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

The function $f$ is continuous at $x_0$ ifThe choice of $\delta$ may depend on $\epsilon$ and $x_0$, but must work for every $x$ in that neighbourhood. This is the epsilon-delta definition of continuity.

The function $f$ is differentiable at $x_0$ if there exists a finite real number $L$ such thatThen $f^{\prime }(x_0)=L$. The puncture $0<|x-x_0|$ makes the difference quotient defined; requiring a finite real $L$ distinguishes an ordinary derivative from an infinite slope.

Along the first sequence, $\sin (1/x_n)=\sin (2n\pi +\pi /6)=1/2$, whereas along the second sequence, $\sin (1/x_n^{\prime })=\sin (2n\pi )=0$. The outer sine therefore givesBoth sequences approach zero through positive values. For arbitrary $a$, the same calculation gives $f_a(x_n)=x_n^a$ and $f_a(x_n^{\prime })=0$.

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.

For positive integers $n$, the summand is $a_n=\sin (\pi \sin (1/n))/n$. Since $0<\sin (1/n)<1$, these terms are positive. Apply the limit comparison test with the convergent p-series $\sum n^{-2}$:Each sine ratio tends to one. The limit comparison test says that two positive-term series whose term ratio tends to a finite positive number are either both convergent or both divergent. ThereforeThere is also a direct comparison test for series: $0\le \sin (\pi \sin (1/n))\le \pi \sin (1/n)\le \pi /n$, giving $0\le a_n\le \pi /n^2$.