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Theorem: If $f: D \rightarrow \mathbb{R}$ a function and $a$ an accumulation point of $D$, then $\lim_{x\rightarrow a} f(x)$ exists and it is unique.

I assume that $\lim_{x\rightarrow a} f(x)$ is not unique when $a$ is not an accumulation point. Can you explain why it is true? Can you show an example where the limit is not unique when $a$ is not an accumulation point?

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    $\begingroup$ A limit requires an accumulation point, otherwise it is not defined. But without conditions on $f$, the theorem is false. $\endgroup$
    – user65203
    Mar 31, 2021 at 13:36
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    $\begingroup$ What's stated as a theorem is clearly not true: There are plenty of functions that have no limits anywhere (e.g., the function that's $0$ at all rational numbers and $1$ at all irrational numbers). Perhaps you meant to say that if $\lim_{x\to a}f(x)$ exists, then it is unique. $\endgroup$ Mar 31, 2021 at 14:18

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The issue is that the limit does not exist if $a$ is not an accumulation point. When a limit exists it is always unique.

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It depends on your definition of a limit. There are three different versions. One says:

$L$ is the limit of $f(x)$ as $x\to a$ if for every $\varepsilon>0$, there is a $\delta>0$ such that $\vert f(x)-L\vert<\varepsilon$ if $\vert x-a\vert<\delta$.

The second definition says the same, except that "and $x\neq a$" is added at the end. And the third and most common definition also says that $a$ has to be an accumulation point of the domain in the first place.

In the first case, if $a$ is not an accumulation point, then the limit always exists, but it may or may not be unique. If $a$ is part of the domain, but not an accumulation point, then the limit is $L=f(a)$, and it's unique. If $a$ is not in the domain and not an accumulation point, then every number is vacuously a limit, since there is a $\delta>0$ such that no $x$ in the domain satisfies $\vert x-a\vert<\delta$, but then "$\vert f(x)-L\vert<\varepsilon$ if $\vert x-a\vert<\delta$" is always true, since the condition is never satisfied in the first place. So then the limit would not be unique.

In the second case, the limit always exists (caveat: if $a$ is not an accumulation point of the domain), but is never unique, using essentially the argument from the first case where $a$ is not in the domain.

In the third case, the limit does not exist by definition, since by definition it requires an accumulation point. The point of this third definition is exactly to exclude the possibility of a non-unique limit as it can occur with the other two definitions.

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    $\begingroup$ I think it's standard to specify $0\lt|x-a|\lt\delta$ in the definition of limit, not just $|x-a|\lt\delta$. $\endgroup$ Mar 31, 2021 at 13:55
  • $\begingroup$ @BarryCipra: Yeah, I thought so, too, after writing this down. Which is why I added that that the third definition is the most common just before your comment. $\endgroup$ Mar 31, 2021 at 13:57

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