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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.

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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 $z\to-\infty$.

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We have that

$$\left( \frac{ (1-\phi) x}{\phi (k-1)}+1\right)^ {1-k}=e^{(1-k)\log\left( \frac{ (1-\phi) x}{\phi (k-1)}+1\right)}$$

and

$$(1-k)\log\left( \frac{ (1-\phi) x}{\phi (k-1)}+1\right)=-\frac{\log\left( \frac{ (1-\phi) x}{\phi (k-1)}+1\right)}{\frac 1{k-1}}=\\=-\frac{ (1-\phi) x}{\phi }\frac{\log\left( \frac{ (1-\phi) x}{\phi (k-1)}+1\right)}{\frac{ (1-\phi) x}{\phi (k-1)}}\to \frac{ (1-\phi) x}{\phi }$$

since, by standard limits, as $t \to 0$ we have $\frac{\log (1+t)}{t} \to 1$.

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