Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 10E / d / 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 $|x\sin (1/x)|\le x\to 0$, $f_2$ is continuous at zero and hence everywhere.

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