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What does this mathematical sentence mean?

"Let $f(x)$be a function defined on an interval that contains $x=a$, except possibly at $x=a$."

How can a function contain $x=a$ "except possibly at $x=a$"? That is so illogical to me.

This is an excerpt out of the definition for $\lim$.

I know there are questions here that have discussed this already, however I think my question is a bit different as it focuses more on the definition as to what it says in relation to limits. My main question, again, is:

How can a function contain $x=a$ "except possibly at $x=a$"?

Blue
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Math Noob
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    Example: $f(x) = \sin(x)/x$ is defined on the interval $I=(-1, 1)$, except at $x=0$. – Martin R Jun 14 '20 at 13:46
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    "that contains $x=a$" describes the interval. To restate: "Consider an interval that contains $x=a$. Let $f(x)$ be defined on that interval except possibly at $x=a$." Does that help? – Blue Jun 14 '20 at 13:48
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    The interval contains $a$. $f(x)$ is defined for all $x$ in the interval, except possibly if $x=a$. No reason to write that the interval contains $x=a$. – Jens Renders Jun 14 '20 at 13:53
  • So when we say „that contains $x = a$“. Lets take a = 1 for example. Now lets say we have an interval $]-1,1[$. According to the definition that would mean that the interval contains $a = 1$, except at $a = 1$.Did I understand correctly? – Math Noob Jun 14 '20 at 14:01
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    I think the statement is meaningful without any ambiguity. If you want formalism then $f$ is a function of the form $f:I\setminus {a} \to\mathbb {R} $ where $I$ is an open interval with the property $a\in I$. Sometimes one uses the phrase "$f$ is a real valued function defined in a certain deleted neighborhood of $a$". The statement in your question also allows for $f:I\to\mathbb {R} $ where $a\in I$. – Paramanand Singh Jun 14 '20 at 14:02
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    Considering your example of $a=1$ we need to choose interval like $(0,2)$ or $(0,1.1)$ since we need $1$ as a point in this interval. And we can consider $f:(0, 2)\to\mathbb {R} $ as well as functions like $f:(0, 2)\setminus{1}\to\mathbb {R} $. – Paramanand Singh Jun 14 '20 at 14:07
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    @MathNoob: The interval under consideration should definitely contain $x=a$, but the function may-or-may-not be defined there. The idea is that we don't care what happens with the function at $x=a$; we're only concerned about what happens with it around $x=a$. This is a concept I try to convey with the image in this answer. – Blue Jun 14 '20 at 14:09
  • Yup, I now see the mistake in my example. Thanks for taking the time guys you really helped me out here. – Math Noob Jun 14 '20 at 14:12
  • This whole statement is poorly written throughout. "Let $f(x)$ be a function..." Is already wrong. I would write this statement as simply "Consider a function $$f: I \backslash {a} \rightarrow \mathbb{R}$$ given $a \in I$". – K.defaoite Jun 14 '20 at 14:24
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    @K.defaoite But $f$ is only possibly not defined at $a$, so it could also be $I\to\mathbb R$. – Oscar Cunningham Jun 14 '20 at 20:25
  • Ok, fair enough. – K.defaoite Jun 14 '20 at 21:06
  • @OscarCunningham We might want to apply the definition of limit to find $\lim_{x\to 1} f(x)$ where $f: \mathbb R \setminus{0}\to \mathbb R$ such that $f(x) = 1/x.$ In that case the domain of $f$ is neither an interval containing $1$ nor such an interval with $1$ deleted. But $f$ is defined on the interval $(0,\infty),$ which contains $1,$ as long as we do not interpret "is defined on $I$" to mean "is defined only on $I.$" – David K Jun 14 '20 at 22:58

2 Answers2

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It means the same as "let $I$ be an interval containing $a$, and let $f$ be a function to $\mathbb R$ from either $I$ or $I\smallsetminus\{a\}$". Nothing specifies whether $I$ has to be open, closed, or finite, so it should allow any of these possibilities. In particular $I$ could be a closed interval with $a$ as an endpoint.

Oscar Cunningham
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It means $f(x)$ is defined everywhere in the interval, except that it may or it may not be defined at $x=a.$

Allawonder
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