Difference of two infinite reciprocals

Idea. As $x\to0^+$ both $\frac1{\sin x}$ and $\frac1x$ tend to $+\infty$. Their difference is an "$\infty-\infty$" form: we do not know in advance who wins. A common denominator makes the competition visible.

Computation.
$$\frac1{\sin x}-\frac1x=\frac{x-\sin x}{x\sin x}.$$
We know $x-\sin x=\dfrac{x^3}6+o(x^3)$, and $x\sin x=x^2+o(x^2)$. Hence
$$\frac{x-\sin x}{x\sin x}=\frac{\frac{x^3}6+o(x^3)}{x^2+o(x^2)}=\frac{x}{6}\cdot\frac{1+o(1)}{1+o(1)}\longrightarrow0.$$

Conclusion. The limit is $0$: the two infinite quantities are so close to each other that their difference vanishes.