= Solution
Since $A>0$ and $\omega>0$ are fixed, the standard <sine> limit gives
$$
f(x)=A\omega\,\operatorname{sgn}(x)|x|^{p+1}\frac{\sin(\omega x)}{\omega x}\sim A\omega\,\operatorname{sgn}(x)|x|^{p+1}\qquad(x\to0).
$$
If $p>-1$, this tends to the assigned value zero. At $p=-1$, the right and left limits are $+A\omega$ and $-A\omega$; if $p<-1$, they diverge with opposite signs. Hence
$$
\boxed{f\text{ is continuous at }0\iff p>-1,\quad\text{for every allowed }A,\omega>0.}
$$
The cancellation comes from the simple zero of $\sin(\omega x)$ at the origin, not rapid oscillation there.
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