Is $\frac{1}{x}$ on [0,$\infty$] continuous at zero?

Question

Taking the definition of continuity, two of three conditions are met, i.e.

a) We would have to define $f(0)=\infty$, but normally division by zero is not well-defined.

b) The limit $\lim_{x\to 0}f(x)=\infty$ exists.

c) The value of the limit equals the value $f(0)$.

Questions:

1) I that correct?

2) Does anything change if $f(x)=x^2$, or if we take $$\frac{1}{|x|} \text{ or } \frac{1}{x^2}$$ on the domain $[-\infty,\infty]$?

3) As for the participating notions that are needed to come to an answer, are there any common definitions which would lead to another conclusion? Something like "we say that a limit exists if its value is a finite number" etc.

Answer 1

I would argue at least b) is wrong. "$\lim f(a) = \infty$" is just a shorthand people use to mean "the limit is undefined" (at least in my experience).

Answer 2

I don't understand people that say that conditions b) is wrong. Are you telling me that you don't know the difference between a function that admits no limit like $x \sin x$ and one that just grows without limit? (like $x$? )

$\lim_{x \to c} f(x) = + \infty$ means that $$\forall N > 0\ \exists \delta > 0 : |x-c|<\delta \implies f(x) > N$$

So I would argue that conditions b) is met. The point is that the function is not defined at $x=0$, and there is no point saying that in that point it equals $\infty$ as $\infty$ is not a number.

The correct way to think about it is with the limit definition.

Answer 3

If you equip $\mathbb R\cup \{-\infty,\infty\}$ with the right topology (namely such that the $(a,\infty]$ with $a\in\mathbb R$ are open neighbourhoods of $+\infty$), then your eextension of $x\mapsto \frac1x$ to the interval $[0,\infty]$ by letting $f(0)=\infty$ and $f(\infty)=0$ is indeed continuous. The coninuoty at $x=0$ is precisely a rephrasing of the statement that $\lim_{x\to0^+}\frac1x=+\infty$ (i.e., that for any $a\in\mathbb R$ we find $\delta >0$ such that $\frac1x>a$ for all $x$ with $0<x<\delta$).

The same works (with the same argument) with the obvious extensions of $x\mapsto \frac1{|x|}$ or $x\mapsto \frac1{x^2}$.

You could make it work with $x\mapsto \frac1x$ on all of $\mathbb R$ if you used the one-point compactification of $\mathbb R$ instad of the two-point compactification above. In it, neighbourhoods of $\infty$ contain both all large positive and small negative numbers. Visually, you bend $\mathbb R$ to a cicle and close the gap with a single point.