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

Question

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.

Answer 1

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 $\lim_{x\to 0}f(x)=1$.

Answer 2

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

Answer 3

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 $\lim_{x\to0} f(x)=1$.

Answer 4

This is tricky (Edit: And so tricky that this answer -six upvotes- is wrong. )

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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)

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(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)

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(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.)

Answer 5

Yes your guess is correct indeed, according to the more general definition of limit, excluding from the domain the isolated points where the expression is not defined, we have

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

Of course the answer depends upond the definition we are assuming for the limit and with reference to the standard definition, often used in high school level, we should conclude that limit doesn't exists. Anyway this approach is really unsatisfactory when we face with more advanced calculation of limits.

See also the related

• Why are we allowed to cancel fractions in limits?

Answer 6

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 $1.$

Answer 7

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 $0<|x-0|<\delta \to |f(x)-1|<\epsilon$.