Some basic question: The first variation of a functional $J(y)$ is defined to be (see here, f.e.) $$ \delta J(y,h)=\lim_{\varepsilon\to 0}\frac{J(y+\varepsilon h)-J(y)}{\varepsilon}=\frac{d}{d\varepsilon}J(y+\varepsilon h)_{|\varepsilon=0}. $$
Hope this question is not too silly, but why do we have that $$ \lim_{\varepsilon\to 0}\frac{J(y+\varepsilon h)-J(y)}{\varepsilon}=\frac{d}{d\varepsilon}J(y+\varepsilon h)_{|\varepsilon=0}? $$
This is just an alternative definition. Consider the function $J_{x,v}\colon (-\varepsilon,\varepsilon)\to\mathbb{R}, t\mapsto J(x+tv)$. Then $J_{x,v}(0)=J(x)$ and $$ J_{x,v}'(0)=D_vJ(x). $$
You would need to subtract the perturbed functional's first two elements found in its Taylor series from those of the original functional's own Taylor series. Taking the limit of your difference as \varepsilon approaches 0 will give you the aforementioned result
