Logic of the precise definition of limits? I'm a college freshman studying calculus from Thomas. Right now I'm learning the precise definition of limits using the epsilon delta method to use in proofs etc.
I notice that when proving the limit of a function the limit is given first and then the epsilon delta method is used. But isn't this like putting the cart before the horse.
For e.g say I need to prove the limit of 5x-3 at x->1. It would seem that the prescribed order is just use the substitution rule for polynomials, get 2 as the Limit, then use the epsilon delta method to show the limit is indeed at 2. This confuses me because shouldn't the limit be an 'unknown' first and a proof be used to derive the limit as 2. How do I know in the first place that the limit 2 should be used as the starting point? The logical order appears backward to me.
 A: You first need a suitable candidate/educated guess for what the limit ought to be. Then, only after that, you can use the precise definition to PROVE that your initial guess is indeed the case. Also, you can see that this is the best you can do simply from how the definition of limits is given:

Definition.
Let $f:\Bbb{R}\to \Bbb{R}$ be a function, $a\in\Bbb{R}$. We say $f$ has a finite limit at $a$ if there exists $l\in \Bbb{R}$ such that for every $\epsilon>0$, there exists $\delta>0$ such that for all $x\in\Bbb{R}$, if $0<|x-a|<\delta$ then $|f(x)-l|< \epsilon$.
(In this case, we can prove that $l$ is unique and we denote it as $\lim_{x\to a}f(x)$)

Notice how the definition starts with "there exists $l\in \Bbb{R} \dots$" Just from the way it is phrased, it suggests that before even checking the $\epsilon,\delta$ criterion, you need to have a candidate value for the limit $l$. Nowhere does the definition tell you what $l$ is or how to go about guessing this (such "guess work" is something you pick up along the way as you learn more).
For example, if you had two functions $f$ and $g$, with $\lim\limits_{x\to a}f(x) = l_1$ and $\lim\limits_{x\to a}g(x) = l_2$, then if all you do is stare at the definition of limits, there's no way you can tell that $f+g$ also has a limit and that the limit equals $l_1+l_2$. The only natural guess would be that if $f+g$ had a limit, then it had better be $l_1+l_2$.
Then, once you have this guess, you then proceed to prove this using the precise $\epsilon,\delta$ definition (where the crux of the proof is the triangle inequality).
