Why does $\sin(\frac{1}{x})$ give reciprocals?

Question

I was preparing a calculus lesson for my students where we examine the continuity of various functions. One such function was

$$g(x) = \begin{cases} 0 & x = 0\\ x \sin(\frac{1}{x})) & x \neq 0 \\ \end{cases}$$

Putting the topic of continuity completely aside, I want to point out that one of the things that was observed in the textbook I was using is that the end behavior of the graph of $g$ is described by a horizontal asymptote of $y=1$. This can only be the case if as $x \to \infty$, then $x \sin(\frac{1}{x}) \to 1$ which means that $\sin(\frac{1}{x}) \approx \frac{1}{x}$ for $x \geq 2$, i.e. sin$(\frac{1}{x})$ gives approximate reciprocals of $x$ for any given $x \geq 2$.

In an attempt to understand this, I Googled this phenomenon, but could not find anything close to what I've portrayed above. Is there any explanation / deeper examination for why it just so happens that $\sin(\frac{1}{x})$ is approximately the reciprocal of $x$ for $x \geq 2$?

(Note: I'm now realizing that this might just be analogous to observing that $\sin(\frac{1}{x}) \approx \frac{1}{x}$ as $x \to \infty$. Maybe the actual question is to ask why sine leaves $\frac{1}{x}$ "mostly invariant")

This is called small angle approximation, where $\sin z \approx z$ for small $z$ — Ninad Munshi, Dec 01 '22 at 15:11
Your observation is equivalent to $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. — Martin R, Dec 01 '22 at 15:13
Oh, wait I see it now. Lurking in the background here is the fact that $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$. ^ What Martin said — Grigor Hakobyan, Dec 01 '22 at 15:13
The standard argument for the derivative of $\sin$ when you get there is that $\lim{{ x \to 0} (\sin x)/x = 1$. — Ethan Bolker, Dec 01 '22 at 15:14

Why does $\sin(\frac{1}{x})$ give reciprocals?

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