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?
 A: It might be easier to use the Ratio Test:
$$
\begin{align}
\lim_{n\to\infty}\frac{f_{n+1}(x)}{f_n(x)}
&=\lim_{n\to\infty}\frac{(n+1)^2}{n^2}(1-x)\\
&=1-x
\end{align}
$$
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,
$$
\lim_{n\to\infty}f_n(x)=0
$$
A: Hint: Use L'hopital rule to find $\lim_{n \to \infty} f_n(x)$, where
$$f_n(x) = x  \frac{n^2}{(1-x)^{-n}}.$$
A: 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?
A: Denote $y=\dfrac{1}{x}$. Since $0<x<1$,then $y>1.$ Hence
$$\lim_{n \to \infty}n^2 x^n=\lim_{n \to \infty}\frac{n^2}{y^n}=\lim_{n \to \infty}\frac{2n}{y^n\ln y}=\lim_{n \to \infty}\frac{2}{y^n(\ln y)^2}=0.$$
A: 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)
&=n^2(x(1-x)^n)'\\
&=n^2((1-x)^n-nx(1-x)^{n-1})\\
&=n^2(1-x)^{n-1}((1-x)-nx)\\
&=n^2(1-x)^{n-1}(1-(n+1)x)\\
&=0
\qquad\text{for } x=1/(n+1)
\end{array}
$
At $x=1/(n+1)$,
$\begin{array}\\
f_n(x)
&=n^2(1/(n+1))(1-1/(n+1))^n\\
&=n^2(1/(n+1))(n/(n+1))^n\\
&=n\dfrac1{(1+1/n)(1+1/n)^n}\\
&\approx n\dfrac1{e}\\
&\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 $0 \le x \le 1$.
