Set of valid values for expansion $$f(x)=1+\frac{1}{x-3}-\frac{1}{x+2}$$
Expand this equation up to the term$\frac{1}{x^3}$.
Here's my answer:
$$f(x)=1+\frac{5}{x^2}+\frac{5}{x^3}$$
My question is what is the valid set of values for the expansion? 
Using $$|\frac{3}{x}|<1$$
I get the valid set is $$|x|>3$$
(If you try to expand the equation, you will know why I am using $|\frac{3}{x}|<1$.)
However, when substituting the value, it seems like the valid set is not $|x|>3$. I don't know whether I evaluate the wrong answer or this is the nature of the expansion.
 A: The function
\begin{align*}
 f(x)&=1+\frac{1}{x-3}-\frac{1}{x+2}\\
\end{align*}
 has two simple poles at $-2$ and $3$.

We look at the poles $-2$ and $3$ and see they determine three regions.
\begin{align*}
 |x|<2,\qquad\quad
 2<|x|<3,\qquad\quad
 3<|x|
 \end{align*} 
  
  
*
  
*The first region $ |x|<2$ is a disc with center $0$, radius $2$ and the pole $-2$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-2$ and $3$  admit a representation as Taylor series at $x=0$.
  
*The second region $2<|x|<3$ is the annulus with center $0$, inner radius $2$ and outer radius $3$. Here we have a representation of the fraction with pole $-2$ as principal part of a Laurent series at $x=0$, while the fraction with pole at $3$ admits a representation as power series.
  
*The third region $|x|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $x=0$.

Since the problem is asking for an expansion up to $\frac{1}{x^3}$ we have to consider the third region and the expansion is according to OP
\begin{align*}
f(x)&=1+\frac{1}{x-3}-\frac{1}{x+2}\\
&=1+\frac{5}{x^2}+\frac{5}{x^3}+\cdots\tag{1}
\end{align*}
Hint: It is essential to write at least dots ($\cdots$)  in (1) otherwise the equal sign ($=$) is not correct.
