I am trying to expand $\sqrt{x(1-x)}$ in order to find the order of it as $x\to0$.

Using Taylor expansion to find the derivative of the equation as $x\to0$ gives each item $0$.

Then how do we expand it? Thanks,
['https://math.stackexchange.com/questions/1623321', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/206901/']
In general
$$\lim\_{x \to 0} (1+x)^{\alpha} = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2} x^2 + \binom{\alpha}{3} x^3 + \ldots = \sum\_{i = 0}^{n} \binom{\alpha}{i} x^{i} + O(x^n)$$
thus as $x \to 0$
\begin{align}
\sqrt{x(1+x)}&=\sqrt{x} \cdot \sqrt{1+x} \\
&=x^{1/2} \cdot (1+x)^{1/2}\\
&=x^{1/2} \cdot \left( \sum\_{i = 0}^{n} \binom{\frac{1}{2}}{i} x^{i} + O(x^n) \right) \\
&=x^{1/2} \cdot \left( 1 + \frac{x}{2} - \frac{x^2}{8} + \ldots \right)
\end{align}
so, as $x \to 0$
\begin{align}
\sqrt{x(1-x)}&=\sqrt{x} \cdot \sqrt{1-x} \\
&=x^{1/2} \cdot (1-x)^{1/2}\\
&=x^{1/2} \cdot \left( \sum\_{i = 0}^{n} \binom{\frac{1}{2}}{i} (-x)^{i} + O(x^n) \right) \\
&=x^{1/2} \cdot \left( 1 - \frac{x}{2} + \frac{x^2}{8} + \ldots \right)
\end{align}
my answer: 

$\sqrt{x(1-x)}$

$=x^{1/2}(1-x)^{1/2}$ 

expand for $x\to 0$, 

$=x^{1/2}(1+\frac 12(-x)+\frac{\frac12(1-\frac{1}{2})}{2!}(-x)^2+...........)$ 

$=x^{1/2}-\frac 12x^{3/2}-\frac{1}{8}x^{5/2}+...........$