Curious about changing the definition of limit. The definition of the limit is: $\forall \varepsilon \gt 0,\,  \exists\delta \gt 0$ such that,
$$
0\lt |x-c| \lt \delta \implies \lvert f(x)-L\rvert \le \varepsilon.
$$
I am curious about:


*

*If I let $\varepsilon=0$ or $\delta=0$, what will happen?

*If I change the position of $\varepsilon$ and $\delta$? In other words
if $\forall \delta \gt 0,\,  \exists \varepsilon \gt 0$ such that,
$$
0\lt |x-c| \lt \delta \implies |f(x)-L| \le \varepsilon.
$$

 A: *

*If you let $\delta=0$, then you will always find the conditions true (there is nothing to check in the interior of a ball of radius 0)

*Consider the step function that is discontinuous at 0 (0 if $x \leq 0$ and 1 if $x>0$).  If you take $\epsilon=2$, then your modified definition of continuity will be met for all $\delta$ even though the function is not continuous in the normal sense.

A: If you let $\delta=0$, then you get $0<|x-c|<0$, which is FALSE for all $x$. So  $0<|x-c|<\delta \implies |f(x)-L| < \epsilon$ will be TRUE yet meaningless. It is a useless requirement. The reason for the $0$ in $``0<|x-c|<\delta"$ is because the limit of the function $f(x)$ as $x$ approaches $c$ must have nothing to do with the value or even the existence of $f(c)$. The limit is a property of the values of $f(x)$ near $x=c$ not at $x=c$. That means that we are specifically not interested in what happens when $\delta = 0$.
If you let $\epsilon = 0$, then $|f(x)-L| \le \epsilon$ implies that $f(x)=L$ for all $x$ in the intervals $(c-\delta, c) \cup (c, c+\delta)$. This will always be TRUE when $\delta = 0$ but very seldom otherwise. Again, it is a useless requirement.
If you use "$\forall \delta \gt 0,\,  \exists \varepsilon \gt 0$", then every bounded function, no matter how evil, is continuous
A: *

*The idea of limit is "$f(x)$ is near of something (the limit) when $x$ is near $c$ but not in $c$?". $\delta = 0$ means $x = c$ (excluded by the "not in $c$"). $\epsilon = 0$ requires $f(x) = L$ (too strong, less is enough).

*As general rule, you can't exchange the order of $\forall$ and $\exists$:
$$\forall\epsilon>0\ \exists n\in\Bbb N:1/n<\epsilon$$
is true, but
$$\exists n\in\Bbb N\ \forall\epsilon>0:1/n<\epsilon$$
is false.
