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}
Find the derivatives of $$f(x)=\sqrt{1-x}$$
$$=\sqrt{1-0}+\frac{\frac{d}{dx}\left(\sqrt{1-x}\right)\left(0\right)}{1!}x+\frac{\frac{d^2}{dx^2}\left(\sqrt{1-x}\right)\left(0\right)}{2!}x^2+\frac{\frac{d^3}{dx^3}\left(\sqrt{1-x}\right)\left(0\right)}{3!}x^3+\ldots \:$$
Then

$\sqrt{1-0}\quad :\quad 1$

$\frac{d}{dx}\left(\sqrt{1-x}\right)\left(0\right)\quad :\quad -\frac{1}{2}$

$\frac{d^2}{dx^2}\left(\sqrt{1-x}\right)\left(0\right)\quad :\quad -\frac{1}{4}$

$\frac{d^3}{dx^3}\left(\sqrt{1-x}\right)\left(0\right)\quad :\quad -\frac{3}{8}$

$\frac{d^4}{dx^4}\left(\sqrt{1-x}\right)\left(0\right)\quad :\quad -\frac{15}{16}$

So
$$=1+\frac{-\frac{1}{2}}{1!}x+\frac{-\frac{1}{4}}{2!}x^2+\frac{-\frac{3}{8}}{3!}x^3+\frac{-\frac{15}{16}}{4!}x^4+\ldots $$