Approximation of non-analytic function I have a function which is of the form
\begin{equation}
f(x) = \frac{1 - x^{1/2} + x - x^{3/2} + \ldots}{1+x^{1/2} - x + x^{3/2} - \ldots}.
\end{equation}
Intuitively, I would assume that for small $x$, it holds
\begin{equation}
f(x) \approx \frac{1-x^{1/2}}{1+x^{1/2}}
\end{equation}
and then, furthermore, 
\begin{equation}
f(x) \approx 1 - a x^{1/2} + \ldots 
\end{equation}
where $a$ is some factor. My question is: How can I determine $a$ and the range of $x$ for which this approximation is valid? Obviously, I cannot use a Taylor approximation since $f$ is not analytic and the derivative diverges in the origin.
Let me point out that I am not so much interested in the specific example above, which I have just invented. Much rather, I would like to know what is the general theory and methods behind this type of fractional functions. 
 A: Note, that the numerator, is 
$$ \sum_{k=0}^\infty (-x^{1/2})^k = \frac{1}{1 + x^{1/2}}, \quad |x| <  1 $$
and the denominator equals 
$$ 2 -\sum_{k=0}^\infty (-x^{1/2})^k = 2 - \frac 1{1 + x^{1/2}} $$
Hence 
\begin{align*}
    f(x) &= \frac{\frac 1{1 + x^{1/2}}}{\frac{2(1 + x^{1/2}) - 1}{1 + x^{1/2}}}\\
   &= \frac 1{2(1 + x^{1/2}) - 1}\\
   &= \frac 1{1 + 2x^{1/2}}\\
   &= \sum_{k=0}^\infty (-2x^{1/2})^k\\
   &= 1 - 2x^{1/2} + 4x \mp \cdots
\end{align*}
That is, $a = 2$ seems like a good choice.

Addendum: If we write $f$ as $g(\sqrt x)$ with 
$$ g(y) = \frac{1 - y + y^2 \mp}{1 + y - y^2 \pm } $$
then we can Taylor expand $g$ nicely, giving 
$$ g(y) = 1 - 2y + 4y^2 \mp $$
and hence
$$ f(x) = 1 - 2x^{1/2} + 4x \mp $$
and the range of convergence can be obtained by that of $g$.
A: I would suggest $$\frac{1-\sqrt{x}}{1+\sqrt{x}}\approx 1-2\sqrt{x}$$
because of $(1+\sqrt{x})(1-2\sqrt{x})=1-\sqrt{x}-2x$
A: Another way to treat this particular sort of problem.  It is an analytic function of, say, $z$, where $z=x^{1/2}$.
