I'm having difficulty understanding the following example:
If $f(\epsilon) = \sin(\epsilon)$ then, using Taylor's theorem for $f$ around $\epsilon_0 = 0$, $f = \epsilon - \dfrac{\epsilon^2}{2}\sin(\xi)$. Thus, $\lim_{\epsilon\downarrow 0}(f/\epsilon) = 1$, and from this it follows that $f = \mathcal{O}(\epsilon)$ as $\epsilon\downarrow 0$.
Question: Why do we have $f = \epsilon - \dfrac{\epsilon^2}{2}\sin(\xi)$? And what is meant with $\xi$ in this expression? If I were to calculate the taylor series around $\epsilon_0 = 0$ I would get something like $$ f(\epsilon) = f(0) + \epsilon f'(0) + \dfrac{\epsilon^2}{2!}f''(0)+ ... $$ with $f(0) = \sin(0) = 0, f'(0) = \cos(0) = 1, f''(0) = -sin(0) = 0$ so that $$ f(\epsilon) = e - \dfrac{\epsilon^2}{2!}*0\, + .... $$ I would also conclude that $f = \mathcal{O}(\epsilon)$ but based on a different Taylor series.
See 'Mean value form of remainder' in https://en.wikipedia.org/wiki/Taylor%27s_theorem
Once you get $\xi$ use the fact that $|sin(\xi)| \leq 1$.