Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 19 / a

Continuity at $x_0$ requires that $f(x_0)$ be defined and that ${\lim }_{x\to x_0}f(x)=f(x_0)$. In the epsilon-delta definition, for each $\epsilon >0$ there must be a $\delta >0$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ within the domain. Differentiability requires the finite two-sided limitIn particular, merely having a punctured limit of a function does not imply continuity unless the assigned value agrees with that limit.

For every $n\ge 1$, $|x^n\sin (1/x)|\le |x{|}^n\to 0$, so the squeeze theorem gives limit zero. For $n=0$, sequences $x_k=(\pi /2+2k\pi {)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2k\pi {)}^{-1}$ give function values $1$ and $-1$, so no limit exists. HenceThe value assigned at zero when $n=4$ does not affect this punctured limit.

For $n=1,2,3$ the limit zero equals the assigned value at zero. For $n=4$ the limit is zero but the assigned value is one, giving a removable discontinuity. For $n=0$ there is no limit. Thus continuity at zero holds precisely for

For $n=1,2,3$, the difference quotient at zero is $x^{n-1}\sin (1/x)$. It oscillates without a limit for $n=1$, while the squeeze theorem gives limit zero for $n=2,3$. The other two cases fail continuity and therefore cannot be differentiable. Hence