Why is $n^2x(1-x^2)^n$ not uniformly bounded in $I=[0,1]$?

Question

I have proved that $f_n=n^2x(1-x^2)^n$ is not uniformly convergent in $I=[0,1]$.

As $f_n \to 0$ then $\int_0^1f_ndx=\frac{n^2}{2n+2}\to \infty$ but $\int_0^10dx=0$. So $\{f_n\}$ is not uniformly convergent then $\exists \epsilon>0$ s.t $\forall N \in\Bbb N$ $\exists x_N\in I$ s.t $f_n(x_N)\geq \epsilon$ for some $n>N$. Do I get any help from here?

Answer 1

Hint:

$$f_n(\frac{1}{\sqrt{n}}) \to \infty$$

as $n \to \infty.$

Answer 2

When $x=\frac 1{\sqrt n}$ we get $n^{2}x(1-x^{2})^{n} =n^{3/2} (1-\frac 1 n )^{n} \to \infty $ since $(1-\frac 1 n )^{n} \to \frac 1 e$.

Answer 3

Note that$$(\forall n\in\Bbb N):f_n\left(\sqrt{\frac1{2n+1}}\right)=2^n n^2 \sqrt{\frac{1}{2 n+1}} \left(\frac{n}{2 n+1}\right)^n>n^2\sqrt{\frac1{2n+1}}\to\infty.$$So, any $\varepsilon>0$ will do.