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Why is it wrong to say that $1/x$ is not continuous at $0$. Because $1/0$ is not defined??

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It is also true that the singularity at $x=0$ is not a removable singularity - i.e. there is no value which can be defined for $x=0$ (where the function is undefined) which would make the function continuous. –  Mark Bennet Nov 1 '11 at 22:35
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Doesn't sound wrong (as in false) to me, either the version in the title or the version in the post. –  André Nicolas Nov 1 '11 at 22:41
    
Its from a book an anecdote that there are school teachers who say that $1/x$ is discontinous at $x=0$, and this is wrong. And that the correct formulation is $f(x)=1/x$ is continuous $\forall x \in \mathbb{R} \backslash \{0\} $ –  VVV Nov 1 '11 at 22:44
    
@AndréNicolas: I guess "continuous everywhere but at $0$" implyies that $0$ is in the domain of the function. I took it as VVV said in the last comment. –  Ross Millikan Nov 1 '11 at 23:43
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@VVV: Again, please do not "sign your posts". Also, at least to me, "Tell me. Please." just sounds wrong. –  Arturo Magidin Nov 2 '11 at 3:37
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4 Answers 4

up vote 6 down vote accepted

I note that in one of the comments, VVV writes:

Its from a book an anecdote that there are school teachers who say that $1/x$ is discontinuous at $x=0$, and this is wrong. And that the correct formulation is $f(x)=1/x$ is continuous $\forall x\in\mathbb{R}\setminus\{0\}$.

It really depends on the precise definition of "discontinuous".

Sometimes there is a subtle difference between discontinuous and "not continuous". I've seen some sources where "discontinuous" means "defined, but not continuous"; whereas "not continuous" means "either not defined, or defined but not continuous". If this is the case, then "$\frac{1}{x}$ is discontinuous at $0$" would be false, but it would not be the same as "$\frac{1}{x}$ is not continuous at $0$". That is, under those conventions,

$f(x)$ is discontinuous at $a$.

is not synonymous with

$f(x)$ is not continuous at $a$.

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Right: this is related to the comments I made to The Chaz's answer. –  Pete L. Clark Nov 2 '11 at 3:45
    
@Pete: Quite so. What I wanted to highlight is that VVV used one turn of phrase in the comments, and the other in the question, as if they were identical. It's entirely possible the two phrases are not synonymous (and I've certainly seen sources where they are not synonymous). –  Arturo Magidin Nov 2 '11 at 4:24
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Probably more egregiously, is that you can't hope to extend the function $f(x)=\frac{1}{x}$ on $\mathbb{R}-\{0\}$ to a (right, left) continuous function on $\mathbb{R}$ because neither of the limits $\displaystyle \lim_{x\to0^\pm}f(x)$ exist.

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I guess by continuous you mean 'continuous function'.

Then to define a function you need to specify a domain and codomain. But $f:\mathbb{R} \to \mathbb{R}: f(x)=1/x$ is not a function, since it is not well defined at $x=0$.

If you specify, for example $f:(0,\infty)\to\mathbb{R}: f(x)=1/x$, then this is well-defined (and continuous)

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It is impossible to extend $f(x) = 1/x$ to a continuous function on all of $\mathbb{R}$ because $\lim_{x \to 0^-} f(x) = -\infty$ and $\lim_{x \to 0^+} f(x) = +\infty$


(Leftover from before)

$x = 0$ is not in the domain of the function, so it is not clear what it would mean for $f$ to be discontinuous there.

Also, the left- and right-sided limits are $- \infty$ and $+ \infty$, respectively, and this would poses even more problems even if you had your function defined on (say) all of $\mathbb{R}$.

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Now, if we consider $f(x) = \frac{\sin x}{x}$, we still have that $x = 0$ is outside our domain, but (springing off of Mark Bennet's comment) this is a removable singularity. In other words, if we (which we can) define $f(0) = 1$, then we will have a continuous function everywhere (including at $x = 0$). –  The Chaz 2.0 Nov 1 '11 at 22:38
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This is incorrect. You can't talk about a real-valued function $f:S\subseteq\mathbf{R}\rightarrow\mathbf{R}$ being discontinuous at a point outside of $S$. It would be correct to say that the function $f:\mathbf{R}\rightarrow\mathbf{R}$ defined by $f(x)=1/x$ for $x\neq 0$ and $f(0)=2$ is discontinuous at zero. –  Keenan Kidwell Nov 1 '11 at 23:25
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I agree with Keenan. There is no such thing as a "maximal domain" for a function, so are we supposed to say that every function $f: X \rightarrow Y$ between topological spaces is discontinuous at all the points at which it is not defined? E.g. that every function I have ever seen written down is discontinuous at me (i.e., at the "point" Pete L. Clark)? This seems like an unnecessarily weird position to take... –  Pete L. Clark Nov 2 '11 at 1:09
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@Chaz: What I'm saying is that the idea that a function is "discontinuous" at all points which are not in its domain looks weird when you move beyond functions having as their domain some subset of the real numbers. It's still just terminology: the really mathematical issue here is the following: suppose you have a topological space $X$, a subspace $Y$ and a continuous function $f: Y \rightarrow Z$. Now let $x \in X$. The question is whether $f$ can be continuously extended to $Y \cup \{x\}$. (This is only interesting if $x \in \overline{Y}$.) ... –  Pete L. Clark Nov 2 '11 at 2:26
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<s>Like this</s>. –  Pete L. Clark Nov 2 '11 at 12:57
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