Sine over x at infinity

Idea. The numerator oscillates and has no limit, but it is *bounded*: $-1\le\sin x\le1$. The denominator grows without bound, which crushes the quotient. The tool for this is the squeeze (or sandwich) theorem.

Computation. For every $x>0$,
$$-\frac1x\le\frac{\sin x}{x}\le\frac1x.$$
Both bounds tend to $0$ as $x\to+\infty$, so by the squeeze theorem $\dfrac{\sin x}{x}\to0$.

Remark. "Bounded divided by something tending to infinity tends to $0$" is a rule worth remembering.