I would like you guys to verify my proof for the following exercise:
Let $$f_n:[-1,1] \to \mathbb R; \quad f_n(x) := nx(1-x^2)^n$$ Is $f_n$ pointwise (uniformly) continuous? And in case it is, specify $\lim_{n\to \infty} f_n.$
My solution:
$f_n$ converges pointwise towards the $0$-function, i.e. $f_n \to 0$ pointwise. Proof:
$f_n(1) = f_n(-1) = 0$, so let $x\in ]-1,1[$. For fixed $x$ with $|x| < 1$ we have that $$|f_n(x) - 0| = n|x||1-x^2|^n \leq n |1-x^2|^n \to 0$$ since $nq^n \to 0$ when $|q| < 1$. Therefore $f_n \to 0$ pointwise.
$f_n$ does not uniformly converge. Pick $x_n = 1/n$. We then have $$ \lim_{n\to \infty}f_n(x_n) = \lim_{n\to \infty} n \frac{1}{n} \left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n = e^{-1}e = 1 \neq 0$$ as desired.
EDIT: Changed $[0,1]$ to $[-1,1]$.