$\lim_{x \to 0}\sin(\frac{1}{x})$=? What is the limit:
$\lim_{x \to 0}\sin(\frac{1}{x})$?
I plotted this function on https://www.desmos.com/
And this was the result:

I searched the web for this function and it was straight forward stated as "The limit doesn't exist" without any theoretical proof. Could anyone provide a theoretical proof or something a bit more concrete? 
 A: Recall that when a limit 
$$\lim_{x\to x_0} f(x)=L$$
exists it is unique and it is the same for all the subsequences, that is
$$\forall x_n \to x_0 \implies f_n=f(x_n) \to L$$
Therefore to prove that a limit doesn't exist it suffices to show that at least two subsequences exist with different limit.
In this case let consider
$$x_n=\frac2{\pi n}\to 0^+$$
then
$$\sin\left(\frac{1}{x_n}\right)=\sin\left(n\frac \pi 2\right)$$
What can we conclude form here? (try for example with $n=4k$ and $n=4k+1$)
A: Suppose that $\ell=\lim_{X\to 0}\sin(1/x)$. Using the definition of the limit with $\varepsilon=1/2$, we get the existence of a positive $\delta$ such that if $0\lt \left\lvert x\right\rvert\lt\delta$, then $\left\lvert \sin(1/x)-\ell\right\rvert\lt 1/2$. 
In particular, if $0\lt s,t\lt \delta$, 
$$
\left\lvert \sin\left(\frac1s\right)-\sin\left(\frac1t\right)\right\rvert
\leqslant\left\lvert \sin\left(\frac1s\right)-\ell\right\rvert+\left\lvert \ell-\sin\left(\frac1t\right)\right\rvert\lt 1 $$
hence we reach a contradiction by choosing $$s=\frac1{2\pi n+\pi/2}\mbox{ and }t=\frac1{2\pi n-\pi/2}$$
where $n\geqslant 1$ is such that $2\pi n+\pi/2>\delta^{-1}$. 
A: The reason is essentially because the function "oscillates infinitely back and forth and does not settle on a single point". If this does not satisfy you, we may prove this formally with the following theorem

$\lim_{x\to c}f(x)=L$ if and only if, for every sequence $(x_n)\in\mathbb R$ tending to $c$, it is true that $(f(x_n))$ tends to $L$.

This is sometimes known as the sequential criterion for the convergence of a function. But in our case, this condition means that the limit does not exist. This is because we can just pick two sequences: the first so that the sequence $(f(x_n))$ corresponds to the points at the "top" of the graph; i.e., $f(x_n)=1$ for all $n$. You can find an explicit formula if you wish. Similarly pick our second sequence so that $f(y_n)=-1$ for all $n$. If we make sure both sequences are, say, positive, and decreasing to $0$, then we now have two sequences converging to $0$, but giving a different limit: $1$ and $-1$ respectively. If the limit existed, then by the theorem, $1=-1$, which is obviously absurd. We are done.
(BTW, if even this isn't rigorous enough for you, you can prove $(x_n),(y_n)\to 0$ by the monotone convergence theorem, and that $(f(x_n))\to1$, $(f(y_n))\to-1$ by using the formal definition for the limit.)
A: I will use the property that for all $N\in\mathbb{R}$ there exists $x,y>N$ such that $\sin(x)=1$ and $\sin(y)=-1$. If you also want a proof of this, just tell me.
Assume for the contrary that $\lim_{x\to0}\sin(\frac1x)$ exists, so it equals some $L\in\mathbb{R}$. Then, by definition of the limit, for $\varepsilon:=\frac12>0$ we get some $\delta>0$ such that for all $x$ with $|x|<\delta$ we have $|\sin(\frac1x)-L|<\varepsilon=\frac12$. Now use the beforementioned property that for $N:=\frac1{\delta}$ there exist $x',y'>N$ such that $\sin(x')=1$ and $\sin(y')=-1$. Define $x:=\frac1{x'}$ and $y:=\frac1{y'}$. Then $|x|,|y|<\delta$ and $\sin(\frac1x)=1$ and $\sin(\frac1y)=-1$. However, we then find by the triangle inequality $2=|(-1-L)-(1-L)|\leq|\sin(\frac1y)-L|+|\sin(\frac1x)-L|<2\varepsilon=1$. A contradiction.
A: Let’s look at half the limit: approaching from the right, say. Note that $1/x\to\infty$ as $x\to 0^+$, so we may rewrite the limit as 
$$\lim_{x\to 0^+} \sin(1/x) = \lim_{x\to\infty} \sin(x).$$
Can you see why $\sin(x)$ doesn’t have a limit in this case?
