My textbook writes that there are several equivalent relations of epsilon delta proof, i.e., $p \iff q$.
Let the normal epsilon delta definition of $\lim_{x \to a} f(x) = b$,
$$\forall\,\varepsilon > 0, \quad \exists\, \delta > 0 \;\text{ s.t. }\quad 0 < \left|x - a\right| < \delta \implies \left|f(x) - b\right| < \varepsilon$$
be labelled as proposition $p$. Then the proposition $q$ is
$$\forall\, \varepsilon > 0, \quad \exists\,\delta > 0 \;\text{ s.t. }\quad 0 < \left|x - a\right| < \delta \implies \left|f(x) - b\right| \le \varepsilon$$
Can someone explain why $p$ is equivalent to $q$? I ask this because for $p$, the range $f(x)$ may fall in is the interval $\left(b−\varepsilon, b+\varepsilon\right)$. For $q$, however, the range f(x) may fall in contains $b−\varepsilon$ and $b+\varepsilon$. So while $p \implies q$ is valid, $q \implies p$ is invalid. Can someone explain to me how $q \implies p$ can be valid?
Let's start with proposition $q$.
Let $\epsilon > 0$ be any positive real number. We know that $\dfrac{\epsilon}{2}$ is also a positive number. So we can apply proposition $q$ on the positive real number $\dfrac{\epsilon}{2}$ (proposition $q$ is a statement that starts by saying something about every positive real number) by saying that $\exists \delta > 0$ such that $ 0 < | x - a | < \delta \implies | f(x) - b | \leq \dfrac{\epsilon}{2}$.
Since $\dfrac{\epsilon}{2} < \epsilon$, this, in particular, means that this $\delta$ which was such that $ 0 < | x - a | < \delta \implies | f(x) - b | \leq \dfrac{\epsilon}{2} < \epsilon $ is also such that $ 0 < | x - a | < \delta \implies | f(x) - b | < \epsilon$ (since $ | f(x) - b | \leq \dfrac{\epsilon}{2} \implies | f(x) - b | < \epsilon $).
So what we have done was: we picked any $\epsilon > 0$ and by using proposition $q$ we proved that $\exists \delta > 0$ such that $ 0 < | x - a | < \delta \implies |f(x) -b| < \epsilon$. But this is exactly what proposition $p$ says. So this proves $q \implies p \, \blacksquare$.