Order of statements in the Delta-Epsilon Limit definition So the definition of Limit I see is
$$\lim_{x \to a}f(x) = L$$ means:
for all $\epsilon >0$, there exists a $\delta >0$ such that $$0<|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon $$
I was wondering if modifying the definition to: Limit exists when for all $\delta >0,$ there exists a $\epsilon > 0$ such that
$0<|f(x) - L| < \epsilon $ $\Rightarrow |x - a| < \delta$
would cause any problem.
Because to me it seems like this definition should also work except it does seem a lot harder when it comes to the actual proving part.
Can this flipped version of the limit definition also work? Why did mathematicians define in this order?
 A: That is not equivalent to the usual definition. For instance, if $f\colon\Bbb R\longrightarrow\Bbb R$ is the constant function $L$, then $\lim_{x\to a}f(x)=L$ (for any $a\in\Bbb R$), according to the usual definition. But not according to your definition.
A: Huge problems.
Basic issue.  We hone in on the function by honing in on the image of the function by honing in on the domain.  We can't hone in on the domain by honing in on the function because the function needn't be invert-- it may fail the horizontal line test-- and we have point in the image close together that are miles apart in the domain.
Example a constant function.  Or the periodic $\sin$ function where $|f(x) - L| < \epsilon$ would not mean $|x-a| < \delta$ as $x$ may be multiples of $2\pi$ distant for $a$.  So $|f(x)-L| < \epsilon \not \implies |x-a| < \delta$.
... Now that counter-intuitive aspect that confuses every student (except the students who lie) is that it seems if we are honing in one the domain ($\delta$) to hone in on the image, should we start with the $\delta$??? Doesn't starting with the $\epsilon$ seem backwards.
And .... yes, it seems that way, but if you draw enough pictures and practice enough you will see that we must work backwards by establishing "acceptable error" range in the final image and finding the intial input range to produce the acceptable error.  That's the only way it'll work.  The wording sound backwards... but it's how it must be.

We start with the "acceptable error" range in the end result, to retrofit and finish with ant initial input range.

That's why it's $\epsilon$ first, then $\delta$ and not the other way around.
And if you are confused, rest assured every student before you either was equally confused at one time or are is lying about it.
