Imagine you are deriving the Euler-Lagrange Equations. Starting with the functional: $$ S[F] = \int_{x_0}^{x_1} {F(x, y, y') \; dx} $$ Then, you add a variation to your functions: $$ \bar{y}(x) = y(x) + \varepsilon \eta(x) \\ \bar{y}'(x) = y'(x) + \varepsilon \eta'(x) $$ After that, you plug these varied functions onto your equation, and seek for an $\varepsilon$ that extremizes the functional (basically $dS/d \varepsilon = 0$)
The thing is, I was watching this video, and on minute 14:34 it is mentioned that you correct value of $\bar y (x)$ arises when $\varepsilon$ tends to zero.
This feels contradictory to me. As I mentioned before, if you want to extremise the functional, you basically do $dS/d \varepsilon = 0$, but first of all, if $\varepsilon \approx 0$, then $\varepsilon$ is not a variable, because you already know it must have a value near to zero (if not equal), no?
Secondly, suppose for a moment you were on a Calculus class at high school, if we do $dS/d \varepsilon = 0$, we will find a value or values of $\varepsilon$ (assuming the function has at least one extremal point), that would give us the location of these extrema; but imagine (as the video says) that our extremal point arises when $\varepsilon \approx 0$, then we already have our solution, what's the point on doing the derivative? Take this as an analogy. What I was trying to say is that if $\bar{y}(x) = y(x) + \varepsilon \eta(x)$ and $\varepsilon \approx 0$, then you already have your solution: $\bar y(x) = y(x)$ (I know this does not give you any analytical result, but these are the conclusions you arrive to); why on earth allowing $\varepsilon$ to be a variable (when you know is $\approx 0$) and varying your functions like this, would give you a correct result?
What I assume you are seeking for when you do $dS/ d \varepsilon = 0$ are its implications (which are the Euler-Lagrange Equations), not values of epsilon which extremizes the functional. In other words, what $F(x, y, y')$ and $y(x)$ must satisfy.
I know my line of reasoning is probably not correct, so feel free to correct me. I hope I made my explanation clear.
I agree that it seems wrong to say that the "correct" value of $\bar{y}(x)$ arises as $\varepsilon \to 0$.
The motivation for the variational argument using $\varepsilon$ is that we want to discover what conditions a minimizer of the functional must satisfy. So we assume that $y$ is a minimizer of the functional. Then we know that the function $g(\varepsilon) = S(y + \varepsilon \eta)$ has a minimum at $\varepsilon = 0$. Then we use calculus to deduce that $y$ satisfies the Euler-Lagrange equations.
To actually find the minimizer then, you would solve the Euler-Lagrange equations and then prove that the solution is indeed a minimizer of the functional. A good example of the entire procedure is the geodesic equation, where the critical point of the functional need not be a minimizer.