Prove $f_n(x) = n^2 x (1-x)^n$ does not converges uniformly

Question

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?

Answer 1

$f_n(x)$ is given by $n^2x(1-x)^n$ for $x\in J:=[0,1]$. Then $$ \begin{aligned} \|f_n-0\|_\infty &=\max_{x\in J}|f_n(x)| \\ &\ge f_n(1/n) \\ &=n^2\frac 1n\underbrace{\left(1-\frac 1n\right)^n}_{\to 1/e} \\ &\to \infty\ . \end{aligned} $$ So the sequence of functions $(f_n)$ does not converge uniformly (i.e. in the supremum = infinity norm) to the pointwise limit, the only chance of a limit. (It is not even bounded.)