Let $\text{⨍ : R→R}$ be defined as 

$\text{⨍ (x)=}\left\{\begin{array}{ll}\frac{{\mathrm{x}}^3}{{\left(1-\cos 2\mathrm{x}\right)}^2}{\text{log}}_{\mathrm{e}}\left(\frac{1+2{\mathrm{xe}}^{-2\mathrm{x}}}{{\left(1-{\mathrm{xe}}^{-\mathrm{x}}\right)}^2}\right)& \mathrm{x}\ne 0\\ \mathrm{\alpha }& \mathrm{x}=0\end{array}\right.$ 

If $⨍$ is continuous at x=0 , then  $\alpha$ is equal to:

$$\begin{gathered} \mathrm{f}: \mathrm{R}\to \mathrm{R} \\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{{\mathrm{x}}^3}{{\left(1-\cos 2\mathrm{x}\right)}^2} \log \left(\frac{1+2{\mathrm{xe}}^{-2\mathrm{x}}}{{\left(1-{\mathrm{xe}}^{-\mathrm{x}}\right)}^2}\right) ; \mathrm{x}\ne 0\\ \mathrm{\alpha } ; \mathrm{if} \mathrm{x}=0\end{array}\right. \\ \mathrm{is} \mathrm{continuous} \\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{{\mathrm{x}}^3}{4 {\sin }^4\mathrm{x}}\left(\log \left(1+2{\mathrm{xe}}^{-2\mathrm{x}}\right)-2 \log \left(1-{\mathrm{xe}}^{-\mathrm{x}}\right)\right) ; \mathrm{x}\ne 0\\ \mathrm{\alpha } ;\mathrm{x}=0\end{array}\right. \\ \mathrm{f}\left(0\right)=\underset{\mathrm{x}\to 0^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{\alpha }=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{4}\frac{{\mathrm{x}}^3}{{\sin }^3\mathrm{x}}\left(\log \frac{\left(1+2{\mathrm{xe}}^{-2\mathrm{x}}\right)}{\mathrm{x}}-\frac{2\log \left(1-{\mathrm{xe}}^{-\mathrm{x}}\right)}{\mathrm{x}}\right) \\ \mathrm{\alpha }=\underset{\mathrm{x}\to 0}{\lim }\frac{1}{4}\left(2+2\right)\Rightarrow \mathrm{\alpha }=1 \end{gathered}$$