Summary
How would I prove that
$$ fQ(u) = \frac{1}{a} \left( 1 - \Phi^{-1}(au+b)^2 \right)\varphi\left(\Phi^{-1}(au+b)\right) \sim (1 -u), $$
as $u \to 1$, where $\Phi, \varphi$ are the CDF and PDF of the normal distribution, respectively, and $a = \Phi(1) - \Phi(-1)$, $b = \Phi(-1)$?
Note: $a, b$ are such that $\Phi^{-1}(au+b) \to 1$ as $u \to 1$.
Context
I'm interested in studying the tail behaviour of the distribution obtained by applying a $\tanh^{-1}$ transform to a truncated normal distribution. That is, $Y \sim \tanh^{-1}(X)$ where X is distributed according to $\mathcal{N}(0,1)$ truncated to $[-1, 1]$.
I'm using the classification scheme laid out in Parzen, 1979 (Section 9). This studies the density-quantile function $fQ(u)$ (the composition of the quantile function $Q$ and density function $f$) of the distribution and categorises the tail behaviour based on the asymptotic behaviour of $fQ(u)$ as $u \to 1$.
Specifically, we categorise the tail behaviour by finding the $\alpha >0$ such that
$$ fQ(u) \sim (1-u)^\alpha, \quad u \to 1. $$
When $\alpha=1$ we refine the tail behaviour by finding $0 \leq \beta \leq 1$ such that
$$ fQ(u) \sim (1-u) \left\{ \ln\frac{1}{1-u} \right\}^{1-\beta}, \quad u \to 1. $$
The density-quantile function for this distribution (verified numerically) is
$$ fQ(u) = \frac{1}{a} \left( 1 - \Phi^{-1}(au+b)^2 \right)\varphi\left(\Phi^{-1}(au+b)\right) $$
where $\Phi, \varphi$ are the CDF and PDF of the non-truncated normal distribution, respectively, and $a = \Phi(1) - \Phi(-1)$, $b = \Phi(1)$.
From plotting this function, it looks like $fQ(u) \sim (1-u)$ as $u \to 1$, but I am struggling to prove this.
Attempts
I'm aware that Blair, 1976 provides an asymptotic expansion for the squared inverse error function, but this is as its argument tends to $1$, not $1 - b$ as in this truncated case.