Determining if a sequence of functions is not uniformly continuous.

Question

We let $f_n(x) = x^{n}\sin(nx)$ and let $S=(-1,1)$.
The limit function seems to be $0$ on $S$.
I want to show that $f_n(x)$ is uniformly convergent on any compact subset of $S$ but not on $S$.
One strategy is to find a sequence $x_k$ converging to $1$, such that $|f_n(x_k) - f(x_k)| \geq \epsilon$.
But I am not sure quite sure how to go about it.

Answer 1

You wrote "uniformly continuous ", but I think it should read "uniformly convergent".

If $C$ is a compact subset of $S$, then there is $0 \le q <1$ such that $|x| \le q$ for all $x \in C.$

Then: $|f_n(x)| \le |x|^n \le q^n$ for all $n \in \mathbb N$ and all $x \in C.$

This shows that $(f_n)$ converges uniformly on $C$.

Let $x_n:=1- \frac{1}{n}.$ It is your turn to show that the sequence $(f_n(x_n))$ does not converge to $0$. This shows that $(f_n)$ does not converges uniformly on $S$.