When $\alpha \to \infty$, I have a series of equations (Similar, but there is no connection between them.): $${\frac{\cosh\alpha}{\cosh^{2}\alpha+\alpha^{2}}}=2\mathrm{e}^{-\alpha}-\mathrm{e}^{-2\alpha}-0.345(\alpha^{2}\:\mathrm{e}^{-2\alpha}+100\alpha^{5}\:\mathrm{e}^{-6.5\alpha})$$ $$\frac{\sinh\alpha}{\cosh^{2}\alpha+\alpha^{2}}=2e^{-\alpha}-2(1+\alpha)\:\mathrm{e}^{-2.5\alpha}-5\alpha^{4}\:\mathrm{e}^{-4\alpha}$$ $$\frac{\cosh\alpha}{3\cosh^{2}\alpha+\sinh^{2}\alpha+\alpha^{2}-\frac{3}{4}}=\frac{1}{2}\:\mathrm{e}^{-\alpha}-\frac{1+8\alpha}{18}\:\mathrm{e}^{-17\alpha}-\frac{\alpha^{2}}{2}\:\mathrm{e}^{-3\alpha}+7\alpha^{3}\mathrm{e}^{-4\alpha}$$ $$\frac{\cosh\alpha}{2\cosh^2\alpha+\sinh^2\alpha+\frac\alpha3-\frac23}=\frac23\mathrm{e}^{-\alpha}+\frac{1+10\alpha}{12}\mathrm{e}^{-2\alpha}-7\alpha^3\mathrm{e}^{-4\alpha}$$ But I have no idea whether this is a limit, an expansion, or an asymptotic analysis. For my attempt: I guess this is asymptotic, so I try for example the second equation, $$ \sinh \alpha=\frac{e^{ \alpha }-e^{ -\alpha}}{2} $$ $$ \cosh \alpha = \frac{e^{ \alpha }+e^{ -\alpha }}{2} $$ $$ \frac{\sinh \alpha}{\cosh ^{2}\alpha+\alpha^{2}}= \frac{2(e^{ \alpha }-e^{ -\alpha })}{(e^{ \alpha }+e^{ -\alpha })^{2}+4\alpha^{2}} $$ When $\alpha \to \infty$, I can easily get the leading term $$ \frac{2e^{ \alpha }}{e^{ 2\alpha }}=2e^{ -\alpha } $$ If I do the right, for the next order, I have to subtract the leading term from the original equation, $$ \frac{2(e^{ \alpha }-e^{ -\alpha })}{(e^{ \alpha }+e^{ -\alpha })^{2}+4\alpha^{2}}-2e^{ -\alpha }=\frac{-8\alpha-4-2e^{ -2\alpha }}{e^{ 4\alpha }+4\alpha^{2}e^{ \alpha }+2e^{ \alpha }+e^{ -\alpha }}=-8\alpha^{2}e^{ -3\alpha } $$
This is what I can do so far. I would be very grateful if you have any tips or answers.