Find the limit function $f$ of $\{F_n\}$. $F_n=nx^n(1-x^2) $
My solution:
For $x=\pm1, 0, F_n=0 \implies f(0)=f(1)=f(-1)=0$.
For $|x|>0$. we have $$(1-x^2)\lim_{n\to\infty} nx^n=\infty\text{ , if }x>1$$
and $$(1-x^2)\lim_{n\to\infty} nx^n=-1^n\infty\text{ if }x<-1$$
For $0<x< 1$:we can find $n_0$ such that: \begin{align} &\dfrac{1}{n_0+1}\leq x\leq\dfrac{1}{n_0}\\ &\dfrac{n}{(n_0+1)^n}\leq nx^n\leq\dfrac{n}{(n_0)^n}\\ &\lim_{n\to\infty}\dfrac{n}{(n_0+1)^n}=0\leq \lim_{n\to\infty}nx^n\leq\lim_{n\to\infty}\dfrac{n}{(n_0)^n}=0\\ \end{align} By the squeeze theorem, $\lim_{n\to\infty}nx^n=0$
Hence $\{F_n\}$ converges pointwise to $f=0$ on the interval $[-1,1]$
Is my solution correct?