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How is it possible to define $\lim_{x \rightarrow a} f(x)$ if $f(x)$ is undefined at $a$? There is an infinite amount of $x$-values on either side of $x=a$, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?

I am using this definition of a limit [1].

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x-a| < \delta $ we have $|f(x) - A|< \epsilon$. That is, when $x$ is near $a$ (within a distance $\delta$ from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it). In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

[1] David V. Widder. Advanced Calculus. Dover 1989.

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  • $\begingroup$ This question was asked zillion times. $\endgroup$ – Yves Daoust Dec 30 '20 at 17:17
  • $\begingroup$ @YvesDaoust I guess it is a question a new student learning real analysis would always ask. I can see the struggle between intuitive understanding and logical thinking (daily language and object language, in fact). $\endgroup$ – Ziqi Fan Dec 30 '20 at 17:24
  • $\begingroup$ @ZiqiFan: OPs are deemed to do some research among the questions before asking. $\endgroup$ – Yves Daoust Dec 30 '20 at 17:36
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A limit uses the values of $f(x)$ everywhere except at $x=a$ ! Whether $f(a)$ is defined or not and whether $f(a)$ coincides with the limit or not is irrelevant. This is expressed in the definition by $0<|x-a|$.

The goal of a limit is precisely to "guess" what $f(a)$ is should be.

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The limit of $f$ at a point $a$ is defined as \begin{equation} \lim_{x\to a}f\left(x\right) = L \longleftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-a \rvert < \delta \longrightarrow \lvert f\left(x\right) - L \rvert < \epsilon. \end{equation} You can see that the condition $0 < \lvert x-a \rvert < \delta$ does not require the definition of $f$ at $a$. Whenever you would like to tell if $L$ exists, you may first consult the definition. If it is not straigtforward to tell from the definition, you may use the Cauchy theorem of limit.

Your question is based on intuitive understanding of math, which is the type of math you learned at high school. Unfortunately, contemporary math is based on bottom-level logical systems. Using rigorous logical language is the gap between math in high school and math in college.

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  • $\begingroup$ @math I guess not. Otherwise, $f$ has to be defined on $a$, right? $\endgroup$ – Ziqi Fan Dec 30 '20 at 17:17
  • $\begingroup$ @math: not a all ! That would make the limits useless. $\endgroup$ – Yves Daoust Dec 30 '20 at 17:18
  • $\begingroup$ @math: $\delta$ is an upper bound ! $\endgroup$ – Yves Daoust Dec 30 '20 at 17:20
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I present you with the grap of the function $\;f(x)=\cfrac{\sin x}x\;$ :

enter image description here

Clearly the functions isn't defined at $\;x=0\;$ , yet the limit if the function when $\;x\to0\;$ is $\;1\;$ .

In words: the limit $\;\lim\limits_{x\to x_0}f(x)\;$ doesn't depend at all on whether $\;f(x_0)\;$ is defined or what its possible value could be, but only depends on the behavior of the function on an open punctured neighborhood of $\;x_0\;$ , meaning: without the point $\;x_0\;$ itself.

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  • $\begingroup$ Yeah, this is definite a good example. A trend to a point does not require the existence at that point. $\endgroup$ – Ziqi Fan Dec 30 '20 at 17:22
  • $\begingroup$ The open circle could be put anywhere on the y-axis. Why put it at 1, and not, for example, at 1.0000001? $\endgroup$ – Christina Daniel Dec 30 '20 at 17:22
  • $\begingroup$ @ChristinaDaniel As for the definition of the function you could choose any value at all...or even none whatsoever. It doesn't make any difference at all for the limit to exist and be eqial to $\;1\;$ $\endgroup$ – DonAntonio Dec 30 '20 at 17:24

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