If $x>0$, $\,x^{1/n}$ tends to $1$ as $n\to\infty$
Solution 1:
Suppose $x>1$. Then, let $x^{1/n}=1+h_n$.
We will show that $h_n \rightarrow 0$. $x=(1+h_n)^n \geq 1+nh_n$. Then, $0 \leq h_n \leq (x-1)/n \rightarrow 0$. Thus the result follows by Sandwich theorem.
For $0<x<1$, just consider $\dfrac{1}{x}$
Solution 2:
An argument avoiding Bernoulli's inequality: Suppose first that $ x\ge1$. It suffices to prove that $ n^{1/n}\to1$, since $ x <n $ if $ n $ is large enough. For this, just check that $ n <(1+\sqrt {2/n})^n $, by directly expanding the right hand side.
Finally, if $0 <x <1$, apply the above to conclude that $(1/x)^{1/n}\to1$, and the result follows by considering inverses.