<s>[INST] Taylor series for $\sqrt{1+x^2}$ I want to expand 
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$ 
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example : 
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So  I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
 [/INST] If you want to take the limit when $x\to 2$ it is easier to set $x=2+u$ with $u\to 0$ because you will get it easier to make the expansions in variable $u$.
Thus $f(x)=\sqrt{1+x^2}=\sqrt{1+(2+u)^2}=\sqrt{5+4u+u^2}$
The proper expansion in zero is $\displaystyle \sqrt{1+v}=1+\frac v2-\frac{v^2}8+\frac{v^3}{16}-\frac {5v^4}{128}+\cdots$
We need first to factorize $5$ out of the square root.
$f(x)=\sqrt{5}\sqrt{1+\frac 45u+\frac 15u^2}\quad$ and then substitute $v=(\frac 45u+\frac 15u^2)$ in the expansion.
For instance let's limit ourselves to $o(u^3)$
$$f(x)=\sqrt{5}\left(1+\frac 12\left(\frac 45u+\frac 15u^2\right)-\frac 18 \left(\frac 45u+\frac 15u^2\right)^2 + \frac 1{16} \left(\frac 45u+\frac 15u^2 \right)^3 + o(u^3)\right)$$

We will do the calculation while ignoring all terms smaller than $o(u^3)$ (i.e terms in $u^4, u^5, \ldots$).
\begin{align}
f(x) & =\sqrt{5}\left(1+\frac 12\left(\frac 45u+\frac 15u^2\right)-\frac 18 \left(\frac {16}{25}u^2+2\frac 45\frac 15u^3\right)+\frac 1{16} \left(\frac 45u\right)^3 + o(u^3)\right) \\[8pt]
&=\sqrt{5}\left(1+\frac 25 u+\frac 1{50}u^2-\frac 1{125}u^3+o(u^3)\right)
\end{align}
In the end replace $u$ by $(x-2)$ to have the desired expansion:
$$f(x)=\sqrt{5}\left(1+\frac 25 (x-2)+\frac 1{50}(x-2)^2-\frac 1{125}(x-2)^3+o((x-2)^3)\right)$$
</s>