Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 19 / a / i / Solution

Since $A>0$ and $\omega >0$ are fixed, the standard sine limit givesIf $p>-1$, 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. HenceThe cancellation comes from the simple zero of $\sin (\omega x)$ at the origin, not rapid oscillation there.