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

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

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

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