In the epsilon-delta proof for $\lim\limits_{x \to 0} \frac{\sin(x)}{x}$, we can show that for $|x| < 1$, $|\frac{\sin(x)}{x}| < \frac{x^2}{6}$. We can set $\delta = \sqrt{6\epsilon}$ and complete the proof. During the proof, $\delta$ is taken to be the minimum of $\sqrt{6\epsilon}$ and $1$. Why is $\delta$ defined this way, and why is 1 included in the definition of $\delta$? What would go wrong if $\delta$ is just $\sqrt{6 \epsilon}$? Can we use another number other than $1$?
Update:
So now I understand that in the proof, $\delta=\sqrt{6\epsilon}$ is guaranteed to work when $|x|<1$. So we need to deal with the case when $|x| \geq 1$. We want to choose a $\delta$ such that $|x| < \delta$ still yields $$\left|\frac{\sin(x)}{x} - 1\right| = |\frac{x^2}{3!} - \frac{x^4}{5!} + \frac{x^6}{7!} - \frac{x^8}{9!} + \frac{x^{10}}{11!} - \cdots | < \frac{x^2}{6} < \epsilon$$
My question is how do we deal with this case? If there is an upper bound on $\delta$, how do we find it? What is a formal proof?
Also, for $|x| \geq 1$ are we just saying that $\delta = 1$ works or is it actually the case that we are taking $\delta = \min(1, \sqrt{6\epsilon})$? If we set $\delta=1$, note that $|x| < 1$ would be false, so the whole statement $$0<|x|<\delta \Rightarrow \left|\frac{\sin(x)}{x}-1\right| < \epsilon$$ would be vacuously true.