Find $\lim_{n \to \infty}f_n(x)$ Find $$\lim_{n \to \infty}f_n(x)$$ where $$f_n(x)=n^2x(1-x)^n$$ $0 \lt x \lt 1$
My try:
By symmetry $$\lim_{n \to \infty}f_n(x)=\lim_{n \to \infty}f_n(1-x)=\lim_{n \to \infty}n^2(1-x)x^n=(1-x)\lim_{n \to \infty}n^2 x^n$$
Now $$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{x^n}{\frac{1}{n^2}}$$
Now can we use L'hopital's rule here?

An option:
For $0< x <1$,  show that 
$\lim_{n \rightarrow \infty} n^2x^n =0.$
Set $x=e^{-y} , y>0$, and consider $\dfrac{n^2}{e^{ny}}$.
$e^{ny} =$
$ 1+ ny +(ny)^2/2! + (ny)^3/3! +.. \gt (ny)^3/3!$.
Hence :
$\dfrac{n^2}{e^{ny}} \lt \dfrac{(3!)n^2}{n^3y^3}= (\dfrac{3!}{y^3})(\dfrac{1}{n}).$
The limit $n \rightarrow \infty$ is?
