Convergence of $f_n(x)=x^n(1-x)^n$

Question

Does $f_n(x)=x^n(1-x)^n$ converges uniformely on $[0,1]$ and why?

If $x=0$ or $x=1$ we have $f_n(x)=0$. If $x\in (0,1)$ we have $x^n\rightarrow 0$ and $(1-x)^n\rightarrow 0$. So $f_n\rightarrow 0$. To check uniform convergence, we have to check that $\forall\varepsilon>0\ \exists N\in\mathbb{N}$ such that $\forall n\geq N, x\in [0,1]: |f_n(x)|<\varepsilon$. How do I show that?

Answer 1

Note that $x^n(1-x)^n$ has its maximum at $x=1/2$.

The convergence at $x=1/2$ Implies the uniform convergence on $[0,1]$

Answer 2

Hint: The maximum of the $f_n$'s in the segment $[0,1]$ is at $x=1/2$.