Q:

checking slope = $0$ at a point for a function using $\epsilon $, $\delta $ definition

From the continuity definition, a function is continuous at a point $a$ if  : $$\forall \epsilon \gt 0 \exists \delta \gt 0  : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon$$
If I change the order of quantifiers like below do I get a definition for checking slope = $0$ at $x = a$ ? 
$$\exists \delta \gt 0   \forall \epsilon \gt 0 : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon$$

If my interpretation of second quantifier order is flawed, could you please tell me how to interpret it correctly in this context ? I realized that order matters by going through other mse posts already.

A:

The order of the quantifies matters a lot. The statement
$$\exists \delta \gt 0, \forall\epsilon \gt 0 : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon$$
implies that there exist an interval around $a$ where $f(x)$ is constant. If we change the order of the quantifiers to
$$\forall \delta \gt 0, \exists\epsilon \gt 0 : |x-a| \lt \delta \implies |f(x)-f(a)|\lt \epsilon$$
then for example any function which is bounded will satisfy it since if $|f(x)| < M$ then take $\epsilon = 2M$ to get $|f(x)-f(a)| < \epsilon$ for all $\delta$.

