What is the power series of $1/(D-h(x))$ if $h(x)\ll D$? I have a problem, which I do not conceptually understand. I need to approximate an arbitrary function 
$$\frac{1}{D-h(x)}$$
where $h(x)$ is arbitrary, $h(x)\ll D$ and $D$ is a constant. Friends say to me that it is just equal to 
$$\frac{1}{D-h(x)}\approx \frac{1}{D} \left(1+\frac{h(x)}{D}\right)$$
up to first linear term, which is basically a power series. I know that this would work for specific function where $h(x)=x$ and then the power series expansion would be 
$$\frac{1}{D-x}\approx \frac{1}{D} \left(1+\frac{x}{D}\right),$$
but I do not see how we can just replace $x$ by $h(x)$.
Also, it seems that it is not possible to get expansion by using Taylor series if we use $h(x)$. Why power series would work and Taylor series not?
 A: Here it is sufficient to consider the expansion of a geometric power series
\begin{align*}
\frac{1}{1-x}=1+x+x^2+\cdots\qquad\qquad|x|<1
\end{align*}
which is convergent for $|x|<1$.
The only crucial aspect here is the convergence criterion $|x|<1$, which enables us  to do a series expansion. 

When considering a function $h(x)$ with $\left|h(x)\right| \ll D$ we know that $\left|\frac{h(x)}{D}\right|<1$ and the geometric series expansion is admissible. We obtain
  \begin{align*}
\frac{1}{D-h(x)}&=\frac{1}{D}\cdot\frac{1}{1-\frac{h(x)}{D}}\\
&=\frac{1}{D}\left(1+\frac{h(x)}{D}+\frac{h(x)^2}{D^2}+\cdots\right)\tag{1}\\
\end{align*}
Since $\left|h(x)\right| \ll D$ it follows that  higher powers of $\frac{h(x)}{D}$ are small and we can use the approximation
  \begin{align*}
\color{blue}{\frac{1}{D-h(x)}\approx \frac{1}{D} \left(1+\frac{h(x)}{D}\right)}\tag{2}
\end{align*}

Note, that we do not require analytical properties of $h(x)$ in order to deduce (2). But if $h(x)$ is sufficiently often differentiable at $x=0$ we could also do a Taylor expansion in order to get (2).
