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This question is from my calculus 1 worksheet about limits. I am given the function $$f:[0,1]\cup\{2\}\to\mathbb{R}:x\mapsto\begin{cases} \arctan(x), &x\in[0,1] \\ 8, & x=2 \end{cases}$$ I am supposed to use the $\epsilon$-$\delta$-definition of a limit to show that the limit of $f$ when $x\to2$ is equal to any real number $b$. In other words $$\lim_{x\to 2}f(x)=b$$ $$b\in\mathbb{R}$$ I know that the $\epsilon$-$\delta$-definition of that limit is $$\forall\epsilon>0,\exists\delta>0,\forall x\in \operatorname{dom}(f):0<\vert x-2\vert<\delta \Rightarrow \vert f(x)-b \vert<\epsilon$$ And I am inclined to say that the limit of $f$ as $x\to2$ does not exist. Still, I am very confused by this question. My question is how to prove that the limit of $f$ as $x\to2$ is equal to any real number.

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    $\begingroup$ Recall that any universally quantified statement over an empty domain is trivially true. $\endgroup$ Oct 22, 2021 at 18:57
  • $\begingroup$ It depends on your definition, but it is vacuous. If you take $\delta = 1$ then the set of $x$ in the domain that satisfy $0<|x-2|< \delta$ is empty, so the conclusion is (vacuously) true. $\endgroup$
    – copper.hat
    Oct 22, 2021 at 18:58

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Take any $\varepsilon>0$ and take $\delta=1$. Then there is no element $x\in\operatorname{Dom}(f)$ such that $0<|x-2|<\delta$, and therefore is indeed true (actually, vacuously true) that$$\bigl(\forall x\in\operatorname{Dom}(f)\bigr):0<|x-2|<\delta\implies\bigl|f(x)-b\bigr|<\varepsilon.$$

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