Derivation of Euler-Lagrange equation

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Here is a simple (probably trivial) step in the derivation of the Euler-Lagrange equation. If we denote $Y(x) = y(x) + \epsilon \eta(x) $, I want to know why is

$\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \dfrac{\partial f(y,x)}{\partial y} $

Could someone justify the steps involved in justifying this ? I am sure am missing something elementary.

EDIT: Could someone comment on correctness of the 'proof ' ? $\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta + H,x) -f(y + \epsilon \eta,x)}{H} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $

EDIT 2: The answers provided still leave me confused. Here is another attempt at a "proof". $\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{\epsilon\rightarrow 0} \dfrac{\partial f(Y,x)}{\partial Y} = \lim_{\epsilon \rightarrow 0} \left(\lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta + H,x) -f(y + \epsilon \eta,x)}{H} \right)$

Now if I could interchange the limits, then it would make sense that I get $\lim_{H\rightarrow 0} \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $.

So is the interchange of limits allowed. What hypothesis must be satisfied for that to happen ?

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The formula for $\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}$ is not correct: you should write

$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}:= \lim_{\epsilon\rightarrow 0}\frac{ f(y+\epsilon\eta,x)-f(y,x)}{\epsilon}= \lim_{\epsilon\rightarrow 0}\frac{ \langle \nabla f(y,x),h_\epsilon\rangle+O(\|h\|^2)}{\epsilon}, $$

where we suppose that $f$ is differentiable at $(y(x)+\epsilon\eta(x),x)$ for all $x$, denoting by $h_\epsilon$ the increment

$$h_\epsilon=( \epsilon\eta(x), 0)$$

and by $\nabla f(y,x)=\left(\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\right)$ the gradient of $f$ at $(y(x),x)$.

Then

$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}= \lim_{\epsilon\rightarrow 0}\frac{\frac{\partial f}{\partial y}\epsilon\eta(x) +O(\epsilon^2)}{\epsilon}=\frac{\partial f}{\partial y}(y(x),x)\eta(x), $$ as claimed (for all small variations $\eta(x)$). What we are missing here are the boundary conditions on the increment $\eta$ of $y$ in order to talk about a "variational problem" and the explicit form of the functional $f$.