$\pm\infty$ versus neutral $\infty$ in extensions (compactifications) of $\mathbb R$ One context in which a distinction between $+\infty$ and $-\infty$ is important is in things like
$$
\lim_{t\,\to\,+\infty} \frac 1 {1+e^t} = 0, \qquad \lim_{t\,\to\,-\infty} \frac 1 {1+e^t} = 1.
$$
However, with rational functions $f(x)$ one can write
\begin{align}
& \lim_{x\,\to\,\infty} f(x) = \ell \\[8pt]
\text{and } & \lim_{x\,\to\,c} f(x) = \infty
\end{align}
and one should make no distinction at all between $+\infty$ and $-\infty$. This makes these functions continuous everywhere including $\infty$ and $c$ (the point where there is a pole).
Similarly
$$
\lim_{\theta\,\to\,\pi/2} \tan \theta = \infty
$$
and there's no $\text{“}{\pm}\text{''}$ involved, so $\tan$ is continuous everywhere in $\mathbb R/\pi$ (the reals modulo $\pi$, a space in which there is no $\infty$ that $\theta$ could approach).
My question is: What amount of agreement or disagreement about the above exists among mathematicians?  (I find this disagreed with in a comment under this question.)
 A: As you may know you can make $\mathbb R$ and $\mathbb C$ compact by using the Alexandroff compactification (i.e. called Riemann sphere is the case $\mathbb C$). On this compactification you can define rational/meromorphic functions. In this construction we have $+\infty = -\infty$. 
Like you noticed above you get $\lim_{t\,\to\,+\infty} \frac 1 {1+e^t} \neq \lim_{t\,\to\,-\infty} \frac 1 {1+e^t}$. So how does this fit in this construction? 
The answer isn't too obvious, but well known in function theory. Let $\hat{\mathbb C} = \mathbb C \cup \{\infty\}$. If a function $f: \hat{\mathbb C} \to \hat{\mathbb C}$ has a pole in $c \in \mathbb C$ it holds that $\lim_{z \to c} f(z) = \infty$ with respect to the chordal metric. But why do we still get $\lim_{t\,\to\,+\infty} \frac 1 {1+e^t} \neq \lim_{t\,\to\,-\infty} \frac 1 {1+e^t}$?
The function $f: \hat{\mathbb C} \to \hat{\mathbb C}, z \mapsto \dfrac 1 {1+e^z}$ has no pole in $\infty$. If you consider the Laurent series of $f$ you will see, that $f$ has an essential singularity in $\infty$. But what do we know about these singularities? The Casorati–Weierstrass theorem tells us that for every $c \in \hat{\mathbb C}$ there exists a sequence $(z_n)_{n\in \mathbb N}$ with $\lim_{n \to \infty} f(z_n) = c$. So you basicly just found sequences for $c = 0,1$. That is nothing to worry about, there is just no limit in $\hat{\mathbb C}$ at $\infty$. 
Hope that makes sense and helps you :)
