I want to confirm the logic within the $\epsilon$-$\delta$ argument when proving the limit exists. When we are proving $\displaystyle\lim_{x \to a} f(x)=L$ is true, we are actually trying to prove a "if-then" statement: $ \exists L\in \mathbb R, \forall \epsilon > 0, \exists \delta >0,\forall x\in domf : 0<|x-a|<\delta \implies |f(x)-L|< \epsilon $. Is this correct
Then when proving the "if-then" statement is true. We use a property of the "if-then" statement, which is if this "if-then" statement is true, then assume the "if" part is true, then the "then” part must be true. (i.e. if $0<|x-a|<\delta \implies |f(x)-L|< \epsilon $ is true, then if $ 0<|x-a|<\delta $ is true, $|f(x)-L|< \epsilon $ must be true.
Then because $ a $ is a limit point of $ domf$ then the "if" part is always true(we don't even need to assume it is true). Therefore, we need to test whether the $ |f(x)-L|< \epsilon $ is true.
However, we test $|f(x)-L|< \epsilon $ part by using a slightly inverse method. Because we know the $ 0<|x-a|<\delta $ is always true(we can find any positive number to be our $ \delta $, therefore, we're trying to find a "suitable" $ \delta $ within all positive numbers to make then part :$ |f(x)-L|< \epsilon $ be true. That's why we always firstly calculate $|f(x)-L|< \epsilon $ this part, then make a suitable $ \delta $ value.
My question is what parts in my above words are wrong, or do I misunderstand something? Thank you!