Find the set S on which $\{F_n\}$ converges pointwise, and find the limit function $F_n(x)=nx^n (1-x^2)$ 
Find the set S on which $\{F_n\}$ converges pointwise, and find the limit function $$F_n(x)=nx^n (1-x^2)$$

Running some graph with different $n$ values $F_n$ seems to converge pointwise on $S=[-1,1]$ where the limit function points to zero. if correct. How do you justify this formally? 

The limit function will be $$F(x) = \lim\limits_{n \rightarrow \infty} nx^n (1-x^2) $$
its derivative
$$F'_n(x)=n^2x^{n-1}\left(1-x^2\right)-2nx^{n+1}=-nx^{n-1}\left(\left(n+2\right)x^2-n\right)$$
3 critical points:
$$x_1=0, x_2=\sqrt{\dfrac{n}{n+2}}, x_3=-\sqrt{\dfrac{n}{n+2}}$$ where $F_{x_2}$ and $F_{x_3}$ are two maximums and $F_{x_1}$ is a local minimum. 
As $n \rightarrow \infty$, the critical points: $ x_2 \rightarrow 1$, and $x_3 \rightarrow -1$ . 
It follows that when $n$ is large, $\lim\limits_{n \rightarrow \infty}F_{x_2}=\lim\limits_{n \rightarrow \infty}F_{x_3}=\lim\limits_{n \rightarrow \infty}F_{x_1} = 0$
Therefore, $\{F_n\}$ converges pointwise on $[-1,1]$ and the limit function on $S$ is  $F(x) =0$ 

.
Is this argumentation appropriate for my conclusion? Is there a more efficient or different way to proceed?
Much appreciated 
 A: You are not required to find whether the convergence is uniform so
it is unnecessary to find maxima/minima.
There are three cases:
(i) $|x|<1$. Then $nx^n\to 0$ and so $nx^n(1-x^2)\to0$.
(ii) $|x|=1$. Then $f_n(x)=0\to0$.
(iii) $|x|>1$. Then $1-x^2\ne0$. Also $(nx^n)$ diverges, so $(nx^n(1-x^2))$
also diverges.
A: Your argument is incorrect.  For one thing, you have not given any reason why $F_n(x)$ cannot converge if $x\not\in[-1,1]$.  For another, your calculations of critical points simply do not prove that $F_n(x)$ actually converges to $0$ for any $x\in (-1,1)$.  If you knew that the values of $F_n$ at the critical points $x_2$ and $x_3$ converged to $0$, then it would be possible to bound $F_n(x)$ between $F_n(0)=0$ and $F_n(x_2)$ and $F_n(x_3)$ to make such an argument, but you have not said anything about the values of $F_n$ at these critical points.  In any case, this argument would require more detailed justification than what you have said, since you need to explain exactly how you are bounding $F_n(x)$.
I would suggest that instead of thinking about $F_n$ as a function, you just fix a single value of $x$ and think about the single number $F_n(x)$.  For instance, take $x=1/2$: you want to prove that the sequence of numbers $F_n(x)=n\cdot\frac{1}{2^n}\cdot\frac{3}{4}$ approaches $0$ as $n\to\infty$.  Can you prove that?  Can you generalize the argument to show $F_n(x)=nx^n(1-x^2)$ approaches $0$ for any fixed $x\in (-1,1)$?
For the case $x\not\in [-1,1]$, you might similarly start with an example.  Why does the sequence of numbers $F_n(2)=n\cdot 2^n \cdot (-3)$ diverge as $n\to\infty$?  Can you generalize to show that $F_n(x)$ diverges for any fixed $x\not\in [-1,1]$?
