CDF approximation - L'Hopital's rule The following CDF,
\begin{equation}
F_{y}(x)  = 1- \Big(  \frac{ (1-\phi) x}{\phi (k-1)}+1\Big)^ {1-k} e^{- \frac{x}{\phi y}}
\end{equation}
is approximated for $k \rightarrow \infty$ as follows
\begin{equation} 
F_{y}(x)  \approx 1- e^{- (\frac{1-\phi}{\phi}+ \frac{1}{\phi y})x}
\end{equation}
I would like to obtain the approximation and tried to apply L'Hopital's rule (I am not sure whether it is the right approach) and I was not successful.
Can anyone guide me to find the approximation please? Thank you in advance.

Note that it suffices to approximate
$$\left(1+\frac{(\phi-1)x/\phi}{1-k}\right)^{1-k}$$
by $e^{\frac{(\phi-1)x}{\phi}}$.
With the substitution $n=1-k$ and $r=(\phi-1)x/\phi$, this follows immediately from the fact that
$$\lim_{n\to-\infty}\left(1+\frac{r}{n}\right)^n=e^r$$
You can prove this by L'Hospital applied to $\frac{\ln(1+rz^{-1})}{z^{-1}}$. Note that differentiation on top and bottom gives
$$\frac{r}{1+rz^{-1}}$$
which tends to $r$ as $z\to-\infty$.
