Taylor expansion in derivation of Euler-Lagrange equation

Question

I got this from QFT Demystified in the author attempt to derive the Euler Lagrange equation. But isn't the Taylor expansion for $f(x+a)$ supposed to be: $$f(x+a)=f(a)+x\frac{df(a)}{dx}+...(1)$$ Even if I exchange $x\rightarrow \epsilon, a \rightarrow x$, it should have been: $$f(\epsilon+x)=f(x)+\epsilon\frac{df(x)}{d\epsilon}+...(2)$$

I understand that in (1), $a$ is the point where we start or "nail" the fitting process, and $x$ is the independent variable. But what are $\epsilon$ and $x$ in (2)? Feel free to be rigorous if there's no intuitive way to answer, cause I really don't understand Taylor expansion at all and I need any answer. :((

Thank you! :D

Answer 1

$$ \frac{\mathrm d V(x)}{\mathrm d x}=\lim_{\varepsilon\to 0}\frac{V(x+\varepsilon)-V(x)}{\varepsilon} $$ so for $\varepsilon\ll 1$ we can write

$$ \frac{\mathrm d V(x)}{\mathrm d x}\approx \frac{V(x+\varepsilon)-V(x)}{\varepsilon} $$ that is $$ V(x+\varepsilon)\approx V(x) + \varepsilon \frac{\mathrm d V(x)}{\mathrm d x} $$

Answer 2

The confusion arises because of the notation you are using. Write (1) as $$f(x+a)=f(a)+xf'(a).$$ Now proceed as you suggest and replace $x$ by $\epsilon$ and $a$ by $x$. This yields $$f(\epsilon+x)=f(x)+\epsilon f'(x).$$ Is it clearer now?