Why is it possible to neglect higher order terms in the variation of the action? In order to get the Euler-Lagrange equations, we should find the variation of the action $\delta S$ and to neglect higher-order terms:
$$\delta S=\int L(q+\delta q,\,q'+\delta q',\,t)dt-\int L(q ,\,q',\,t)dt+O[(\delta q)^2]$$
I have two questions:

*

*Why is it legal to neglect the higher-order terms?


*If we get the Euler-Lagrange equations from a first-order approximation, doesn't it means that the equations themselves are only an approximation?
 A: The following is not mathematical rigorous, but a sketch of the central ideas.
To start, let us consider the case of a (differentiable) real scalar function $f:   \mathbb R \longrightarrow \mathbb R$. We say that $x$ is a stationary point if and only if $f^\prime(x)=0$. The Taylor expansion of $f$ around a point $x$ is given by:
$$f(x+h) = f(x) + f^\prime(x)  \, h + \mathcal O(h^2) \quad. \tag{1} $$
Now if $x$ is a stationary point, we have $f^\prime(x) =0$, which regarding equation $(1)$ yields
$$ f(x+h)-f(x) = \mathcal O(h^2)\tag{2} \quad . $$
What this means is that small changes of $x$ induce only changes to second (and higher) order in $f$ if $x$ is a stationary point.
Conversely, if equation $(2)$ holds, then $x$ is a stationary point, which can be seen by dividing $(2)$ with $h$ and taking the limit $h \to 0$, yielding $f^\prime(x)=0$.
This also means that you can find the stationary points of $f$ by (Taylor) expanding it in terms of $h$ and set the terms proportional to $h$ to zero. But we don't neglect any terms or approximate anything.

The very same line of thought applies to functionals: Indeed, consider a functional $S:F\longrightarrow \mathbb R$, with a suitable chosen space of functions $F$.
We define the $n$-th functional derivative as
$$ \delta^{n} S[f][\eta]:=\frac{\mathrm d^n S[f+\epsilon \, \eta]}{\mathrm d \epsilon^n} \big\vert_{\epsilon=0}\tag{3} \quad ,$$
where $\epsilon \in  \mathbb R$ and $\eta$ denotes a suitable function. We can further define a Taylor series as follows:
$$ S[f+h\eta] = S[f] + \sum\limits_{n=1} \frac{h^n}{n!}\, \delta^{n} S[f][\eta] \tag{4}$$
We say that $S$ is stationary at $f$ if the first functional derivative vanishes for all $\eta$, or, equivalently, $S$ does not change to first order for a small change of $f$, i.e. for all $\eta$ we have
$$ S[f+h \eta] - S[f] = \mathcal O(h^2)\tag{5} \quad .$$
Again, note that $(5)$ is equivalent to say that the functional derivative of $S$ at $f$ vanishes.
