Show that $\sin(\mathcal{o}(x)) = \mathcal{o}(x)$ as $ x\to 0$

Question

So I want to show that $\sin(\mathcal{o}(x)) = \mathcal{o}(x)$ as $ x\to 0$.

So far I have thought that my result will come from showing $ \displaystyle \Big|{\frac{\sin(f(x))}{x}}\Big| \to 0$ as $x\to 0 $, where $f(x) = \mathcal{o}(x)$.

However, the only useful thing I know is that $\displaystyle\Big|{\frac{f(x)}{x}}\Big| \to 0$ as $x\to 0 $, and i'm not sure how to employ this!

I'm sure this is fairly straight forward, but I'm just not sure of the method.

Answer 1

Hint: Use the fact that $|\sin(x)|\le|x|$ for all $x$.

Answer 2

This follows from a rule of asymptotic analysis: $\,o(o(x))=o(x)$, and $\,\sin x=o(x)$.