Proof of $x^n(1-x)$ uniformly converges to $0$ on $[0,1]$

Question

I attempted to prove $x^n(1-x)$ converges uniformly to $0$ on $[0,1]$ but I'm not sure if my proof is right.

My proof is:

Let $\epsilon >0$ be given. Then we can find $N\in \mathbb N$ such that $N > 2/\epsilon$. For all $x \in [0,1)$ if $n\ge N$ we have $$|f_n(x)-0| = |x^n(1-x)| \le |x|^n(1+|x|) \le 2x^n\le2x^N\le2N<\epsilon$$ and if $x=1$, for all $n\ge N, \quad |f_n(x)-0| = |0-0| = 0\le\epsilon$. So $f_n \rightarrow 0$ uniformly on $[0,1]$.

Is my proof correct? Thank you!

Answer 1

Hint: $|f_n(x) - 0| = x^n(1-x)$ attains a maximum at $x^* = n/(n+1)$ where $$f_n(x^*) = \frac{1}{(n+1/n)^n(n+1)}$$

Answer 2

Your proof has a mistake. $2x^N < \varepsilon$ is not given since $\lim_{x\to1}x^N = 1$ for any $N\in\mathbb{N}$.