n}\to x^{0}=1\forall x>0$
A: For $x>1$ Bernouilli's inequality yields $$\left(1+\frac{(x-1)}{n}\right)^n\ge1+n\frac{(x-1)}{n}=x,$$ so $$ \left(1+\frac{(x-1)}{n}\right)\ge x^{\frac1n}\ge1.$$
Now let $n$ tend to infinity.
A: The case where $a=1$ is trivial.
We know that $\lim_{n\to\infty}\frac{a}{q^n}=0$ when $q>1$
Let $a>1$ and let $q=1+\varepsilon$
Thus, for some $N$, $1<a<(1+\varepsilon)^n$ for all $n>N$
Then $1<a^{\frac{1}{n}}<1+\varepsilon$ for all $n>N$
Thus, by the definition of the limit $\lim_{n\to\infty}a^\frac{1}{n}=1$
Now, if $a<1$, then letting $b=\frac{1}{a}$
$\lim_{n\to+\infty} a^\frac{1}{n}=\frac{1}{\lim_{n\to\infty}b^\frac{1}{n}}=1$
QED.

As for the proof of $\lim_{n\to\infty}\frac{a}{q^n}=0$, I shall write it here.
Let $n>log_q(\frac{a}{\varepsilon})$
Then $q^n>\frac{a}{\varepsilon}$
And thus, $\frac{a}{q^n}<\varepsilon$
Also, if you can't use logarithms, this proof is evident from the principle(or axiom) of Archimedes
A: $$\log(x^{1/n})=\frac{\log(x)}n\to0$$
and as the exponential is a continuous function,
$$\lim_{n\to\infty}e^{\log(x)/n}=e^{\lim_{n\to\infty}\log(x)/n}.$$
