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I'm puzzled by quote from "Calculus and Analytic Geometry" by Stein and Barcellos that I found in this answer: If a function is undefined at a point, is it also discontinuous at that point?

Unfortunately I don't have access to said book to clarify, so I ask here.

According to this definition any polynomial is continuous. So is each of the basic trigonometric functions, including $y = \tan x$... You may be tempted to say, 'But $\tan x$ blows up at $π/2$ and I have to lift my pencil off the paper to draw the graph.' However, $π/2$ is not in the domain of the tangent function... If $a$ is not in the domain of $f$, we do not define either continuity or discontinuity there.

I don't see any logical foundation for exclusion of $π/2$. The only thing that comes to my mind is "The text implies that a function can't have number "c" in its domain if f(c) is undefined. For example, division by zero is undefined, so if there is such "c" that f(c) leads to division by zero, then "c" can't belong to the domain of said function." But it's hard to believe that such interpretation is true because it has absurd consequences:

Namely, considering that functions are discontinious only over their domains (i.e. it's meaningless to talk about discontiniuty of a function outside its domain), it would mean that all functions are continious. After all, how is it possible to have discontiniuty if we exclude x's each time they "blow up" the function?? The concept of discontinuity would just stop making any sense.

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    $\begingroup$ To answer your last question a function can be discontinuous without "blowing up". Take for example $$f(x)=\begin{cases} 1 \text{ if } x \geq 0 \\ 0 \text{ if } x<0 \end{cases}$$. Then the function is well defined but discontinuous at $x=0$. $\endgroup$ – Delta-u Oct 11 '18 at 16:09
  • $\begingroup$ There is a difference between being continuous and being continuous over a domain. We commonly restrict functions to where they are continuous because it is easier to work with them where they are continuous over their entire domain. For example, if we restrict the function $\tan x$ over the domain $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$, this is a well-understood function. It is not the entire tangent function, but a restriction to a location where tangent is continuos. $\endgroup$ – InterstellarProbe Oct 11 '18 at 16:10
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    $\begingroup$ @InterstellarProbe "There is a difference between being continuous and being continuous over a domain". As far as I'm aware it doesn't make sense to even talk about continiuty outside domain of a function. In your example we don't partly use the domain of tan(x), we create whole new domain (although narrower one). So this is entire tangent function, despite narrower domain. $\endgroup$ – user161005 Oct 11 '18 at 16:31
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    $\begingroup$ @user161005 You are correct :-), it is incorrect to say that $1/x$ is discontinuous at $x=0$. In fact $x \mapsto 1/x$ is continuous on its domain of definition $\mathbb R \backslash \{0\}$ and undefined at $x=0$. $\endgroup$ – Delta-u Oct 11 '18 at 16:41
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    $\begingroup$ Related: math.stackexchange.com/questions/1087623/… $\endgroup$ – Hans Lundmark Oct 11 '18 at 18:01

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