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While self learning about the definition of limit from Calculus Early Transcendentals by James Stewart I could understand few portions of it. The statement:

"Let $f$ be a function defined on some open interval that contains the number $a$, except possibly at $a$ itself. Then we can say that the limit of $f(x)$ as $x$ approaches $a$ is $L$"

I could not understand what the author has meant to refer to by "open interval" and "contains the number $a$, except possibly at $a$ itself"

By far I assumed that he meant that the function has to be defined for values as close as the point at which the limit is to be determined but it might not be defined exactly at that point. But I am completely clueless about the "open interval" portion. By open interval, I used to understand an inequality like $a<x<b$ or $(a,b)$, If author actually meant this then does that mean we cannot determine limits for functions which have closed interval as domain or such domain which does not extend from negative infinity to positive infinity?

N.B: I am not a math major student

MSKB
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    It is also worth noting that Stewart is not saying that the function's domain is an open interval. He is only saying that it is defined on (at least) an open interval around $a$. It might also be defined on a larger set. – Xander Henderson Jun 24 '22 at 12:52
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    If the function is defined on the closed interval $[p,q]$ then it is also defined on the open interval $(p,q)$. The author just lists standard minimal assumptions required to define the limit, which is exactly what you think: "the function has to be defined for values as close as the point at which the limit is to be determined but it might not be defined exactly at that point" (end quote). Nothing stops you from having a function defined on all real numbers. – Michal Adamaszek Jun 24 '22 at 12:54
  • ohh I get it... for simplification we are basically taking a portion of the actual domain of the function such that this portion be it closed or interval it is involving all the values closer to point a. Isn't it? – MSKB Jun 24 '22 at 12:58
  • My feeling is that this answer more-or-less addresses this question (the relevant sentence is "An open interval that contains also contains all points sufficiently close to $a$". I am going to close this question as a duplicate---if this does not answer the question, please edit to indicate where confusion remains. – Xander Henderson Jun 24 '22 at 16:55
  • @XanderHenderson As someone who tried to answer this question, I agree with the choices, particularly the second one on modifying the definition. It shows that open intervals aren't exactly "sacrosanct" in any sense, and that there is a sensible, more general way of doing things. – Sarvesh Ravichandran Iyer Jun 25 '22 at 08:54

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