Let's take $f(x)=a^x+b$ where $a\in\mathbb R$ and $b\in\mathbb R^+$. clearly $L=\lim_{x\to\infty}f(x)>0$, but is the interval written $L\in(0,\infty)$ because limits only approach infinity or is $L\in(0,\infty]$ because infinity is one of the values $L$ can take? I can't find an example of this one way or another.
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You mean starting from $0$ (including it) to $\infty$ ? – Spectre Aug 26 '20 at 05:08
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@Spectre, I didn't really pay attention to the $0$ side, I'm more worried about whether $\infty$ is able to be expressed as explicitly a part of the interval without having to write it elsewhere. – Jacob Claassen Aug 26 '20 at 05:15
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Look at @BrianMoehring's answer. – Spectre Aug 26 '20 at 05:16
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What he said is true. – Spectre Aug 26 '20 at 05:17
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If you want to be able to write $L = \lim_{x\to\infty} f(x) = \infty$, then it would be false to state $L \in (0,\infty)$.
You would need to be able to write $L \in (0,\infty]$, but we also need to keep in mind that $L = \infty$ isn't a real number, so this only makes sense in the context of the extended real number line.
Brian Moehring
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Technically the answer is neither.
When we say $x \in (0, \infty)$, that notation is only valid when $x$ is a number.
When we say $\lim_{x\to\infty} = \infty$, that does not mean the limit is the number $\infty$. It means $\forall N> 0: \exists M> 0:$ if $x > M,$ then $f(x) > N$.
However you could extend the definition of the bracket $[a,b]$ notation to allow for such limits that do not take any value.