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Does
$$\lim_{x \to 0}\;\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$$ exist?

I believe the limit should be $1$. Because function being defined at the point is not a condition for limit to exist.

This question came in my test and the answer given is limit does not exist.

But if we see the graph , it is quite clear the function is exact 1 as $x \to 0$, so the limit should be 0.

Even wolfram alpha gives the limit to be 1.

But we are playing with infinity, so who knows? Maybe I am missing out on something?

So what exactly is the limit and why?

Edit:

Wolfram alpha's widget (the link to which I have posted above) says the limit is 1.

But here wolfram alpha says that the limit doesn't exist on the real line.

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    $\begingroup$ @AvnishKabaj: See this answer of mine. $\endgroup$ – Blue Aug 20 '18 at 17:52
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    $\begingroup$ This is kind of a stupid question for an exam (as opposed to an OP or a math question), because the function the professor presumably wants is the constant function $X \to \mathbb{R}$ for $X = \mathbb{R}\setminus \pi \mathbb{Z} \cup \{0\}$. It's silly to write the latter as $\sin (1/x)/\sin (1/x)$ if that's really what's intended; the latter has a continuous extension across all of $\mathbb{R}$. As an exam question, it would make more sense to look at the function $f(x) = 1$ for $x\not\in \pi \mathbb{Z}$ and $f(x) = 0$ for $x\in \pi \mathbb{Z}$ $\endgroup$ – anomaly Aug 20 '18 at 18:03
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    $\begingroup$ What a surprisingly good question. $\endgroup$ – Randall Aug 20 '18 at 18:08
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    $\begingroup$ Cool question, so I upvoted, no matter how tall it is. $\endgroup$ – Randall Aug 20 '18 at 18:20
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    $\begingroup$ Wolframalpha claims that the limit does not exist: Wolframalpha. It says there are infinitely many singularities in every neighborhood about $x=0$. It also shows the limit to be 1 (if you restrict to a subset of the domain). So, it gives both answers $\endgroup$ – InterstellarProbe Aug 20 '18 at 18:32
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I quote Walter Rudin's Principles of Mathematical Analysis for the definition of the limit of a function:

Let $X$ and $Y$ be metric spaces; suppose $E\subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $\lim_{x\to p}f(x)=q$ if there is a point $q\in Y$ with the following property: For any $\epsilon>0$, there exists a $\delta>0$ such that $d_Y(f(x),q)<\epsilon$ for all points $x\in E$ such that $0<d_X(x,p)<\delta$.

The symbols $d_X, d_Y$ refer to the distances in $X$ and $Y$, respectively.

In our case, $X=Y=\mathbb R$ with the metric $d(x,y)=|x-y|$. The function $f(x)=\frac{\sin \frac1x}{\sin \frac1x}$ maps the set $$ E=\mathbb R\setminus (\{\tfrac1{k\pi}:k\in \mathbb Z\setminus \{0\}\}\cup \{0\}) $$ into $\mathbb R$, and $0$ is a limit point of this set. We would conclude $\lim_{x\to 0}f(x)=1$ if for all $\epsilon>0$, we could find a $\delta>0$ so whenever $x\in E$ and $0<|x|<\delta$, then $|f(x)-1|<\epsilon$. But any $\delta$ suffices, since $f(x)=1$ for all $x\in E$.

Therefore, we do conclude that $\lim_{x\to 0}f(x)=1$.

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  • $\begingroup$ This is the clearest answer which settles the sticky part. $\endgroup$ – Randall Aug 20 '18 at 18:22
  • $\begingroup$ @mfl I think exact precision is appropriate, as the devil in the details for this problem. $\endgroup$ – Mike Earnest Aug 20 '18 at 18:37
  • $\begingroup$ How would you explain this answer to a high school student? $\endgroup$ – Archer Aug 20 '18 at 19:32
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    $\begingroup$ @Abcd I would explain it just the way gimusi did. Whenever $x$ is close to $0$, and $f(x)=\frac{\sin \frac1x}{\sin \frac1x}$ is defined, it follows $f(x)$ is close to $1$. The error in $f(x)$ can be made arbitrarily small by making the error in $x$ small enough (indeed, the error in $f(x)$ is always zero in this case!). This means $\lim_{x\to 0}f(x)=1.$ If you are not comfortable with the exact definition of a limit as presented in my post, you will not be able to get a deeper understanding than that. $\endgroup$ – Mike Earnest Aug 20 '18 at 19:39
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    $\begingroup$ @Abcd You ask how one might explain this to a high school student. The first thing that I stated in my answer to this question is that you have to start with a definition. How has this high school student defined a limit? Once you have defined a limit, it is possible to discuss how to analyze limits like the one you provide. $\endgroup$ – Xander Henderson Aug 21 '18 at 11:54
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In mathematics, it is very important to start with a good definition. In Rudin's Principles of Mathematical Analysis, the following definition is given:

