Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 10E (Analysis I)

The statement ${\lim }_{t\to x}f(t)=\ell$ means that for every $\epsilon >0$ there is $\delta >0$ such that $t\in [0,1]$ and $0<|t-x|<\delta$ imply $|f(t)-\ell |<\epsilon$.

The function is continuous at $x$ exactly when ${\lim }_{t\to x}f(t)=f(x)$.

If $x_n\to x$, continuity of $f$ gives $f(x_n)\to f(x)$, and continuity of $g$ at $f(x)$ then gives $g(f(x_n))\to g(f(x))$. The sequential criterion proves continuity of $g\circ f$ at $x$.

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.

Let $S=\{0\}\cup \{1/n:n\ge 1\}$ and let $D$ be the Dirichlet function, equal to $1$ on rationals and $0$ on irrationals. Then $f(x)=d(x,S)D(x)$ is continuous exactly on $S$: the distance factor squeezes it to zero on $S$, while away from $S$ it is a positive continuous factor times an everywhere-discontinuous function.

Using the preceding $f$, define $g(0)=f(0)+1$ and $g(x)=f(x)$ for $x>0$. Changing the value only at zero makes zero discontinuous while preserving continuity at every $1/n$ and discontinuity everywhere else. Thus the continuity set is precisely $\{1/n:n\ge 1\}$.