Consider the following example on Taylor series.
Let's consider the difference quotient function of center $x_0$: $$f'_h(x_0)=\frac{f(x_0+h)-f(x_0)}{h}$$ For $h$ sufficiently small, the difference quotient function approximates the derivative: $$f'_h(x_0)=\frac{f(x_0+h)-f(x_0)}{h}\approx f'(x_0)$$ Now my textbook says:
We want to evaluate how $f'_h(x_0)$ approaches $f'(x_0)$. Let's use Taylor's Theorem, with $n=1,x=x_0+h$ in order to represent $f(x_0+h)$: $$f(x_0+h)=f(x_0)+hf'(x_0)+\frac{h^2}{2}f''(\xi_h)$$ so $$f'_h(x_0)=\frac{f(x_0+h)-f(x_0)}{h}=\\\frac{f(x_0)+hf'(x_0)+\frac{h^2}{2}f''(\xi_h)-f(x_0)}{h}=\\f'(x_0)+\frac{h}{2}f''(\xi_h)$$
I understand all the steps but it's not clear to me why that's done this way, especially:
...with $n=1,x=x_0+h$ in order to represent $f(x_0+h)$...
For context, the textbook is about Numerical Calculus and the paragraph is an introduction to discretization errors.
Could you help me clarify why each step of the above procedure was done?
We are considering the first order expansion ($n=1$) for Taylor's series with Lagrange remainder around the point $x_0$ ($x=x_0+h$) that is indeed
$$f(x_0+h)=f(x_0)+hf'(x_0)+\frac{h^2}{2}f''(\xi_h)$$
Refer also to the related