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The question is about $ \lim_{x\to0}\frac{\sin\frac{\pi}{x}}{\sin\frac{\pi}{x}}$ It is obvious that for all $x$ in the domain the function value is $1$. In DESMOS it shows a horizontal line $y=1$. But...when you look closer around $x=0$, the number of holes in the graph also increases. And when you look at an infinite small neighborhood around $x=0$, there are infinitely many holes in the graph. That's a lot of discontinuities. So with that observation, does the limit then exist? I would like to hear your input on this. Of course, through a piece wise function one can "remove" these discontinuities, but if you do that in an infinite close neighborhood around $0$ one can wonder, "what's left"? Thanks for your input.

Bernard
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imranfat
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    I'd say $0$ is a cluster point for the domain of the function. And for any point in the domain, as you say, the function is equal to $1$. Therefore the limit does exist and it is equal to $1$. I'm also wondering what others in the community will say about this. – dfnu Feb 28 '20 at 22:19

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There are basically two views depending on what definition of limit you use. The first view is that to talk about the limit as $x\to 0$, the function has to be defined in a sufficiently small neighborhood around $x=0$. As you said, your function has "holes" around $x=0$ no matter how closely you approach. So according to this definition, the limit is undefined.

The second view, on the other hand, only requires that $0$ be a limit point for the domain $D$ (true for your example) and adds the requirement $x\in D$ to the definition of limit (see this one, for instance). So according to this definition, the limit is trivially $1$ since $f(x)=1$ for all $x\in D$. You usually find the first definition in introductory calculus textbooks, and the second one in analysis textbooks such as Rudin's.

bjorn93
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