<s>[INST] Continued fractions for $\sqrt{x} $ and beyond, valid formula? For $x > 0$, is this trick valid?
I use $$ ( \sqrt{x}-1)(\sqrt{x}+1)=x-1 $$
then $$  \sqrt{x}+1 = \frac{x-1}{\sqrt{x}+1-2} $$
so I can use iterations to get the rational approximant
$$ \sqrt{x} \approx \frac{a(0)+a(1)x+...a(n)x^{n}}{b(0)+b(1)x+..+b(m)x^{m}} $$
By repeating this procedure I can use this method to get a rational approximant to $ x^{2^{-n}} $ for integer $n$ valid for positive $x$.
 [/INST] It is possible to use ratio of polynomials from continues fraction expansion:
$$\sqrt x = 1 + \cfrac{x-1}{2+ \cfrac{x-1}{2
          + \cfrac{x-1}{...}}}
$$
so $f_{k}=(2+(x-1)\cdot f_{k-1})^{-1}$ and $y_k=1+(x-1)\cdot f_k$ with initial condition $f_0=1$ and $k\in\mathbb{N}$. For example:
$$
y_{6}=
{x \cdot {{x^3+21 \cdot x^2+35 \cdot x+7}\over{7 \cdot x^3+35 \cdot x^2+21 \cdot x+1}}}$$ $$
y_{10}=
{x \cdot {{x^5+55 \cdot x^4+330 \cdot x^3+462 \cdot x^2+165 \cdot x+11}\over{11 \cdot x^5+165 \cdot x^4+462 \cdot x^3+330 \cdot x^2+55 \cdot x+1}}}
$$
The main property is exact zero value at $x=0$ and ${d\over dx }y_k(x)|_{x=0}=k+1$ so for infinitely large $k$ the derivative trends to infinity.
Note that different properties of $y_k$ for even and odd $k$ is obtained: for even $k$ and $x\to\infty$ function $y_k$ trends to slant asymptote and for odd $k$ trends to constant.
Square root approximation
Green lines is for even $k$ and blue for odd, red line is for $\sqrt x$.
Open question is closed form for polynomial coefficients. As can be noted the coefficients at the $y_k$ function's numerator and denominator are in reverse order. This function can be used for $|x|$ approximation via substitution $x=t^2$ with important features $y_k(t)|_{t=0}=0$ and $y_k(t)|_{t=1}=1$. 
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