In a certain text, it was given that a function is given by $$\varphi_{n}(t)= \sqrt{\frac{(\eta)_n}{n !}} \tanh^{n} (\alpha t) \operatorname{sech}^{\eta}(\alpha t)$$ where $\eta$ and $\alpha$ are some constant number, $t$ is time. Further, it was written that at any fixed time $t$ and large $n$ the wavefunction has the form $$\left|\varphi_n(t)\right|^2 \sim e^{-n / \xi(t)}$$ where $\xi(t)$ is "delocalization length" that grows exponentially in time: $\xi(t) \sim e^{2 \alpha t}$ for $\alpha t \gg 1$.
How to get this behaviour for the it at large $n$? I tried to take the function and look at the asymptotic behaviour for large $t$ but then got stuck with the $\tanh$ part how to take the large limit of $n$ for it? Instead, I tried to get the delocalization length as a function of time with the limit $\alpha t \gg 1$ but couldn't get so as with this limit, the $\tanh$ part will be simply one. As a follow-up can someone also suggest a simple strategy to do such asymptotic analysis for other cases without messing up?
First, for fixed $\eta$, $$ \frac{{(\eta )_n }}{{n!}} = \frac{{\Gamma (n + \eta )}}{{\Gamma (\eta)\Gamma (n + 1)}} \sim \frac{{n^{\eta - 1} }}{{\Gamma (\eta)}} $$ as $n\to+\infty$. Second, for $x>0$, $$ \tanh (x) = 1 - 2{\rm e}^{ - 2x} + 2{\rm e}^{ - 4x} + \ldots , $$ whence $$ \log \tanh (x) \sim \log (1 - 2{\rm e}^{ - 2x} ) \sim - 2{\rm e}^{ - 2x} $$ for large positive $x$. This gives $$ \tanh ^{2n} (\alpha t) = \exp (2n\log \tanh (\alpha t)) \sim \exp ( - 2n\exp ( - 2\alpha t)) $$ for large positive $\alpha t$. Third, for $x>0$, $$ \operatorname{sech}(x) = 2{\rm e}^{ - x} - 2{\rm e}^{ - 3x} + \ldots , $$ whence $$ \operatorname{sech}^{2\eta} (\alpha t)\sim 4^{\eta} {\rm e}^{ - 2\eta \alpha t} $$ for large positive $\alpha t$. Accordingly, $$ \left| {\varphi _n (t)} \right|^2 \sim \frac{{4^\eta }}{{\Gamma (\eta )}}{\rm e}^{ - 2\eta \alpha t} n^{\eta - 1} \exp ( - 2n\exp ( - 2\alpha t)) $$ for large positive $n$ and $\alpha t$ with fixed $\eta$.