$ f(x) = \frac{1}{x} $, $ g(x) = \sum_{n=1}^{\infty} \left( 2^{n-1} \sin \frac{x}{2^{n-1}} \sin^2 \frac{x}{2^n} \right)$
What is the limit of $ f(x)g(x) $ at $ x\to 0 $?
Can I exchange lim and sigma to reach the following conclusion?
$ \lim_{x \to 0} f(x)g(x) = 0$
I don't think g(x) is uniformly convergent but it seems so at around $ x = 0 $, right?
Write
$$h(x)=f(x)g(x)=\sum_{n=1}^\infty \frac{2^{n-1}}{x} \sin \frac{x}{2^{n-1}} \sin^2 \frac{x}{2^n}$$
We have $|\sin(x)|\le |x|$. Estimating the summand in the definition of $g$ accordingly gives
$$\left|\frac{2^{n-1}}{x}\sin \frac{x}{2^{n-1}} \sin^2 \frac{x}{2^n}\right|\le \frac{1}{2^{2n}}$$
Therefore, the series for $h$ converges uniformly in $|x|<4$ (Weierstrass $M$-test). Note that the same estimate shows that the series for $g$ is uniformly convergent around $0$, answering your last question. In particular, we can interchange limit and sum to get
$$\lim_{x\rightarrow 0} f(x)g(x)=\lim_{x\rightarrow 0} h(x)=0$$