# 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: 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. – Crostul Aug 5 '14 at 9:58

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) – Ufuk Can Bicici Aug 5 '14 at 12:26
• Yes. The point $c$ is in the closure of $(c,d)$. – Amitai Yuval Aug 5 '14 at 16:52