Let $X$ and $Y$ be metric spaces; suppose $E\subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $f(x) \to q$ as $x\to p$, or $$ \lim_{x\to p} f(x) = q $$ if there is a point $q\in Y$ with the following property: For every $\varepsilon > 0$ there exists a $\delta > 0$ such that $$ d_Y(f(x),q) < \varepsilon $$ for all points $x \in E$ for which $$ 0 < d_X(x,p) < \delta.$$ The symbols $d_X$ and $d_Y$ refer to the distances in $X$ and $Y$, respectively.

There is a lot going on here, and I am not going to parse through all of it. To give some grounding, note that a metric space is (very roughly speaking) a set of "points" together with a way of measuring the "distance" between those points. We don't really need to fuss the details of that here: the space $(\mathbb{R}, |\cdot|)$ is a metric space (the points are real numbers, and the distance between two points $x$ and $y$ is given by $|x-y|$). Indeed, we can make any subset of $\mathbb{R}$ into metric space with the same distance function.

What is important is to note that the metric spaces involved are very important. In particular, we need to correctly understand the domain of the function with which we are working. In the case of $$ f(x) := \frac{\sin\left( \frac{1}{x} \right)}{\sin\left( \frac{1}{x} \right)}, $$ the implication is that $f : E \to \mathbb{R}$, where $$E = \mathbb{R} \setminus \left(\{0\}\cup \left\{\frac{1}{k\pi} : k\in\mathbb{Z}\setminus\{0\}\right\}\right)$$ with the distance measured by the absolute value. We cannot take $X$ to be a larger subset of $\mathbb{R}$, as $f$ is not defined on a larger set. But for all $x\in X$, we have $ f(x) = 1$, thus for any $\varepsilon > 0$, we can take $\delta = 1$ (or, really, anything else we like). Then if $0 < |x| < \delta$, we have $$ d_X(f(x),1) = | f(x) - 1 | = |1-1| = 0 < \varepsilon. $$ Therefore the limit exists, and is equal to 1. That is $$ \lim_{x\to 0} \frac{\sin\left( \frac{1}{x} \right)}{\sin\left( \frac{1}{x} \right)} = 1. $$

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  • $\begingroup$ What is metric space? $\endgroup$ – Archer Aug 20 '18 at 18:19
  • $\begingroup$ A metric space is a set together with a way of measuring distances between points in that space. For example, $\mathbb{R}$ with the usual absolute value function is a metric space (the points of the space are real numbers, and the distance between two points $x$ and $y$ is given by $|x-y|$). $\endgroup$ – Xander Henderson Aug 20 '18 at 18:20
  • $\begingroup$ I do not know who down-voted this post; it is by far the does the best job of combining accuracy with accessibility. (+1) $\endgroup$ – Mike Earnest Aug 20 '18 at 18:39
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    $\begingroup$ @MikeEarnest I saw that you and I had the same idea at the same time. I upvoted your answer, too---it is far more concise than mine, much better capturing the spirit of Rudin. :) $\endgroup$ – Xander Henderson Aug 20 '18 at 18:40
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If a function $f$ is defined by an "analytical expression" then by convention its domain $D$ is the set of $x$ for which this expression can be evaluated without asking questions. In the case at hand this is the set $$D:=\left\{x\in{\mathbb R}\biggm| x\ne 0\ \wedge \ x\ne{1\over k\pi} \ (k\in{\mathbb Z}_{\ne0})\right\}\ .$$ This $D$ is a subset, hence a relative space, of ${\mathbb R}$. The point $0$ is a limit point of $D$, in the same way as the point $1$ is a limit point of the interval $(0,1)$. Since at all points $x\in D$ the function $f$ assumes the value $1$ we can safely say that $\lim_{x\to0} f(x)=1$.

