Let $f_n (x) = n^2 x (1-x^2)^n$ on $[0,1]$. Explain why convergence is not uniform on $[0,1]$

Question

I'm not sure what to do here. I tried $x = \frac{1}{\sqrt n}$ and then $f_n = n^\frac{3}2 (1-\frac{1}n)^n$ which looks eerily close to $n^\frac{3}2 e$, which obviously goes to infinity and therefore would not converge uniformly. Also its a 2 parter- are there intervals such that $f_n$ does uniformly converge? I was thinking it converges for any interval $[0,a]$ where $0 < a < 1$. Since $$|f_n - 0| = |f_n| = |n^2 x (1-x^2)^n| \le |n^2 a(1-a^2)^n|$$ and for any $\epsilon > 0$ we can find an $N \in N$ s.t. $|n^2 x (1-x^2)^n| < \epsilon$ And this would satisfy the definition of uniform convergence right?

Answer 1

For each fixed $x$, $f_n(x) \to 0$. You have already proved that the convergence is not uniform: $f_n(\frac 1 {\sqrt n}) \to \infty$. [ Use the fact that $(1-\frac 1 n)^{n} >\frac 1 {2e}$ for $n$ sufficiently large]. This also proves that convergence is not uniform on $[0,a]$ for any $a>0$. However, $f_n \to 0$ uniformly on $[a,1]$ for any $a >0$ [because $n^{2}(1-a^{2})^{n} \to 0$].