A property of the function $\frac{\sin x}{x}$ How can one prove, that $0$ is the only value of $\frac{\sin x}{x}$ taken infinitely often?
What I tried:
To see how the graph looks like https://www.wolframalpha.com/input/?i=%28sin+x%29%2Fx
The function is continuous and has infinitely many positive and negative values, so by Darboux it has infinitely many zeroes.
Also, the line $y=0$ is an asymptote both in $\pm\infty$, but this thing only, doesn't imply the result. What else should I use?
 A: Let $0\ne r\in [-1,1].$ Let $f_r(x)=-rx+\sin x.$
Since $\frac {\sin x}{x}\to 0$ as $|x|\to \infty,$ take $M>0 $ such that $|x|>M\implies \left|\frac {\sin x}{x}\right|<|r|\implies f_r(x)\ne 0.$
Now $f_r'(x)=-r+\cos x$ so the set $S=\{x\in [-M,M]: f'_r(x)=0\}$ is finite. So let $S\cup \{-M,M\}=\{x_j: 1\le j\le n\}$ for some $n\in \Bbb N,$ where $x_j<x_{j+1}$ for each $j<n.$
Now $f_r$ is strictly monotonic on each interval $[x_j,x_{j+1}]$ for $j<n$ because $f'_r$ is continuous and non-$0$ on $(x_j,x_{j+1}).$ So there is at most one $x\in [x_j,x_{j+1}]$ such that $f_r(x)=0.$
We could also say there is a member of $(f'_r)^{-1}\{0\}$ between any 2 members of $f_r^{-1}\{0\}$ so if $[-M,M]\cap f_r^{-1}\{0\}$ was infinite then $[-M,M]\cap (f_r')^{-1}\{0\}$ would be infinite.
A: Since it's an even function, this is equivalent to other values being achieved by only finitely many $x>0$. As @AnginaSend and @JCAA noted, you just need the limit at $\infty$. Since $-\frac1x\le\frac{\sin x}{x}\le\frac1x$, by the squeeze theorem $\lim_{x\to\infty}\frac{\sin x}{x}=0$. This means$$\forall\epsilon>0\exists N>0\forall x>N\left(\left|\frac{\sin x}{x}\right|<\epsilon\right).$$In particular,$$\forall\epsilon>0\exists N>0\forall x>0\left(\frac{\sin x}{x}-\epsilon=0\implies x\in(0,\,N]\right).$$Since $\frac{\sin x}{x}$ only has finitely many turning points on any set of the form $(0,\,N]$ (proof is an exercise), with no intervals of positive width on which it's constant, we're done.
