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?

Denote $y=\dfrac{1}{x}$. Since $0<x<1$,then $y>1.$ Hence
$$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{n^2}{y^n}=\lim_{n \to \infty}\frac{2n}{y^n\ln y}=\lim_{n \to \infty}\frac{2}{y^n(\ln y)^2}=0.$$
