Puiseux Series? WolframAlpha says that $$\sqrt{x^2-1}$$ expanded in Puiseux series near 1 is $\sqrt 2 \sqrt{x-1}$.
I don't know what a Puiseux series is; I have searched on the net but I haven't understood much... can you briefly explain it to me and how I can obtain this result?

If you are familiar with Taylor series, in this case you easily can get the same expansion "for free," without having to sweat too much. Set $y=x-1$, so that you are looking at 
$$
\sqrt{x^2-1} = \sqrt{(x+1)(x-1)} = \sqrt{(y+2)y} = \sqrt{2y}\sqrt{1+\frac{y}{2}}
$$
when $y\to 0$ (i.e., $x\to 1$). Recalling the Taylor expansion of $t\mapsto \sqrt{1+t}$ around $0$, you get
$$
\sqrt{1+\frac{y}{2}} = 1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)
$$
(I only went to order $3$, but you can go much further) so that
$$\begin{align}
\sqrt{x^2-1} &= \sqrt{2y}\left(1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)\right) = \sqrt{2y}+\frac{\sqrt{2}}{4}y^{3/2}-\frac{\sqrt{2}}{32}y^{5/2} + o\!\left(y^{7/2}\right) \\ 
&=\sqrt{2}\sqrt{x-1}+\frac{\sqrt{2}}{4}(x-1)^{3/2}-\frac{\sqrt{2}}{32}(x-1)^{5/2} + o\!\left((x-1)^{7/2}\right).
\end{align}$$
