Why do we call a limit that evaluates to $\displaystyle\pm\infty$ a particular instance of "The Limit Does Not Exist" The $\epsilon$-$\delta$ definition of a limit is stated as:

$\displaystyle \lim_{x \to a}f(x)=L \iff \forall \epsilon \gt 0\  \exists\delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta \rightarrow\lvert f(x) - L \rvert \lt \epsilon \big ] \quad \dagger$

More generally, one would state:
$$\displaystyle \lim_{x \to a}f(x) \text{ exists} \iff \exists L \in \mathbb R \text{ s.t. }  \forall \epsilon \gt 0\  \exists\delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [ 0\lt \lvert x -a \rvert \lt \delta \rightarrow\lvert f(x) - L \rvert \lt \epsilon \big ] $$

The corresponding negation of $\dagger$ is:

$\displaystyle \lim_{x \to a}f(x)\neq L \iff \exists\epsilon \gt 0\  \text{ s.t. }  \forall\delta \gt 0 \ \exists x \in \mathbb R  \text{ s.t. }\big [ 0\lt \lvert x -a \rvert \lt \delta \land \lvert f(x) - L \rvert \geq \epsilon \big ] $

More generally, one would state:
$$\displaystyle \lim_{x \to a}f(x) \text{ does not exist} \iff \forall L \in \mathbb R\   \exists\epsilon \gt 0\  \text{ s.t. }  \forall\delta \gt 0 \ \exists x \in \mathbb R  \text{ s.t. } \big [ 0\lt \lvert x -a \rvert \lt \delta \land \lvert f(x) - L \rvert \geq \epsilon \big ] \quad \dagger \dagger$$

Given the above definitions, consider what one means for a limit evaluating to $\displaystyle \pm \infty$:

$\displaystyle\lim_{x\to a} f(x) = +\infty \iff \forall M \gt 0 \ \exists \delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [0 \lt \lvert x-a \rvert \lt \delta \rightarrow f(x) \gt M \big ]$

or

$\displaystyle\lim_{x\to a} f(x) = -\infty \iff \forall N \lt 0 \ \exists \delta \gt 0 \text{ s.t. } \forall x \in \mathbb R \big [0 \lt \lvert x-a \rvert \lt \delta \rightarrow f(x) \lt N \big ]$

Within the analysis community, when one claims "$\displaystyle\lim_{x\to a} f(x) = \displaystyle  +\infty$" (or $- \infty$), I have seen that it is acceptable to claim "The limit does not exist".
Based on the negated expression described in $\dagger \dagger$, I am having difficulties seeing why this is justified.
Could someone explain what is meant by "the limit does not exist" in the context of infinities? It appears to me that the first order logics do not match.
 A: One answer, of course, is the argument described in the OP—the limit "equaling" infinity satisfies the negation of the definition of the limit existing.
We could just change the definition if we wanted to; but there are lots of reasons we don't want to. Here's one representative one:
It's a theorem, easy to prove from the definition, that: If $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ exist and are equal, then $\lim_{x\to a} (f(x)-g(x))=0$.
That theorem is definitely false in the case $\lim_{x\to a} f(x)=\infty$ and $\lim_{x\to a} g(x)=\infty$.
(The mantra I always teach to my students is: Infinity is not a number. We use the notation $\lim_{x\to a} f(x)=\infty$ to describe a phenomenon, a specific and useful-to-record way that a limit can fail to exist. But the equals sign can fool us into thinking all sorts of false things if we reason by analogy to limits that actually exist. For all intents and purposes, we should pretend that there is a different symbol than $=$ in $\lim_{x\to a} f(x)=\infty$.)
A: Infinity is not a number, so the real number $L$ in the definition of limit cannot “take value $\infty$.” So to verify $\dagger\dagger$, remember that $L$ is a real number.
Assume that $\lim_{x\to a} f(x)=\infty$; that is, as you note, that
$$\forall M\gt 0\exists \delta\gt0 (0\lt|x-a|\lt\delta\implies f(x)\gt M.$$
I claim that this function satisfies the definition of “the limit does not exist”; that is, that it satisfies what you write as $\dagger\dagger$.
(Intuitively: since $f(x)$ is getting “larger and larger, as large as we care to specify”, it can’t possibly be approaching any particular number $L$: it will be much larger than $L$ once we are close enough to $a$... now to formalize it.)
Indeed, let $L$ be any real number. If $L\leq 0$, then let $\delta\gt0$ be such that $0\lt |x-a|\lt\delta$ implies $f(x)\gt 1$. Now let $\epsilon=\frac{1}{2}$. For all $\delta_1\gt 0$, if $0<|x-a|\lt\min(\delta,\delta_1)$ (and such numbers certainly exist), then $f(x)>1$, and therefore, $|f(x)-L|=f(x)-L\geq f(x)\gt 1\gt\epsilon$.
If $L\gt 0$, let $\delta\gt0$ be such that if $0\lt|x-a|\lt\delta$ then $f(x)>L+1$. Again, let $\epsilon=\frac{1}{2}$; for any $\delta_1\gt 0$, if $0\lt|x-a|\lt\min(\delta,\delta_1)$ (and such numbers certainly exist), then
$$|f(x)-L|= f(x)-L \gt 1\gt \epsilon.$$
Thus, for all values of $L$, there exists $\epsilon\gt 0$ such that for all $\delta_1\gt 0$ there is an $x$ such that $0\lt|x-a|\lt\delta_1$, but $|f(x)-L|\gt\epsilon$. Therefore, $\lim_{x\to a}f(x)$ does not exist.
A similar argument holds if $\lim_{x\to\infty}=-\infty$, or just apply the above argument to $g(x)=-f(x)$.
A: $\Bbb{R}$ denotes the set of finite reals, not the set $\Bbb{R} \cup \{-\infty, \infty\}$ of extended reals. If $\lim_{x \to \infty} f(x) = \pm \infty$ under the standard definition that you give, then $f(x)$ has no finite limit as $x \to \infty$, which makes the right-hand side of your definition $\dagger\dagger$ false. I think you are forgetting that the $L$ in $\dagger\dagger$ ranges over $\Bbb{R}$, i.e., over finite reals.
