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Find the limit : $$ \lim_{x\to 0}\frac{\sin(\pi/x)}{\sin(\pi/x)}$$

My try:

$$f(x)=\frac{\sin (\pi/x)}{\sin (\pi/x)}:D_f=\mathbb{R}-{1/k;k \in\mathbb{Z}}$$

So :

don't exit limit at to zero

Is it true ?

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    Which definition of limit you take? One insists that there exists $\varepsilon$ such that $(x_0-\varepsilon,x_0)\cup(x_0,x_0+\varepsilon)$ is in the domain of the function, one does not. – 2017-02-10
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    @TehRod It's not a duplicate unless the OP meant to type the denominator you show. – 2017-02-10
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    @yo' Which one *does* demand that? – 2017-02-10
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    @yo' Which definition of limit does **not** demand the function is defined in some (puntured) neighborhood of the point to which $\;x\;$ tends? – 2017-02-10
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    @DonAntonio The one where it's enough that $x_0$ is a limit point of the domain of the function. – 2017-02-10
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    @yo' Fair enough.... and how many limit points do you know that can exist **without** having a complete neighborhood (and we're talking of the usual euclidean topology of the reals here) around them? By definition, if $\;D\;$ is the domain of a function, a point $\;x_o\;$ is a *limit point* of $\;D\;$ if in each neighborhood of $\;x_0\;$ there's a point of $\;D\;$ **different from** $\;x_0\;$ itself... – 2017-02-10
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    @DonAntonio I agree. In fact, even if $f$ were defined at reciprocal integers as anything different from $1$, the limit fails to exist. – 2017-02-10
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    @Dr.MV Indeed so. I think this may just be some confusion in the definition, but it is important to require (and to check, of course) the function's defined in a punctured neighborhood of $\;x_0\;$ . This doesn't happen in this case for all the point of the form $\;x=\frac1n\;,\;\;n\in\Bbb Z\;$ and thus the function isn't defined in *any* neighborhood of zero, and thus the limit cannot exist. – 2017-02-10

1 Answers 1

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The fraction you want the limit of is identically $1$ whenever it is defined. The points other than $0$ at which it isn't defined (reciprocals of the nonzero integers) approach $0$ as a limit. Whether or not the limit of the function at $0$ is $1$ depends on how you have defined that limit.

If the limit definition says "for every $\epsilon > 0$ there is a $\delta$ such that for all $x \ne 0$ within $\delta$ of $0$ $\ldots$" then the answer is "no".

If the limit definition says "for every $\epsilon > 0$ there is a $\delta$ such that for all $x \ne 0$ within $\delta$ of $0$ for which $f(x$) is defined $\ldots$" then the answer is "yes".

I think the first definition is the most likely.

My original answer, wrong as pointed out by @DrMV in comments, nevertheless upvoted 5 times:

The fraction you want the limit of is identically $1$ when $x$ is near $0$. So the limit is $1$. The fact that both the numerator and denominator are not defined when $x=0$ is irrelevant.

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    The fraction is not defined at reciprocal integers. There are "holes" in the domain of $f$. So, it is not identically equal to $1$. – 2017-02-10
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    @Dr.MV I carefully worded my answer to say that the fraction was well defined (and equal to $1$) near 0. The distant holes in the domain are irrelevant in this context. – 2017-02-10
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    "Distant holes?" The holes are at $1/2, 1/3, 1/4, 1/5, \dots$. Do you see an issue? – 2017-02-10
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    @Dr.MV You are right and I was careless. Will edit appropriately. Thank you. – 2017-02-10
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    No worry. And you're welcome. – 2017-02-10
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    Nicely written (+1) – 2017-02-10