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 $\frac{1}{f}$ ?


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    $\begingroup$ $\frac{1}{abs(x)}$ 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". $\endgroup$ – Tim B. May 11 '15 at 12:27
  • $\begingroup$ First (possibly) condition for continuity: the function must be defined at the point. $\endgroup$ – Timbuc May 11 '15 at 12:29
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    $\begingroup$ @LeBtz it is very typical, especially in the context of introductory calculus, to talk about the implicit domain of a function $\endgroup$ – Omnomnomnom May 11 '15 at 12:30
  • $\begingroup$ 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. $\endgroup$ – Ethan Bolker May 11 '15 at 12:31
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    $\begingroup$ The answer that your calculus teacher is probably looking for is that the function is "discontinuous" at $0$. $\endgroup$ – 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 $x\mapsto \frac{1}{|x|}$.

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    $\begingroup$ Can the downvoter explain why he downvoted? Is there something wrong with this answer? $\endgroup$ – 5xum May 11 '15 at 12:35
  • $\begingroup$ 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". $\endgroup$ – Andrea Mori May 11 '15 at 12:50
  • $\begingroup$ I think "a" is better here then "the" since there is not "the" function defined by $x\mapsto \frac{1}{|x|}$. $\endgroup$ – Tim B. May 11 '15 at 12:51
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    $\begingroup$ @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. $\endgroup$ – 5xum May 11 '15 at 12:55
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    $\begingroup$ @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. $\endgroup$ – Andrea Mori May 11 '15 at 13:03

Adressing your second question:

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.

  • $\begingroup$ More precisely: if $f$ is continuous at $x$ and $f(x)\neq 0$, then $1/f$ is continuous at $x$. $\endgroup$ – Dirk May 11 '15 at 12:44
  • $\begingroup$ @Dirk How is that more precise? $\endgroup$ – Tobias Kildetoft May 11 '15 at 12:46
  • $\begingroup$ You only need continuity at a single point and only get that. Your answer is totally, btw… $\endgroup$ – Dirk May 11 '15 at 12:47
  • $\begingroup$ @Dirk I am not sure what you mean by "need" here. $\endgroup$ – Tobias Kildetoft May 11 '15 at 12:52

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