Does $f_n(x)=n^2x(1-x^2)^n$ converges uniformly on $[0,1]$
$\lim_{n\to \infty} f_n(x) =f(x)=0$ and then I calculated sup of $|f_n(x)-f(x)|$ which came out to be $\frac{n^2}{\sqrt{2n+1}}\cdot \frac{1}{(1+\frac{1}{2n})^{n}}$ and thus $\lim{n\to \infty}$ of sup does not exists thus $f_n(x)$ is not uniformly convergent , I just wanna ask is there any short method to quickly see its non uniform convergence?
On
Idea: The problem is around $0$. We want $x$ to be small enough so that the exponential term doesn't affect anything, but still large enough so that $n^2x$ tends to infinity. This is possible because the $n^2$ term is very large. If the $n^2$ were replaced by $n^{1/2-\epsilon}$ for any $\epsilon>0$,the function $f_n(x)$ would converge uniformely.
Example: Take $x=\frac{1}{\sqrt{n}}$. Then $$\lim_{n\rightarrow\infty} f_n\left(\frac{1}{\sqrt{n}}\right)=\lim_{n\rightarrow\infty} n^{3/2}\left(1-\frac{1}{n}\right)^n=\infty.$$ Consequently, the convergence to $0$ is not uniform.
If $f_n$ converges to $f$ uniformly then $\int f_n \to \int f$. We can calculate that $\int_0^1 n^2x(1-x^2)^n = \frac{n^2}{2(n+1)}$ which doesn't converge to 0 as n tends to infinity. Hence $f_n$ doesn't converge uniformly.