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.