I want to use the Taylor series epansion of $tanh$ to get an approximate solution $tildex(beta)$ for the equation
$$
fractanh^-1(x)beta -2x =0
$$
for $beta >frac12$ such that I can calculate the limit
$$
lim_beta downarrow frac12fractildex(beta)left(beta-frac12right)^frac12.
$$
Unfortunately I have no clue on how to appropriately use the Taylor expansion to solve this. I guess it is possible to use the $O$-notation and break the problem down to a polynomial equation. Thank you very much in advance for any help.

The positive solution $tildex (beta)$ to your equation tends to zero as $beta downarrow frac12$. Therefore, we can rewrite the equation and use the Taylor series $fracoperatornameartanh(x)x = 1 + fracx^23 + mathcalO(x^4) $ to obtain
$$ 2 beta = fracoperatornameartanh(tildex (beta))tildex (beta) = 1 + fractildex^2 (beta)3 + mathcalO(tildex^4 (beta)) , .$$
We solve for the term with the lowest power of $tildex (beta)$ and find
$$ tildex (beta) = sqrtfrac6 left(beta - frac12right)1 + mathcalO(tildex^2 (beta)) = sqrt6 left(beta - frac12right) left[1 + mathcalO(tildex^2 (beta))right] = sqrt6 left(beta - frac12right) left[1 + mathcalOleft(beta - frac12right)right] , .$$
This expansion is sufficient to compute the desired limit.

Using one more term of the above Taylor series leads to the better result
$$ tildex (beta) = sqrt6 left(beta - frac12right) left[1 - frac95 left(beta - frac12right) + mathcalOleft(left(beta - frac12right)^2right)right] , ,$$
which agrees with the first-order expansion of the more accurate Padé approximant provided by Awe Kumar Jha.

You should first see that,
$$
tanh^-1 (x) = frac 12 ln left( frac 1+x1-x right)
$$
Then, you can use the second order Pade approximant,
$$
ln (1±x) ≈ frac ±x(6±x)(6±4x)
$$
I hope it gives desired results.

Edit

The Pade approximant of the original function is,
$$
tanh^-1 (x) ≈ frac x(15-4x^2)(15-9x^2)
$$
The solution to the resulting quadratic is,
$$
x=± sqrt frac 15(2beta -1)2(9beta -2), beta ≠ frac 29
$$
What remains to do is the limit,
$$
lim_beta → frac 12 sqrt frac 159beta -2 = sqrt 6
$$
I hope it helps.

Write the problem first as $$beta=fractanh ^-1(x)2 x$$
Now, expand the rhs using the usual
$$tanh ^-1(x)=sum_n=1^infty frac x^2n+1 2n+1$$ Using a few terms, we then have
$$beta=frac12+fracx^26+fracx^410+Oleft(x^6right)$$ Now, using series reversion
$$tildex (beta)=sqrt6 sqrtbeta -frac12-frac95 sqrt6 left(beta
-frac12right)^3/2+Oleft(left(beta -frac12right)^5/2right)$$

as already given by ComplexYetTrivial.