I have the following definition of the functional derivative $ \frac{\delta S}{\delta\gamma}$, where $S$ is my functional and $\gamma$ is a curve:
$$\tag{1} \int^B_A \frac{\delta S}{\delta\gamma} h(x) dx = \left. \frac{d}{d\epsilon} S[\gamma + \delta\gamma]\; \right|_{\epsilon = 0} $$
where $h(x)$ is an arbitrary function having the same fixed endpoints as $y$, $\epsilon$ is a small constant, and $\delta \gamma = \epsilon h$ is a small variation of the curve $\gamma$.
According to my notes the first variation is
$$\delta S = \int_A^B \frac{\delta S }{ \delta \gamma}\;\; \delta \gamma\;\; dx \tag{2}$$
which is to my eyes is simply the LHS of $(1)$ multiplied by $\epsilon$ (because $\delta \gamma = \epsilon h$).
If that is correct then I could instead multiply the RHS of $(1)$ to find
$$ \delta S = \epsilon \; \cdot \left( \left. \frac{d}{d\epsilon}\; S[\gamma + \delta\gamma]\; \right|_{\epsilon = 0} \right) \tag{3}$$
However, wiki defines the first variation (using $y$ as the function defining $\gamma$) as
$$ \delta S(y,h) = \left.\frac{d}{d\varepsilon} S(y + \varepsilon h)\right|_{\varepsilon = 0} \tag{4}$$
i.e. without the factor of $\epsilon$. I understand that in practice if one were looking for a stationary curve, setting $\delta S = 0$ would mean $(3) = (4) = 0$ since $\epsilon \neq 0$. But shouldn't $\delta S$ have this factor of $\epsilon$?