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  • $\begingroup$ Hi, @Christian. I feel like this answer might be missing something? What if we had a function defined on $\mathbb{R}\setminus A$, where $A$ is an uncountable subset? Or if $A$ were an interval of $\mathbb{R}$ around $0$? $\endgroup$ – Jam Aug 21 '18 at 10:41
  • $\begingroup$ @Jam: All we need is that $D\subset{\mathbb R}$ and that the point $x_0$ is a limit point of $D$ in ${\mathbb R}$. If your $A={\mathbb R}\setminus D$ is an interval containing $0$ in its interior then $0$ is not a limit point of $D$. $\endgroup$ – Christian Blatter Aug 21 '18 at 13:01
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This is tricky (Edit: And so tricky that this answer -six upvotes- is wrong. )


WARNING: This answer is wrong. Please don't upvote (or accept!) it. I've decided against deleting it because it still might be helpful.

The issue is this: the traditional-elementrary $\epsilon,\delta$ definition of the limit of a function of a real variable around $x=a$ involves a "deleted neighbourhood" $ 0 < | x − a | < \delta$, i.e., a punctured open interval: we must find some $\delta$ such that the function evaluated inside that neighbourhood falls near the limit. The question is: do we require that the function is defined in all that (real) interval? Actually we don't (that was my mistake). (If that were the case, then a function defined only on the rationals would not have any limits.) All that we require is the condition is fullfiled for all points of the domain that are inside that neighbourhood. (Actually, if the function is not defined in some deleted neighbourhood, we need to assert at least that the $x=a$ is a limit point of the domain. Without this, $\lim_{x\to 0} \sqrt{x-1}=3$ would be vacuously true)


(WRONG answer begins)

First, notice that the function is defined for all reals except for $x=0$, and for the points where the denominator is zero: $$\sin(1/x)=0 \iff 1/x= k\pi \iff x = \frac{1}{k\pi}$$

for any integer $k$.

Outside these prohibited points, the function equals $1$.

Now, as you correctly guessed, that the function is not defined at $x=0$ does not matter for computing the limit.

But what matters is that the other prohibited points get arbitrarily close to $x=0$, hence you cannot find any neighborhood around $x=0$ where the funcion is defined. Then, the limit does not exist. (WRONG)


(The graph of this function would consist of an horizontal line ($y=1$) with "holes" at $x=0$, and $x=1/k\pi$ . These holes get more and more concentrated around $x=0$... You cannot trust a computer generated graph for this kind of function.)

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    $\begingroup$ I would maintain that a definition that says the limit doesn't exist is inconvenient. A function defined on $\mathbb{Q}$ ought to be able to have limits. $\endgroup$ – Daniel Fischer Aug 20 '18 at 18:03
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    $\begingroup$ According to baby Rudin, $\lim_{x\to 0} \sin(1/x)/\sin(1/x)$ exists and is one. All that is required for the limit $\lim_{x\to x_0}f(x)$ to be well defined is for $x_0$ to be a limit point of the domain of $f$; you do not need $x_0$ to be an interior point of the domain. $\endgroup$ – Mike Earnest Aug 20 '18 at 18:04
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    $\begingroup$ @IceInkberry The issue here is : the traditional $\epsilon, \delta$ limit definition deals with (reduced) neighbourhoods. Do we require that the function is defined for all points on that interval, or not? Actually, we don't. My mistake was in assuming we required it. $\endgroup$ – leonbloy Aug 20 '18 at 18:13
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    $\begingroup$ I don't think you should delete this. The comments gave me more clarity. $\endgroup$ – Avnish Kabaj Aug 20 '18 at 18:23
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    $\begingroup$ For anyone reading this far, this definition of limit (and by extension, continuity) results in some unexpected continuous functions! See, for example, math.stackexchange.com/questions/1402458/… $\endgroup$ – Steve Kass Aug 20 '18 at 18:34
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Yes your guess is correct indeed we have that for any $x\neq 0$ and $\sin(1/x)\neq 0$

$$\dfrac{\sin\left(\dfrac 1x\right)}{\sin \left(\dfrac 1 x\right)}=1$$

therefore, according to the usual definition of limit, excluding the isolated points where the expression is not defined, we have

$$\lim_{x \to 0}\dfrac{\sin\left(\dfrac 1x\right)}{\sin \left(\dfrac 1 x\right)}=1$$

See also the related Why are we allowed to cancel fractions in limits?

