4

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.)

Community
  • 1
Michael Hardy
  • 1
  • Tangentially related: http://math.stackexchange.com/questions/233855/three-point-compactifications-of-mathbbr $\qquad$ – Michael Hardy Aug 02 '16 at 19:34
  • 3
    Personally, since there are two "canonical" compactifications of $\mathbb{R}$, I prefer to be explicit and call $+\infty, -\infty$ if I intend to use the two-ended and $\infty$ if the one-point. However, in a world where $\subset$ or $\subseteq$ is not agreed upon, I don't expect it to be well-agreed upon in this case either, and I think this question may be very difficult to answer precisely. – Aloizio Macedo Aug 02 '16 at 19:47
  • 1
    Note also that, in this issue, there are several kinds of people who use $\infty$ etc in their notations (mathematicians, engineers, calculus students etc) and many of them have different level of knowledge. For a calculus student, writing $f(\infty)$ may be okay and he isn't even aware of any rigorous step he has to take before doing so. For another calculus student, $f(\infty)$ is a blunder. How do you unify notation used across several fields for which interpretations can vary widely, with no hopes of good understanding? – Aloizio Macedo Aug 02 '16 at 19:49
  • Honestly if my answer isn't what you are looking for, I would ask you too make question clearer. I really don't get what you want to know. – Yaddle Aug 02 '16 at 22:22
  • @Yaddle : I posted the question and I up-voted your answer. $\qquad$ – Michael Hardy Aug 03 '16 at 02:36

1 Answers1

1

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 :)

Michael Hardy
  • 1
Yaddle
  • 5,117