Standard terminology for infinite limits with opposite sign on the two sides? Consider the following limits:
$$ \lim_{x\rightarrow0}\frac{1}{x^2}$$
$$ \lim_{x\rightarrow0}\frac{1}{x}$$
As far as I can tell, most authors say as a matter of terminology that these limits don't exist and are not defined, since they don't have real-number values. For the first one, we can write
$$ \lim_{x\rightarrow0}\frac{1}{x^2}=\infty,$$
whereas I think most authors would say that it was false that
$$ \lim_{x\rightarrow0}\frac{1}{x}=\infty,\qquad \text{(FALSE)}$$
since the one-sided limits have opposite signs. What is the most standard way to verbally describe the second situation in contradistinction to the first? It's true that the limit doesn't exist and is not defined, but how do you communicate, without being too cumbersome, the fact that it's even more naughty and bad than the first one? I'm hoping there's some verbalism that's not as clumsy as "a limit that is infinite and has the same sign from both sides" versus "a limit that is infinite and has opposite signs from the two sides."
Several comments and answers have claimed that $\lim 1/x^2$ is described verbally as being defined or existing, while $\lim 1/x$ is described as being undefined or nonexistent. As far as I can tell from the paper and online sources I have handy, this is not standard. Standard terminology seems to be that both of these limits are considered nonexistent and undefined.
 A: I would simply say that the function has opposite-signed infinite unilateral limits at the point, or the function has infinite unilateral limits of opposite signs at the point, where "unilateral" can be replaced with "one-sided" if you wish. Note that in such a case the absolute value of the function has an infinite bilateral (i.e. two-sided) limit at the point. However, it is possible for the absolute value of a function to have an infinite bilateral limit at a point without having either unilateral limit exist (finitely or infinitely), as for example is the case with the function
$$  f(x)=\begin{cases} q & \text{if} & x = \frac{p}{q} \; \text{and} \;\;  q \;\; \text{is even} \\ -q & \text{if} & x = \frac{p}{q} \; \text{and} \;\;  q \;\; \text{is odd} \\ \frac{1}{x} & \text{if} & x \notin \mathbb Q  \\ 0 & \text{if} & x=0 \end{cases} $$
at $x=0,$ where $p$ and $q$ are relatively prime nonzero integers and $q > 0.$
A: We use $+$ and $-$ as subscripts. As in $$\lim_{x \to 0^+} \frac{1}{x} = + \infty \qquad \lim_{x \to 0^-} \frac{1}{x} = - \infty$$
A shorthand for non-existence is the symbol $\nexists$, so $\nexists \lim_{x \to 0} \dfrac{1}{x}$.
