Extreme Value Theory - domains of attraction and techniques for evaluting a limit We consider the gamma uniform G distribution as specified by
Torabi and Montazeri:
$$f(x) = \frac{1}{\Gamma (a)}\frac{g(x)}{[1-G(x)]^2}\left[\frac{G(x)}{1-G(x)}\right]^{a-1}\exp\left[\frac{G(x)}{1-G(x)}\right]$$
where $G$ is a valid c.d.f., $g$ the corresponding p.d.f. and $a > 0$. 
My question is this: 

If $G$ belongs to the Gumbel domain of attraction, that is, if there
  exists $\gamma(t)>0$ such that $$\lim_{t \uparrow
w(G)}\frac{1-G(t+x\gamma(t))}{1-G(t)}=e^{-x},$$
  for all $x\in\mathbb{R}$, and if the upper end point $w(G) = w(F)$, then is it true that $F$ belongs to the same (Gumbel) domain of attraction as $G$?

The sticking point here is the exponential term, as we consider 
$$\lim_{t \uparrow
w(F)}\frac{1-F(t+x\gamma(t))}{1-F(t)}=\lim_{t \uparrow
w(F)}\frac{f(t+x\gamma(t))(1+x\gamma^\prime(t))}{f(t)},$$
where the equality follows from L'Hopital's rule. Now this gives rise to the term $$\lim_{t \uparrow
w(G)}\frac{\exp\left[\frac{G(t+x\gamma(t))}{1-G(t+x\gamma(t))}\right]}{\exp\left[\frac{G(t)}{1-G(t)}\right]}.$$
How do I evaluate this limit?
Any help much appreciated.
 A: Preliminary interpretation: Your question does not clearly specify what you mean by $w(G)$, but you say that this is an "upper end point"of the distribution.  I am going to assume that you mean that $t \uparrow w(G)$ implies $G(t) \uparrow 1$.  My answer proceeds on this basis.

To simplify the notation in this problem we treat $\gamma$ as fixed and denote $G_x(t) \equiv G(t+x\gamma(t))$.  We then define the functions:
$$E_x(t) \equiv \frac{1-G_x(t)}{1-G(t)} \quad \quad \quad \bar{E}_x(t) \equiv \frac{G_x(t)}{G(t)}.$$
These functions are related by:
$$\bar{E}_x(t) = \frac{1- E_x(t)(1-G(t))}{G(t)}.$$
By assumption, for all $x \in \mathbb{R}$ you have $\lim_{t \uparrow w(G)} E_x(t)= e^{-x}$.  By the composition law of limits (and assuming the required continuity for it to apply) we also have:
$$\lim_{t \uparrow w(G)} \bar{E}_x(t) = \lim_{t \uparrow w(G)} \frac{1- e^{-x}(1-G(t))}{G(t)}.$$
We now define and simplify the function:
$$\begin{equation} \begin{aligned}
H(t,x) 
&\equiv \frac{G_x(t)}{1-G_x(t)} - \frac{G(t)}{1-G(t)} \\[6pt]
&= \frac{1-G(t)}{1-G_x(t)} \cdot \frac{G_x(t)}{G(t)} \cdot \frac{G(t)}{1-G(t)} - \frac{G(t)}{1-G(t)} \\[6pt]
&= \frac{G(t)}{1-G(t)} \Bigg[ \frac{1-G(t)}{1-G_x(t)} \cdot \frac{G_x(t)}{G(t)} - 1 \Bigg] \\[6pt]
&= \frac{G(t)}{1-G(t)} \Bigg[ \frac{\bar{E}_x(t)}{E_x(t)} - 1 \Bigg]. \\[6pt]
\end{aligned} \end{equation}$$
Hence, applying the composition law again yields:
$$\begin{equation} \begin{aligned}
\lim_{t \uparrow w(G)} (1-G(t)) \cdot H(t,x) 
&= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{\bar{E}_x(t)}{E_x(t)} - 1 \Bigg] \\[6pt]
&= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{1- e^{-x}(1-G(t))}{e^{-x} G(t)} - 1 \Bigg] \\[6pt]
&= \lim_{t \uparrow w(G)} G(t) \Bigg[ \frac{1- e^{-x} + e^{-x} G(t) - e^{-x} G(t)}{e^{-x} G(t)} \Bigg] \\[6pt]
&= \lim_{t \uparrow w(G)} G(t) \cdot \frac{1- e^{-x}}{e^{-x} G(t)}  \\[6pt]
&= \frac{1- e^{-x}}{e^{-x}} \\[6pt]
&= e^{x}-1. \\[6pt]
\end{aligned} \end{equation}$$
Now, the limit you want to find is $\lim_{t \uparrow w(G)} \exp(H(t,x))$.  This limit is infinity, but there is a related limit that may be useful:
$$\begin{equation} \begin{aligned}
\lim_{t \uparrow w(G)} \exp(H(t,x))^{\exp(1-G(t))} 
&= \lim_{t \uparrow w(G)} \exp( (1-G(t)) \cdot H(t,x)) \\[6pt]
&= \exp(e^x-1). \\[6pt]
\end{aligned} \end{equation}$$
