Limit of discontinuous function

Question

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.

Answer 1

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$$\bigl(\forall x\in\operatorname{Dom}(f)\bigr):0<|x-2|<\delta\implies\bigl|f(x)-b\bigr|<\varepsilon.$$