# Continuity of $\frac{1}{|x|}$ at $x= 0$

The function $|x|$ is continuous at zero. What can I say about the continuity of $\frac{1}{|x|}$ ? I have two counter arguments for it continuity. Please suggest what is right.

1. The function is not continuous at 0 as it is not defined at 0.

2. The function is continuous at 0 as left hand limit, right hand limit and the value of function at zero, all approach to the same value i.e. infinity.

In general, if a real valued function, $f$, is continuous over an interval $(a,b)$ , what can we say about the continuity of $\frac{1}{f}$ ?

Thanks.

• $\frac{1}{abs(x)}$ isn't a function. For it to be a function, it needs a domain. If an element is not in the domain of the function one cannot talk about continuity in this point. So it is neither continuous in $0$ nor is it not continuous in $0$. That's comparable to saying "an apple is continuous in africa". – Tim B. May 11 '15 at 12:27
• First (possibly) condition for continuity: the function must be defined at the point. – Timbuc May 11 '15 at 12:29
• @LeBtz it is very typical, especially in the context of introductory calculus, to talk about the implicit domain of a function – Omnomnomnom May 11 '15 at 12:30
• Your (1) is right. (2) is wrong because "infinity" is not the value of the function. It's just a way you can think about why the function doesn't have a value. – Ethan Bolker May 11 '15 at 12:31
• The answer that your calculus teacher is probably looking for is that the function is "discontinuous" at $0$. – Omnomnomnom May 11 '15 at 12:32

A function can only be continous in a point that is in its domain. $0$ cannot be in the domain of a function defined by the rule $x\mapsto \frac{1}{|x|}$.

• Can the downvoter explain why he downvoted? Is there something wrong with this answer? – 5xum May 11 '15 at 12:35
• I am not the downvoter, but I would edit your second sentence saying "is not" instead of "cannot" and "the function" instead of "a function". – Andrea Mori May 11 '15 at 12:50
• I think "a" is better here then "the" since there is not "the" function defined by $x\mapsto \frac{1}{|x|}$. – Tim B. May 11 '15 at 12:51
• @AndreaMori I believe it's better the way it's written, because technically, a function is defined by two things: the domain and the rule. Technically, the function $f:(1,2)\to\mathbb R$, defined by the rule $f(x)=1/x$, is not the same function than a function $f:(2,3)\to\mathbb R$ defined by the same rule. So my sentence is that you cannot find a function with the rule $1\mapsto\frac1{|x|}$ and with $0$ in its domain. – 5xum May 11 '15 at 12:55
• @5xum: you have a point there, but then I'd still prefer saying "0 does not belong to any domain for a function defined by the rule $x\mapsto\frac1{\mid x\mid}$" ... just to make clear what your point is. – Andrea Mori May 11 '15 at 13:03

If $f:(a,b)\to \mathbb R$ is continuous and $f(x) \neq 0$ for $x\in (a,b)$, then $\frac{1}{f}$ is continuous.
• More precisely: if $f$ is continuous at $x$ and $f(x)\neq 0$, then $1/f$ is continuous at $x$. – Dirk May 11 '15 at 12:44