Let $f_n(x) = n^2 x (1-x)^n$. Prove that $ f_n \rightarrow0$ pointwise in [0,1] but does not converges uniformly (using the definition).
What I've done
$ \forall n\in\mathbb{N}$
$f_n(0) = f_n(1)=0$, so we have
$$ \lim_{n\to\infty} n^2x(1-x)^n = \lim_{n\to\infty} n^2xe^{n \ln(1-x)} =0$$
Since $ \ln(1-x)<0$ for $0<x<1$ we have that $f(x)=0$ and so
$ f_n \rightarrow0 $ in [0,1] pontwise.
But I don't know how to apply the deffinition to prove that $f_n(x)$ does not converges uniformly
My deffinition (uniform convergence): $\forall\epsilon>0, \exists N\in\mathbb{N}$ such that, $\forall x\in\mathbb{R}$
$$|f_n(x)-f(x)|<\epsilon$$
Any thoughts?