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    $\begingroup$ What about $x=1/\pi$ where both numerator and denomiator are undefined? $\endgroup$ – GEdgar Aug 20 '18 at 17:47
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    $\begingroup$ There are infinite many $x$ such that $|x|<\epsilon$ and $\sin(1/x)=0$ . $\endgroup$ – Math Lover Aug 20 '18 at 17:50
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    $\begingroup$ @MathLover That’s mainly a matter of convention and definition, if we exclude those points the limit exists. $\endgroup$ – gimusi Aug 20 '18 at 17:52
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    $\begingroup$ @SteveKass Yes, it does. If $f:(X\setminus \{x_0\})\to Y$ is a function between metric spaces, where $x_0\in X$, then there is an epsilon-delta definition of $\lim_{x\to x_0} f(x)$. In this case, you let $X=\mathbb R\setminus \{\pm 1/\pi, \pm 1/2\pi,\dots\}$ be the region where $\sin x^{-1}$ is nonzero, and consider the limit as $x\to 0$ in $X$. $\endgroup$ – Mike Earnest Aug 20 '18 at 17:55
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    $\begingroup$ I always learned (real) limits as requiring the function to be defined in a whole nbhd of a point (excluding the point). Maybe I'm wrong. Let me pull out my baby Rudin.... $\endgroup$ – Randall Aug 20 '18 at 17:56
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Let $f:S\subset \mathbb{R}\to \mathbb{R}$ and $x_0\in S'.$ It is said that $$\lim_{x\to x_0}f(x)=L$$ if $$\forall \epsilon >0\exists \delta >0 \:\text{such}\:\text{that}\: x\in S \:\text{and}\: 0<|x-x_0|<\delta\implies |f(x)-L|<\epsilon.$$

Since $\dfrac{\sin\left(\dfrac 1x\right)}{\sin \left(\dfrac 1 x\right)}=1$ on $\mathbb{R}\setminus\left(\{1/(n\pi)|n\in\mathbb{Z}\}\cup\{0\}\right)$ and $0$ is a limit point of the domain of $f$ the limit exits and its value is $1.$

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The limit does not exist, because $\sin{1\over x}=0$ for $x={1\over{k\pi}}$ ($k$ any nonzero integer), and so for $x={1\over{k\pi}}$, the function $f(x)={\sin{1\over x}\over\sin{1\over x}}$ is undefined.

Because of this, if you are given an $\epsilon>0$, there is no $\delta>0$ for which $f(x)={\sin{1\over x}\over\sin{1\over x}}$ is even defined for all $x$ such that $0 < x < \delta$, let alone satisfies the implication that $0<|x-0|<\delta \to |f(x)-1|<\epsilon$.

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    $\begingroup$ Sorry but that’s completely wrong! $\endgroup$ – gimusi Aug 20 '18 at 18:08
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    $\begingroup$ What's your definition of the limit of a function at a point? $\endgroup$ – mfl Aug 20 '18 at 18:10
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    $\begingroup$ Why exactly is it completely wrong? It seems there may be more than one definition being considered here. Some sources appear to allow the function in question to have any domain so long as the limiting value of $x$ is a limit point of the domain. I wasn’t aware of this, but if that is allowed, then I guess we can say the limit exists. $\endgroup$ – Steve Kass Aug 20 '18 at 18:14
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    $\begingroup$ Apparently it is wrong if one is rigorous about the definition of the limit! That said, according to many introductory calculus books, it wouldn’t be wrong, since they sometimes (often?) define limits without mentioning accumulation points or requiring $x$ to be in the domain of the function in addition to satisfying $0<|x-x_0|<\delta$. $\endgroup$ – Steve Kass Aug 20 '18 at 18:30
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    $\begingroup$ Maybe you'd get more upvotes and fewer downvotes if you told us what definition of a limit you're using. If there is no reasonable definition of a limit under which your answer is correct... then your answer is incorrect. $\endgroup$ – Tanner Swett Aug 21 '18 at 17:35

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