# The limit point in the definition of a limit of function

Let $$E\subseteq \mathbb{R}$$ and $$f:E\to \mathbb{R}$$ be a function. When we define the limit of $$f(x)$$ as $$x\to x_0$$ we require $$x_0$$ to be the limit point of $$E$$.

Why do we require $$x_0$$ to be the limit point of $$E$$?

The definition says: We say that $$A=\lim \limits_{x\to x_0} f(x)$$ iff $$\forall \varepsilon>0$$ $$\exists \delta=\delta(\varepsilon)>0$$ : $$\forall x\in E$$ with $$0<|x-x_0|<\delta$$ $$\Rightarrow$$ $$|f(x)-A|<\varepsilon$$.

My thoughts: If $$x_0$$ is the limit of point of $$E$$ then $$\{x\in E:0<|x-x_0|<\delta\}\neq \varnothing$$ for any $$\delta>0$$. Probably this is one of the reasons but I may be wrong!

But what if $$x_0$$ is not a limit of $$E$$? For example, let's take some function $$f(x)$$ defined on $$\{0\}\cup (1,2)$$, where $$x_0=0$$.

What difficulties do we have?

Would be thankful if someone provide detailed answer with examples.

• $0$ is a limit point of your $E$ – Exodd Jun 17 at 19:43
• "A" limit point, not "the" limit point. – Stefan Jun 17 at 19:46
• "Probably this is one of the reasons": This is exactly the reason. If we take your definition literally, every function would converge to all values simultaneously at a non-limit-point. – Stefan Jun 17 at 19:49
• @ZFR: There is no right or wrong. It is just not useful. Imagine everytime you in a theorem or proof that you need to exclude or discuss the case that $x_0$ not a limit point. – user251257 Jun 17 at 22:00
• Maybe nothing, but at least the term $\lim_{x\rightarrow x_0}f(x)$ is only well-defined if there is a unique limit. – Stefan Jun 17 at 22:01

The def'n is flawed. It implies that if $$x_0$$ is not a limit point of $$E$$ then $$A=\lim_{x\to x_0}f(x)$$ is true for any and all $$A,$$ e.g. $$0=\lim_{x\to x_0}f(x)=1.$$

If $$S$$ is a sentence and $$\forall y\; (T)\;$$ (or, respectively $$\exists y\; (T)\;$$) occurs in $$S$$, where $$(T)$$ is the rest of the sentence, then in the negation of $$S$$ this part will change to $$\exists y (\neg T)\;$$ (respectively $$\forall y\;(\neg T)\;$$).

So the def'n says $$A\ne \lim_{x\to x_0} f(x) \text { iff } \exists e>0\;\forall d>0\; \exists x\, (0<|x-x_0| Now if $$0<|x-x_0| then $$f(x)$$ does not exist unless $$x\in E,$$ so a consequence of the def'n is

$$A\ne \lim_{x\to x_0}f(x)\implies x_0$$ is a limit point of $$E,$$

equivalently,

if $$x_0$$ is not a limit point of $$E$$ then $$A=\lim_{x\to x_0}f(x).$$

The def'n should be amended by inserting "$$x_0$$ is a limit point of $$E$$ and" just after the bold-face "iff".

• In some contexts it is convenient to also allow $f(x_0)=\lim_{x\to x_0} f(x)$ when $x_0$ is an isolated point of $E$. – DanielWainfleet Jun 17 at 21:21

Probably, Heine's definition (that is equivalent to epsilon-delta definition) of the limit of a function may answer your question:

For functions on the real line, one way to define the limit of a function is in terms of the limit of sequences. In this setting: $$\lim _{x\to x_0}f(x)=A$$ if and only if for all sequences $$x_{n}$$ (with $$x_{n} \not = x_0$$, $$\forall n$$) converging to $$x_0$$ the sequence $$f(x_n)$$ converges to $$A$$.

Therefore, if $$x_0$$ would not be a limit point of $$\mathcal{D}(f)$$, then how would such $$x_n$$ exist?

I like to work with the following definition of the limit of a function between metric spaces:

Let $$(X,d_{X})$$ and $$(Y,d_{Y})$$ be metric spaces such that $$f:X\to Y$$ is a function. Consider as well that we are given a set $$E\subseteq X$$ such that $$x_{0}\in X$$ is an adherent point of $$E$$ and $$L\in Y$$. Then we say that $$f$$ converges to $$L$$ as $$x$$ approaches $$x_{0}$$ along $$E$$ iff for every $$\varepsilon > 0$$, there corresponds a $$\delta > 0$$ such that, for every $$x\in E$$, \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),L) <\varepsilon \end{align*}

It is slightly different from your definition, because we demand that $$d_{X}(x,x_{0}) < \delta$$ instead of $$0 < d_{X}(x,x_{0}) < \varepsilon$$. However your definition can be considered a particular case from the definition I have mentioned. Indeed, it suffices to consider $$E\backslash\{x_{0}\}$$ and both of them are the same. The set $$E$$ tells us how we are approaching $$x_{0}$$.

Still, the answer to your question has not been given. More precisely, why do we need to consider $$x_{0}\in X$$ to be an adherent point of $$E$$? In order to answer it, remember that the set of adherent points of $$E$$ equals $$\overline{E} = \text{int}(E)\cup\partial(E)$$. Consequently, if $$x\not\in\overline{E}$$, then $$x\in\text{ext}(E)$$. This means there exists a positive real number $$r > 0$$ such that $$B(x_{0},r)\cap E = \varnothing$$. In other words, for a $$\delta$$ small enough $$(\delta \leq r)$$, there is no $$x\in E$$ such that $$d_{X}(x,x_{0}) < \delta$$, which makes the definition of limit vacuous.

That's why we require $$x_{0}\in X$$ to be an adherent point of $$E$$.