Using Taylor series expansion to solve the equation $\frac{\tanh^{-1}(x)}{\beta} -2x =0$ I want to use the Taylor series epansion of $\tanh$ to get an approximate solution $\tilde{x}(\beta)$ for the equation
$$
\frac{\tanh^{-1}(x)}{\beta} -2x =0
$$ 
for $\beta >\frac{1}{2}$ such that I can calculate the limit
$$
\lim_{\beta \downarrow \frac{1}{2}}\frac{\tilde{x}(\beta)}{\left(\beta-\frac{1}{2}\right)^{\frac{1}{2}}}.
$$
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.
 A: You should first see that,
$$
\tanh^{-1} (x) = \frac {1}{2} \ln \left( \frac {1+x}{1-x} \right)
$$
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,
$$
x=± \sqrt \frac {15(2\beta -1)}{2(9\beta -2)}, \beta ≠ \frac {2}{9}
$$
What remains to do is the limit,
$$
\lim_{\beta → \frac {1}{2}} \sqrt {\frac {15}{9\beta -2}} = \sqrt {6}
$$
I hope it helps.
A: Write the problem first as $$\beta=\frac{\tanh ^{-1}(x)}{2 x}$$
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
$$\beta=\frac{1}{2}+\frac{x^2}{6}+\frac{x^4}{10}+O\left(x^{6}\right)$$ Now, using series reversion
$$\tilde{x} (\beta)=\sqrt{6} \sqrt{\beta -\frac{1}{2}}-\frac{9}{5} \sqrt{6} \left(\beta
   -\frac{1}{2}\right)^{3/2}+O\left(\left(\beta -\frac{1}{2}\right)^{5/2}\right)$$
as already given by ComplexYetTrivial.
A: The positive solution $\tilde{x} (\beta)$ to your equation tends to zero as $\beta \downarrow \frac{1}{2}$. Therefore, we can rewrite the equation and use the Taylor series $\frac{\operatorname{artanh}(x)}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4) $ to obtain
$$ 2 \beta = \frac{\operatorname{artanh}(\tilde{x} (\beta))}{\tilde{x} (\beta)} = 1 + \frac{\tilde{x}^2 (\beta)}{3} + \mathcal{O}(\tilde{x}^4 (\beta)) \, .$$
We solve for the term with the lowest power of $\tilde{x} (\beta)$ and find
$$ \tilde{x} (\beta) = \sqrt{\frac{6 \left(\beta - \frac{1}{2}\right)}{1 + \mathcal{O}(\tilde{x}^2 (\beta))}} = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 + \mathcal{O}(\tilde{x}^2 (\beta))\right] = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 + \mathcal{O}\left(\beta - \frac{1}{2}\right)\right] \, .$$
This expansion is sufficient to compute the desired limit.
Using one more term of the above Taylor series leads to the better result
$$ \tilde{x} (\beta) = \sqrt{6 \left(\beta - \frac{1}{2}\right)} \left[1 - \frac{9}{5} \left(\beta - \frac{1}{2}\right) + \mathcal{O}\left(\left(\beta - \frac{1}{2}\right)^2\right)\right] \, ,$$
which agrees with the first-order expansion of the more accurate Padé approximant provided by Awe Kumar Jha.
