$\epsilon - \delta$ proof using inequalities which are true only for some intervals?

When I see the $\epsilon - \delta$ proof of trig limits, they usually start with some inequality which holds with restrictions. Take for example the restriction $0\le x \le\pi/2$ on the inequality $\cos\theta<\frac{\sin\theta}{\theta}<1$ here.

Can someone explain why this restrictions on the inequalities is not affecting our proof(in general)?

The only explanation I could come up with that "limit is a local property and we can somehow choose our e to be less than $\pi/2$.

I'll try to explain your example: In the proof of $\lim_{\theta\to 0}\frac{\sin\theta}\theta$ we prove first that $\lim_{\theta\to0^+}$ exists by having a restriction like $0<\theta\leq 1$ (insert your favourite proof here, as long as it's not l'Hopital). Then after that we use a second restriction, like $-1\leq \theta<0$ to prove that $\lim_{\theta\to 0^-}$ exists by exploiting $\sin\theta = -\sin(-\theta)$ and $\cos\theta = \cos(-\theta)$: $$\cos\theta < \frac{\sin\theta}\theta<1\\ \cos(-\theta)<\frac{-\sin(-\theta)}{\theta} < 1\\ \cos(-\theta)<\frac{\sin(-\theta)}{-\theta} < 1\\$$ and then, setting $\phi = -\theta$, we have the restriction $0<\phi \leq 1$ and the inequality $\cos(\phi)<\frac{\sin(\phi)}{\phi} < 1$, which we know is true by the first case.