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?
 A: 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.
A: 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 $N$.
A: Hint. You have $$\sup_{x\in(0,1)}f_n(x)=1.$$
A: Hint If $x_n =1-\frac{1}{n}$ then $f_n(x_n) \to \frac{1}{e}$. 
Use this to contradict $d(f_n(x_n), f(x_n))<\epsilon$.
A: What exponent is needed for $(.9)^n$ to be less than $.1$? Answer: 
$$n > \ln(.1) / \ln(.9)
$$
What exponent is needed for $(.99)^n$ to be less than $.1$? Answer: 
$$n > \ln(.1) / \ln(.99)
$$
What exponent is needed for $(.999)^n$ to be less than $.1$? Answer: 
$$n > \ln(.1) / \ln(.999)
$$
What is the limit of these exponents $n$ as the number of $9$'s increases to $+\infty$? Answer: 
$$\lim \, n = \lim_{x \to 1^-} \ln(.1) / \ln(x) = +\infty
$$
Therefore, the sequence of functions $f_n(x) = x^n$ does not converge uniformly to zero on the interval $(0,1)$.
A: Let $0 < \epsilon < 1$. Put $$x=\frac{1+\epsilon}{2}$$
This gives $\epsilon < x < 1$.
Then, $\forall n$,
$$\begin{aligned}
x^n = \bigl(\frac{1+\epsilon}2 \bigr)^n &=\frac{1}{2^n}(1+\epsilon)^n \\
&= \frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\epsilon^k > \frac{1}{2^n} \epsilon \sum_{k=0}^n \binom{n}{k} =\frac{1}{2^n}  2^n \epsilon=\epsilon ~.\end{aligned}$$
Hence, there is no number $N$ such that $x^N < \epsilon$, when $0 < \epsilon < 1$.
So, uniform convergence fails.
