# About the definition of one sided limits

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

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: 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?

• The definition requires also that $c$ should be a limit point of the domain. Aug 5, 2014 at 9:58
• 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! Oct 14, 2021 at 17:27

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).
• 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) Aug 5, 2014 at 12:26
• Yes. The point $c$ is in the closure of $(c,d)$. Aug 5, 2014 at 16:52