 Solution

The bound $|\cos (1/x)|\le 1$ gives $|f(x)-f(0)|\le |x{|}^p$ when $x\ne 0$. If $p>0$, given any $\epsilon >0$, choose $\delta ={\epsilon }^{1/p}$. Then $0<|x|<\delta$ impliesthe inequality also holds at $x=0$. This proves continuity using the epsilon-delta definition of continuity.

For necessity, take the positive sequence $x_n=1/(2\pi n)\to 0$. Then $\cos (1/x_n)=1$, so $f(x_n)=(2\pi n{)}^{-p}$. For $p=0$ this equals $1$; for $p<0$ it tends to $+\infty$. In neither case does it tend to $f(0)=0$, so continuity fails. ConsequentlyThis is an instance of the general continuity and differentiability thresholds for a damped oscillation.