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If a limit $L$ of a function $f:A\to\mathbb R$ exists at a point $a\in \mathbb R$, where $A\subset\mathbb R$ is a proper subset of the set of real numbers, is there any difference between the statements "the limit exists when $a$ is an accumulation point" and "the limit exists when $f$ is defined on a deleted neighbourhood of $a$"?

My motivation for asking this question is because of an answer to a previous question of mine.

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  • $\begingroup$ No-no-no-no, this question makes no sense as it is: $0$ is an accumulation point for $(0,1]$, the function $\sin \frac 1 x$ is defined on the deleted neighbourhood $(0,1]$ of $0$, yet it does not have a limit in $0$. What exactly do you want to ask, then? $\endgroup$ – Alex M. May 29 '16 at 15:25
  • $\begingroup$ @AlexM. that is a good example of a limit that does not exist, but my question asks you to assume that the limit exists. $\endgroup$ – ahorn May 29 '16 at 20:57
  • $\begingroup$ One more thing: assume $f :(a,b) \to \Bbb R$ (for instance, $f= \frac {\sin \sqrt x} {\sqrt x} : (0, \infty) \to \Bbb R$). A deleted neighbourhood (in $\Bbb R$) of $a$ looks like $(a-\varepsilon, a-\varepsilon)$ - clearly $f$ cannot be defined on the whole of it, not being defined on $(a-\varepsilon, a]$ (in the given example, the argument of the square root cannot go leftwards of $0$). Wouldn't you rather want to say "a deleted neighbourhood in $A$ of $a$", to fix this issue (i.e. you take a deleted neighbourhood in $\Bbb R$ and you intersect it with $A$ insuring that $f$ be defined on this)? $\endgroup$ – Alex M. May 29 '16 at 21:45
  • $\begingroup$ @AlexM. I don't know. I don't know what domain is required in order for l'Hospital's rule to work. I tried. I just haven't encountered the advanced form of l'Hospital's rule, so all I know is the basic version that's taught in the first calculus course at university. (Sorry to other users if this seems obscure.) $\endgroup$ – ahorn May 29 '16 at 22:43
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    $\begingroup$ @ahorn: Good. In this case, you use the usual formulation of l'Hospital's theorem (the one that everybody uses, in fact): $f$ is taken to exist either on some $(a - r, a + r) \setminus \{x_0\}$, or (for one sided limits, including limits toward $\pm \infty$) on some $(a, a + r)$ (the case $(a-r, a)$ is identical), with $r>0$. In all these cases notice that $a$ is an accumulation point for that domain, and that the domain is also a deleted neighbourhood of $a$, on which $f$ exists. Therefore, the two approaches that you mention are synonymous. $\endgroup$ – Alex M. May 30 '16 at 7:15
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You cannot approaches to any $a \in Dom(f)$. $a$ must be a limit point but this not implies that the limit exists. e.g., $f(x)=\frac{1}{x}$, $x \in [-1,1]=A$, $x=0$ is a limit point for $A$ but as $x\to 0$, $f\to \pm\infty$.

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  • $\begingroup$ Assume that the limit exists, as stated in my question. $\endgroup$ – ahorn May 29 '16 at 13:12
  • $\begingroup$ You wrote: ``the limit exists because $a$ is an accumulation point'' this is false statement as i shown. Even we assume the limit exists the two statement are not equivalent. $\endgroup$ – Mohammad W. Alomari May 29 '16 at 13:28
  • $\begingroup$ The first phrase of my question reads, "If a limit $L$ of a function $f:A\to\mathbb R$ exists." Please expand on your claim that, even if we assume the limit exists, then the two statements are not equivalent. $\endgroup$ – ahorn May 29 '16 at 13:33
  • $\begingroup$ Ok there is no difference, you are right. $\endgroup$ – Mohammad W. Alomari May 29 '16 at 13:37

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