Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 4 / b

The limit of a function $f:E\to \mathbb{R}$ at a limit point $p$ of $E$ is $\ell$ whenThe function converges as $x\to p$ if such a finite $\ell$ exists. The puncture excludes any value assigned at $p$.

For $p\in E$, continuity at $p$ meansWhen $p$ is also a limit point, this says that the limit of a function is $f(p)$. At an isolated point of $E$, every function is continuous relative to $E$, even though the punctured limit does not have a unique forced value.

Take $y=f(p)$ with $p\in \overline{E}$. Given $\epsilon >0$, continuity gives $\delta >0$ such that $|x-p|<\delta$ implies $|f(x)-f(p)|<\epsilon$. Since $p\in \overline{E}$, choose $x\in E$ with $|x-p|<\delta$. Thus every neighbourhood of $y$ meets $f(E)$, provingEquality need not hold. Let $f(x)=e^x$ and $E=\mathbb{R}$. The exponential function is continuous, butThe additional point zero comes from arguments escaping to $-\infty$, not from a finite point of $\overline{E}$.