From the book Quantum Field Theory for gifted amateur, pg. 13, eq. (1.13):
It seems that the author performed a Taylor expansion on $g[f(x)+\epsilon\delta(x-x_0)]$. I know how to do Taylor expansion for the usual $g(x) $ around $x=a$ and the composite function $g(f(x))$ around $x=a. $
But what does it mean to Taylor expand $g[f(x)+\epsilon\delta(x-x_0)]$? It looks like a completely different function to $g(f(x)). $
Best Answer
Lets forget the context of the question and just consider an arbitrary smooth real function $f(x)$
We can ask what $f(x+\epsilon y)$ would be where $\epsilon$ is a small constant and $y$ is an arbitrary constant if we recall the identity
$$ f(x+a) = \frac{1}{0!} f(x) + \frac{a}{1!} f'(x) + \frac{a^2}{2!} f''(x) + ... $$
This identity is not obvious if you have not derived it before: its derivation comes from looking at the taylor series center a $a$ for f(x). then apply the substution $x \rightarrow x + a$ (so in the series identity replace all x with x+a) comment if you want this shown more clearly:
Anyways our identity then says that:
$$ f(x+\epsilon y) = f(x) + \epsilon y f'(x)+ \frac{\epsilon^2 y^2}{2!} f''(x) + ... $$
Now the author of the book has a skipped a few steps but is basically a direct application of this identity. Spelling it out it should be the following
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} g[f(x) + \epsilon\delta(x-x_0)] dx - \int_{a}^{b} g[f(x)] dx \right] $$
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} \left[ g[f(x)] + \epsilon\delta(x-x_0)g'(f(x)) + \frac{\epsilon^2}{2!}\delta(x-x_0)^2g''(f(x)) + ... \right] dx - \int_{a}^{b} g[f(x)] dx \right] $$
Writing it as a sum:
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} \left[ g[f(x)] + \epsilon\delta(x-x_0)g'(f(x)) + \sum_{k=2}^{\infty} \frac{\epsilon^k}{k!}g^{(k)}[f] \right] dx - \int_{a}^{b} g[f(x)] dx \right] $$
Recall that Integral of a Sum is a Sum of Integrals:
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} \left[ g[f(x)] dx \right] + \int_{a}^{b} \epsilon\delta(x-x_0)g'(f(x)) dx + \int_{a}^{b} \sum_{k=2}^{\infty} \frac{\epsilon^k}{k!}g^{(k)}[f] dx - \int_{a}^{b} g[f(x)] dx \right] $$
Now its obvious that left most integral and right most integral cancel out leaving:
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} \epsilon\delta(x-x_0)g'(f(x)) dx + \int_{a}^{b} \sum_{k=2}^{\infty} \frac{\epsilon^k}{k!}g^{(k)}[f] \right] dx $$
Now we can divide through by $\epsilon$
$$\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \int_{a}^{b} \delta(x-x_0)g'(f(x)) dx + \int_{a}^{b} \sum_{k=2}^{\infty} \frac{\epsilon^{k-1}}{k!}g^{(k)}[f] dx \right] $$
Now here's the catch, everything in that infinite sum has an $\epsilon$ to a non zero power multiplied with it. So if $\epsilon \rightarrow 0$ then every part of that sum is also going straight to 0 too. So when we take this limit that entire sum gets killed and we are left with.
$$ \int_{a}^{b} \delta(x-x_0)g'(f(x)) dx $$
Which by definition of the delta function is equal to
$$ g'(f(x_0)) $$
Off Topic:
I think thats a really nice book and I've only read a chapter or two from it but have learned an enormous amount. I hope also to finish reading it one day.