aproximación de funciones usando series

$$\begin{align} 1-\sqrt{1-\frac1{1+x/2}} &=\frac{\frac1{1+x/2}}{1+\sqrt{1-\frac1{1+x/2}}}\\ &=\frac1{x+2}\frac2{1+\sqrt{1-\frac1{1+x/2}}}\\ &=\frac1{x+2}\left(1+\frac{1-\sqrt{1-\frac1{1+x/2}}}{1+\sqrt{1-\frac1{1+x/2}}}\right)\\ &=\frac1{x+2}\left(1+\frac{\large\frac1{1+x/2}}{\small\left(1+\sqrt{1-\frac1{1+x/2}}\right)^2}\right)\\ &=\frac1{x+2}\left(1+\frac12\frac1{x+2}\left(\frac2{\small1+\sqrt{1-\frac1{1+x/2}}}\right)^{\!\!\!2}\right)\\[9pt] &=\frac1{x+2}+\frac12\frac1{(x+2)^2}+O\left(\frac1{(x+2)^3}\right) \end{align}$$ Entonces, si lo deseamos, podemos usar $$\begin{align} \frac1{x+2} &=\frac1x\frac1{1+2/x}\\ &=\frac1x-\frac2{x^2}+O\left(\frac1{x^3}\right) \end{align}$$

$$\begin{align} 1-\sqrt{\frac{x}{2+x}}&=\frac{1-\frac{x}{2+x}}{1+\sqrt{\frac{x}{2+x}}}\\ &=\frac{1}{2+x}\,\frac{2}{1+\sqrt{\frac{x}{2+x}}} \end{align}$$ Ahora $$\frac{1}{2+x}=\frac1x+O\Bigl(\frac{1}{x^2}\Bigr)$$ y $$\frac{2}{1+\sqrt{\frac{x}{2+x}}}=1+O\Bigl(\frac{1}{x}\Bigr).$$