For all $x\in\mathbb{R}$, Let's define a set $P_x$ as such:
$$P_x=\{y\in\mathbb{R}\,\mid\,y=x^n \text{ for some } n\in\mathbb{N}\}$$
The question is, for $x=\frac{1}{2}$, does $P_x$ contain $0$? As I understand it, $\lim_{n\to\infty}(\frac{1}{2})^n=0$, but does that mean that zero is equal to $\frac{1}{2}$ raised to some $n$?
On
As far as I know it holds that
$\forall p \in P_{\frac{1}{2}} \exists r \in \mathbb{R}^+ (0 < r < p) $
$$\lim_{n \rightarrow \infty } (\frac{1}{2})^n = 0$$
But every element will be in the interval [$\frac{1}{2}$,0) and never reach 0.
If you think iteratively about it, even if the element of $P_{\frac{1}{2}}$ corresponding to n=m will be very small, the element corresponding to n=m+1 will always be, independent of m, twice as small, but still larger than 0.
The set does not contain zero. One way to think about it is that $\frac{1}{2}^n$ is just the product $ \frac{1}{2}\frac{1}{2}\frac{1}{2}...\frac{1}{2},$ $n$ times. Since a product is only zero if one of its factors is zero, and since $\frac{1}{2} \not= 0$, the product is not zero, no matter how large $n$ is. Thus 0 is not in the set.
I hope that helps!!