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.

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.
