# Does extreme value theorem hold for neighborhoods near $\pm \infty$

Given a continuous function $f: \mathbb R \to \mathbb R$ can we state that $f$ satisfies the conditions of the extreme value theorem at an interval $A=[x_0, x_0 + c], c \in \mathbb R$ as $x_0 \to + \infty$ (or minus infinity)?

Can we state that $f(A) = [m,M]$ where $m,M$ are the extreme values of $f$ at $A$?

There are cases when $m,M$ are e.g. infinitely large but since $\pm \infty$ is a limit point that seems to generate a contradiction. Or must $f$ be defined as $f: \mathbb R \cup \{ \pm \infty \} \to \mathbb R \cup \{ \pm \infty \}$?

With a continujoius extension to the extended reals, you are fine. However, without actually allowing the points o finfinity, even $f(x)=\frac1x$ does not attain its minimum on $[1,\infty)$. Of course it does for every $[1,1+c]$, but that is nothing special
As far as there is no limiting condition , $f(A)=[m,M]$ by extreme value theorem itself . Where do you want to use $x_0 \rightarrow \infty$? It is not clear from your question.