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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 $\pi/2$.

Thanks in advance!

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    $\begingroup$ The proofs which use this technique always deal with the local behavior of functions. To put differently, the result to be proved does not depend on the behavior of the functions outside that interval. In fact the result depends on the behavior of functions in arbitrarily small neighborhood of the point under consideration. $\endgroup$ – Paramanand Singh Sep 30 '17 at 2:30
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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.

In other words, if someone set up restrictions like that, found a limit under that restriction, and then doesn't mention the restriction again, then that's bad practice. But it's also a sign that the other cases are either proven completely analoguously or that they are relatively easily transformed into the proven case. Or at least that the author of the proof thought so.

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