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In the wikipedia page about one sided limits at: http://en.wikipedia.org/wiki/One-sided_limit the rigorous definition of the right sided limit is given as:

$$\forall \epsilon >0, \exists \delta>0, \forall x\in I (0<x-a<\delta)\implies |f(x)-L|<\epsilon$$

It is similar for the left sided one. In the above definition $I$ is defined as an interval within the domain of $f$.

What disturbs me about this definition is that it can allow a weird thing as the following:

enter image description here

In this sketch the function $f$ has the interval $[a,b]$ as its domain. The function $f$ is constant with $f(x)=y'$. Now I think that the above definition of the right sided limit allows the evaluation of the limit $\lim_{x \to c^{-}}f(x) = y'$. Here, for each $\epsilon > 0$, we have for all $x \in [a,b)$ and for $\delta = b - c$ that it is $0 < x - c < \delta \implies |f(x) - y'| < \epsilon $. But this seems illogical as $c$ is not even contained in $f$'s domain. I know that $f$ can be undefined at $c$ in the case of the regular, double sided limit $\lim_{x \to c} f(x)$ but it needs to be contained in the domain of $f$. Since no such containment condition has been given for the definition of the one sided limit, it allows such a weird limit to be evaluated.

Am I wrong with my assessment here? Or is it natural that one sided limits allow such situations?

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    $\begingroup$ The definition requires also that $c$ should be a limit point of the domain. $\endgroup$
    – Crostul
    Aug 5, 2014 at 9:58
  • $\begingroup$ Yes, without the specification/requirement that $c$ be an accumulation point of the domain, the given statement would be vacuously true so that every member of the codomain would be the required limit! $\endgroup$
    – ryang
    Oct 14, 2021 at 17:27

1 Answer 1

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Your observation is correct, but this is just inaccuracy in the definition. The one sided limit at a point $a$ can only exist if $a$ is in the closure of the domain of $f$, i.e. if for any $\delta>0$ there is a point $b$ in the domain of $f$ with $0<b-a<\delta$ (or $0<a-b<\delta$, when the limit is from the other side).

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  • $\begingroup$ Does this mean that for example, in an interval $(c,d)$ when the point $a$ is inside of the interval with $c < a < d$ or when $c = a$, then the point $a$ is in the closure of the interval $(c,d)$ since we can always give a $b \in (c,d)$ for each $\delta > 0$ such that $0 < b - a < \delta$. (Thinking about the right side) $\endgroup$ Aug 5, 2014 at 12:26
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    $\begingroup$ Yes. The point $c$ is in the closure of $(c,d)$. $\endgroup$ Aug 5, 2014 at 16:52

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