Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 19 / c

For $a>0$, the fact that sine is a bounded function gives $|f_a(x)-f_a(0)|\le |x{|}^a$. Given $\epsilon >0$, choose $\delta ={\epsilon }^{1/a}$. If $|x|<\delta$, then $|f_a(x)-f_a(0)|<\epsilon$. This proves continuity directly from its definition.

For $a\le 0$, use the first sequence from part (b), for which $f_a(x_n)=x_n^a$. When $a=0$ these values are one; when $a<0$ they tend to $+\infty$. In either case, for $\epsilon =1/2$ every neighbourhood of zero contains a sufficiently late $x_n$ with $|f_a(x_n)-f_a(0)|\ge 1/2$. This contradicts the definition of continuity. Hence

The difference quotient at zero isFor $a>1$, its absolute value is at most $|h{|}^{a-1}$. Given $\epsilon >0$, set $\delta ={\epsilon }^{1/(a-1)}$. Then $0<|h|<\delta$ makes the difference quotient have absolute value below $\epsilon$. Thus $f_a^{\prime }(0)=0$.

Conversely, along the second sequence the difference quotient is exactly zero. Any finite derivative would consequently have to be zero. Along the first positive sequence, however, the difference quotient equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\epsilon =1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite derivative exists. We obtainIn particular, $0<a\le 1$ gives continuity without differentiability.