I have come upon the equation $$x=|\tau(x)|^{1/\beta-1}(\beta-1)^{1/\beta}\frac{\pi/\beta}{\sin(\pi/\beta)}-|\tau(x)|^{-1}+o(|\tau(x)|^{-1})$$ and the author of the paper states, that it leads to the approximation $$\tau(x)\sim -b(\beta)x^{-\beta/(\beta-1)}, \quad b(\beta):=(\beta-1)^{1/(\beta-1)}\left(\frac{\pi/\beta}{\sin(\pi/\beta)}\right)^{\beta/(\beta-1)} \text{for } x\rightarrow0.$$ For the function $\tau(x)$ holds $\lim_{x\rightarrow 0}=-\infty, \lim_{x\rightarrow \infty}=(\beta-1)^{-1}$, $\tau(1)=0$, and it is strongly increasing.
So far, I have made not alot of progress in comprehending this, and I am wondering, if anybody could help me?
Greetings, Prob
We can re-arrange the initial equation into $$\tag{1} 1 + x\left| {\tau (x)} \right| + o(1) = \left| {\tau (x)} \right|^{1/\beta } (\beta - 1)^{1/\beta } \frac{{\pi /\beta }}{{\sin (\pi /\beta )}}. $$ Taking the power of $\beta/(\beta-1)$ of both sides gives $$ (1 + x\left| {\tau (x)} \right| + o(1))^{\beta /(\beta - 1)} = \left| {\tau (x)} \right|^{1/(\beta - 1)} b(\beta ) $$ which, after further simplification, becomes $$ \left( {\frac{1}{{x\left| {\tau (x)} \right|}} + 1 + o\!\left( {\frac{1}{{x\left| {\tau (x)} \right|}}} \right)} \right)^{\beta /(\beta - 1)} = \left| {\tau (x)} \right|^{ - 1} b(\beta )x^{ - \beta /(\beta - 1)} . $$ Now since $x|\tau(x)|\to \infty$ as $x\to 0$ (this follows from $|\tau(x)|\to \infty$ as $x\to 0$ and $(1)$), we obtain $$ \left| {\tau (x)} \right|^{ - 1} b(\beta )x^{ - \beta /(\beta - 1)} \to 1 $$ as $x\to 0$. Since $\tau(x) \to -\infty$ as $x\to 0$, we must have $$ - (\tau (x))^{ - 1} b(\beta )x^{ - \beta /(\beta - 1)} \to 1 $$ as $x\to 0$.