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?

We can see this directly. If it were uniformly convergent (to $0$), for any $\varepsilon>0$, we could find $N_0$ such that $0<x^n<\varepsilon$ for all $n\ge N_0$ and all $x\in(0,1)$. In particular, $x^{N_0}<\varepsilon$, which is the same as $\,0<x<\varepsilon^{\frac1{N_0}}$ for all $x\in(0,1)$. That is impossible.
