Why is it wrong to say that $1/x$ is not continuous at $0$. Because $1/0$ is not defined??
Why is it wrong to say that $1/x$ is continuous everywhere but at $0$
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1@VVV: Again, please do not "sign your posts". Also, at least to me, "Tell me. Please." just sounds wrong. – 2011-11-02
4 Answers
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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0@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). – 2011-11-02
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.
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)
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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1`
Like this`. – 2011-11-02