How to prove uniform convergence for $f_n = x^n(1-x)$ using definition

Question

I'm aware this question has been posted multiple times before but the proofs given haven't been given via the definition of uniform convergence but using sups and differentiation, and I've only come across the definition of uniform convergence so far.

Let $f_n:[0,1]→\mathbb{R}$ be a function defined by $f_n(x)=x^n(1−x)$

How would you prove that $f_n = x^n(1-x)$ converges to $0$ uniformly using the definition?

I've done the following so far:

Let $\epsilon>0$. Then $N >$_______ implies:

$|x^n(1-x)-0|=$...

And I have no idea where to go from here. I've been given the hint that I may need a different argument for $x$ 'close' to 1 from other values of $x$ but I'm quite confused about this - is it trying to get at the fact that if $x$ is close to $1$, then $(1-x)$ is close to $0$?

Thank you in advance!

Answer 1

Given $1> \epsilon >0$, on the interval $(1-\epsilon, 1]$, we have $$ x^n(1-x) \leq 1^n (1-x) < \epsilon \ .$$

On the other hand there exists an $N$ such that for all $n>N$ $$ (1-\epsilon)^n < \epsilon ,$$ This is easy to see because $\epsilon$ is fixed and $(1-\epsilon)^n \to 0$ as $n \to \infty$. Now, for any $n>N$ and for any $x \in [0,1-\epsilon],$ $$ x^n(1-x) \leq (1-\epsilon)^n \cdot1 < \epsilon \ .$$

So, we succeeded in finding for every $\epsilon$, and $N$ such that if $n>N$, then $|f_n(x)-0| < \epsilon$ no matter where $x$ is in our domain. This is by definition equal to $f_n \to 0$ uniformly.

Answer 2

Hint

The local maximum of the function happens at $x={n\over n+1}$ which is $$\max_{[0,1]} f(x)=\left({n\over n+1}\right)^n\cdot {1\over n+1}$$and use $$\lim_{n\to \infty}\left({n\over n+1}\right)^n={1\over e}$$