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I'm confused with a fundamental concept of calculus about continuity. I'm studying the concept of continuity and discontinuity of a function. The case is the following: "It makes no sense to talk about whether or not a function is continuous at a point that does not belong to its domain. For example, the function $f(x) = 1 / x$ is continuous at all points in its domain. At zero, since it is not in the domain, we cannot talk about whether it is continuous or whether it is not." Suppose now that we are asked to study the continuity of the function $f (x) = \ln (2x^2 + 4x -6)$ at $x = -3$. Clearly the domain of this function is the set $(- \infty, -3) \cup (1, \infty)$, so, based on the above, since $-3$ does not belong to the domain of this function, I could say that it does not you can decide whether the function is continuous or not continuous at $x = -3$. I'm right? Or should I say that the function has an asymptotic discontinuity at this point? why?

As another example, suppose we want to study the continuity of the function $f (x) = \tan x$ at $x = \pi / 2$. Obviously this point is not in the domain of this function, therefore, I do not know if the function has an asymptotic discontinuity at $x = \pi / 2$ or I just can't decide if the function is continuous or not continuous at this point because it is not in its definition domain.

I really need an explanation on this concept. Any good reply is appreciated.

Jack
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  • You seem to be confusing limits with continuity. Intuitively, a continuous function has no breaks. If the function is not defined at a value, then that value breaks the continuity of the function. – John Douma Sep 22 '21 at 20:27
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    You have your answer in the part of your question that says "it makes no sense ...". The definition of continuity at a point requires that the function have a value at that point. – Ethan Bolker Sep 22 '21 at 20:27
  • What do you mean by "asymptotic discontinuity"? – Miguel Sep 22 '21 at 20:28
  • @Miguel Asymptotic discontinuities, from my usage of the term, refer to discontinuities due to the function tailing off to $\pm\infty$ as it approaches the point. – Rushabh Mehta Sep 22 '21 at 20:28
  • @JohnDouma So in my example, can I say that the functions $f (x) = \ln (2x ^2 + 4x -6)$ and $g (x) = \tan x$ are discontinuous at the points $x = -3$ and $x = \pi / 2$ respectively? – Jack Sep 22 '21 at 20:36
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    There are posts on this site with conflicting answers. This one says that any point where a function is not continuous (including points not in the domain) are points of "discontinuity" https://math.stackexchange.com/a/1580539/693577 But this post suggests that some authors would consider the words "continuous" and "discontinous" not applicable to a function at points outside of its domain https://math.stackexchange.com/a/1087642/693577 – Joe Sep 22 '21 at 20:39
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    This question has apparently been asked many times on this site. Here's another saying that "discontinuous" only applies to points in the domain https://math.stackexchange.com/questions/2626069/do-discontinuities-only-exist-strictly-inside-the-domain-of-a-function AND ANOTHER https://math.stackexchange.com/questions/1431796/if-a-function-is-undefined-at-a-point-is-it-also-discontinuous-at-that-point AND one with answers suggesting that it's a common abuse of notation https://math.stackexchange.com/questions/2546258/can-a-function-be-discontinuous-outside-its-domain-of-definition – Joe Sep 22 '21 at 21:02
  • The domain of definition is $(-\infty,-3) \cup (+1,\infty)$ (note the plus) and you are correct, since $-3$ is not in the domain of definition it makes no sense to talk about continuity there. – copper.hat Sep 23 '21 at 00:27
  • @copper.hat Thank you very much for the observation – Jack Sep 23 '21 at 00:41

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A function $f(x)$ is continuous at a given point $a$ if the value of the function and the limit of the function at $x=a$ exist and they are the same. That means $f(a) $ and $$ \lim _{x\to a} f(x)$$ exist and $$\lim _{x\to a} f(x)=f(a)$$ Otherwise, the function is discontinuous at that point. Thus a point $x=a$ which is not in the domain of the function $f(x)$ is a point of discontinuity due to the fact that $f(a)$ does not exist.

Some authors define different types of discontinuity such as jump discontinuity or asymptotic discontinuity and so forth.

Mohammad Riazi-Kermani
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  • So you're allowing continuous functions to have discontinuities (outside their domain)? That 's a bit weird, isn't it? See discussion here: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow. – Hans Lundmark Sep 23 '21 at 04:44
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    @HansLundmark, it is an interesting distinction some seem to make. For example, if we define $f:\mathbb{R} \to \mathbb{R}$ by $$f(x)=\int_0^x \frac{1}{t^2+1}dt$$And a student asks if $f$ is discontinuous at $i$, I can imagine a professor saying "Our function is only defined as an integral over an interval on the real line, so it doesn't even make sense to talk about it anywhere else" And if another student asked, "Is it discontinuous at the center of the Sun", someone would say, "What? That question makes no sense. The domain is $\mathbb{R}$, not points in our solar system." – Joe Sep 23 '21 at 12:03