$\lim_{x \to 0}\sin(\frac{1}{x})$=? 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?
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 $n=4k+1$)
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?