Why is $f_n(x) = x^n$ not uniformly convergent on $(0, 1)$? Definition of uniform convergence:

For all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(f_n(x), f(x)) < \epsilon$ for all $n > N \in \mathbb{N}$ and all $x \in (0,1)$.

It's easy to see that for all $x$, $f_n(x) \to 0$ on $(0, 1)$ which means that $f_n$ is pointwise convergent, but since $f_n$ converges pointwise to $0$ for all $x$, I don't see any counter-examples that we can take $x$ to be in order for $f_n$ to converge to any other limit besides $0$. I also don't see how we can use the definition in order to prove that $f_n$ is not uniformly convergent. In this case, what should I do?

Note that $\lim_{x\to1} f_n(x) = 1$ for all $n$. This breaks uniform convergence because we can get close enough to $1$ such that $f_N(x) > \frac12$ for any fixed $N$.
