# Question about small variation in path in derivation of Euler-Lagrange equation?

Consider the first derivation of the Euler-Lagrange equation shown in this Wikipedia article. We begin by considering the path $$f(x)$$ that yields stationary action. We then consider a new path with the addition of a small perturbation given by:

$$g_\epsilon(x) \ = \ f(x) \ + \ \epsilon \eta(x)$$

Here is my concern: At no point in this proof do we actually make use of the fact that $$\epsilon$$ is small. It would seem as though this proof would hold for any size of $$\epsilon$$. I do understand that $$\epsilon$$ terms of order greater than or equal to $$2$$ in the Taylor expansion of the action (in terms of $$\epsilon$$) will become negligible if $$\epsilon$$ is very small. Along with the absence of a first-order $$\epsilon$$ term, this makes the action at the original and perturbed path essentially the same, but this doesn't seem to have anything to do with the proof itself.

• Someone may correct me but $\epsilon$ in this case can take any value. $g_\epsilon(x)$ defines a set of curves. – Charlie Apr 20 '20 at 0:54
• Oh interesting, would saying that $\epsilon$ is small be incorrect then? (I've seen this on Wikipedia and a few other places) – Jack Ceroni Apr 20 '20 at 1:00
• Someone more qualified could give a better answer than me I'm afraid. – Charlie Apr 20 '20 at 1:42
• – Vivek Apr 20 '20 at 2:20