For $\lim_{x \to a} f(x) = L$, the way I've always thought that $f(x)$ can get "close" to $L$ if $x$ is "close enough" to $a$. So I've worked with the $|f(x)-L|<\epsilon$ first and then match it with the $0<|x-a|<\delta$. This way, for any $\epsilon$ I can show there exists a $\delta$.
But, at least on YouTube, some people go in the opposite way and I'm wondering if there is a reason for this. They seem to conceptualize it like this: if $x$ is close to $a$ then $f(x)$ can get close to $L$"
Which is the correct (or better) way of thinking about it? Also, what are the issues if thought about in the incorrect (or worse) way?
Picking up Mauro ALLEGRANZA's comments, the order of quantifiers is important. The definition requires that for every $\epsilon$ there exists a $\delta \ldots$, so you start with an $\epsilon$ and find a $\delta$ that supports that $\epsilon$. The $\delta$ (almost) always depends on $\epsilon$. To prove the limit is as you say, you then start with $|x-a| \lt \delta(\epsilon)$ and prove $|f(x)-L| \lt \epsilon$.
If you go the other way, you don't know how close you have to get. I can certainly choose a $\delta$, compute $\epsilon$ and prove that $|x-a| \lt \delta \implies |f(x)-L| \lt \epsilon$ but we don't know if that is close enough. Let $f(x)=10^{-9}$ and somebody claim that $\lim_{x \to 1}f(x)=0$ If they start with $\delta=1$ they can prove that $|f(x)-L| \lt 10^{-8}$. It sounds impressive, but they fail if we choose $\epsilon $ smaller than $10^{-9}$. That is why the correct definition starts with "For any $\epsilon \gt 0$"