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Consider the definition of a limit of function. Suppose that $E\subseteq \mathbb{R}$ and the function $f:E\to \mathbb{R}$ and $x_0$ is the limit point of $E$.

Definition: 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 $0<|x-x_0|<\delta$ $\Rightarrow$ $|f(x)-A|<\varepsilon$.

We note that in this definition $x_0$ may not be an element of $E$.

But I have a question: Why do we care that $0<|x-x_0|<\delta$? I guess that even if I take $|x-x_0|<\delta$ it should be OK (in both cases when $x_0\in E$ or $x_0\notin E$) because I am restricting the inequality $|x-x_0|<\delta$ over all $x\in E$.

Can anyone answer am I right or not? Would be very grateful for detailed answer!

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Actually when you define the limit in $x_0$ you want to exclude $x_0$. This is the reason of $|x-x_0|>0$. If you don't do so, then you would need continuity in $x_0$ for the limit to be defined (if the function is defined in $x_0$) which is not convenient in general.

Indeed, consider for instance $f(x) = 0$ for $x\neq 0$ and $f(0) = 1$. Then with your definition of limit the limit for $x\to 0$ of $f(x)$ would not exist.

Why with your definition the limit is not defined?

Let's consider the usual definition of limit. For any $x\neq 0$ you have $f(x) = 0$. In particular you can say that $\lim_{x\to 0}f(x) = 0$ since for $|x|\leq \delta$ (whatever $\delta>0$) $|f(x)|=|0|=0<\epsilon$ (whatever $\epsilon>0$) if $x\neq 0$.

Now your definition requires that this is true even if $x=0$. But this is not the case since $|f(0)|=|1|>\epsilon$ if $\epsilon<1$.

So with your definition the function $f(x)$ does not admit a limit.

Now why should we want it to have a limit?

The point is that the concept of limit wants to measures the behaviour of a function in the neighbourhood of a point, as a separate thing compared to the value of the function in the point itself. Otherwise the definition would work for continuous functions only, so that for each point in which the function is defined it would not really make sense to give such a complicated definition, since the limit would just be the value of the function in that point.

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  • $\begingroup$ 1) I am understanding your first paragraph separately but cannot get the point. Could you clarify it a bit please? 2) You said that when we define the limit in $x_0$ we need to exclude $x_0$. Why do we need to exclude it? What's the motivation? $\endgroup$ – ZFR Jun 17 at 17:49
  • $\begingroup$ Right now it is much better. At least i get the point of the limit of function. So it turns out that we care about the behaviour of $f(x)$ when points $x$ are very close to $x_0$ but $\neq x_0$. Right? $\endgroup$ – ZFR Jun 17 at 18:19
  • $\begingroup$ This is only relevant if $x_0 \in E$. $\endgroup$ – Paul Frost Jun 18 at 10:37
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If you allow $x = x_0$ (which is be possible if $x_0 \in E$ and $\lvert x - x_0 \rvert < \delta$), then you get $\lim_{x \to x_0} f(x) = f(x_0)$. This means that $f$ is continuous in $x_0 \in E$. This a much stronger requirement than the existence of $\lim_{x \to x_0} f(x)$.

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The problem is not when $x_0\notin E$ but rather when $x_0\in E$. Consider the function $f(x)=0$ for $x\in\mathbb{R}\backslash\{0\}$ and $f(0) = 0$. Then you want that $\lim_{x\to 0}f(x) = 1$, so you need to exclude $x=0$ from the elements on which you "test the limit."

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  • $\begingroup$ Probably you mean "then you want that $\lim \limits_{x\to 0}f(x)=0$, right? $\endgroup$ – ZFR Jun 17 at 17:52
  • $\begingroup$ @ZFR Absolutely, corrected the typo. $\endgroup$ – Daniel Robert-Nicoud Jun 17 at 20:25

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