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At $-a$, the prefactor $(x-a)/x$ tends to $2$, so the nonvanishing amplitude leaves the oscillatory discontinuity intact. At zero, use $\sin (\tan x)=x+O(x^3)$ to obtainThe limit is $-a$, whereas the assigned value is zero. This is a removable discontinuity, not a continuous crossing.

At $a$, write $x=a+h$. The estimateproves continuity at the assigned zero. But its difference quotient isThe sequences from part (i) make this approach $1/a$ and $-1/a$, so the derivative does not exist. This illustrates the distinct continuity and differentiability thresholds for a damped oscillation.

Therefore the discontinuities are $x=-\pi /2$ and $x=0$; the nondifferentiability points are $x=-\pi /2,0,\pi /2$. The function is smooth elsewhere.

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.