Solution
= Solution
Both <functions> are continuous at every $x>0$. At zero, $f_1(1/(\pi/2+2\pi n))=1$ while $f_1(1/(3\pi/2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin(1/x)|\le x\to0$, $f_2$ is continuous at zero and hence everywhere.
= Solution
Both <functions> are continuous at every $x>0$. At zero, $f_1(1/(\pi/2+2\pi n))=1$ while $f_1(1/(3\pi/2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin(1/x)|\le x\to0$, $f_2$ is continuous at zero and hence everywhere.