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?
It might be easier to use the Ratio Test:
$$
\begin{align}
\lim_{n\to\infty}\frac{f_{n+1}(x)}{f_n(x)}
&amp;=\lim_{n\to\infty}\frac{(n+1)^2}{n^2}(1-x)\\
&amp;=1-x
\end{align}
$$
For $0\lt x\lt1$, we have $0\lt1-x\lt1$. Therefore, the series
$$
\sum_{n=0}^\infty f_n(x)
$$
converges. Thus, the terms go to zero. That is,
$$
\lim_{n\to\infty}f_n(x)=0
$$
Differentiate to find the max.
If it goes to zero,
that's the limit;
otherwise the limit 
depends on $x$.

If
$f_n(x)=n^2x(1-x)^n$,
then

$\begin{array}\\
f_n'(x)
&amp;=n^2(x(1-x)^n)'\\
&amp;=n^2((1-x)^n-nx(1-x)^{n-1})\\
&amp;=n^2(1-x)^{n-1}((1-x)-nx)\\
&amp;=n^2(1-x)^{n-1}(1-(n+1)x)\\
&amp;=0
\qquad\text{for } x=1/(n+1)
\end{array}
$

At $x=1/(n+1)$,

$\begin{array}\\
f_n(x)
&amp;=n^2(1/(n+1))(1-1/(n+1))^n\\
&amp;=n^2(1/(n+1))(n/(n+1))^n\\
&amp;=n\dfrac1{(1+1/n)(1+1/n)^n}\\
&amp;\approx n\dfrac1{e}\\
&amp;\to \infty\\
\end{array}
$

So
$\lim_{n \to \infty} f_n(0)$
does not exist
since
$f_n(0) = 0$.

If
$f_n(x)
=x(1-x)^n$
then
$\lim_{n \to \infty} f_n(x) = 0$
for $0 \le x \le 1$.