Find $\lim\limits_{A \to \infty} (1 - \frac{\pi \omega^2}{A})^{N-1}$, $N \sim \mathrm{Poisson}(\lambda A)$ This is not a part of my homework, just something came up to my mind when I did it.
Given $N \sim Poisson(\lambda A)$, find $\lim\limits_{A \to \infty} (1 - \frac{\pi \omega^2}{A})^{N-1}$
It is in similar form of $\lim\limits_{n \to \infty} (1 - \frac{k}{n})^{n} = e^{-k}$ but how can I make use of this result to find the limit above where the power is $N$ that is a r.v. with a parameter $A$?
 A: Here is a try:
$N \sim \mathrm{Poisson}(\lambda A)$, where $\lambda \in \mathbb{R}^+$, $A \in \mathbb{R}^+$
$$
E\left(\frac{N}{A}\right) 
= \frac{\lambda A}{A} 
= \lambda
\quad , \quad
Var\left(\frac{N}{A}\right) 
= \frac{\lambda A}{A^2} 
= \frac{\lambda}{A}
$$
Using Chebyshev's inequality, 
$$
Pr\left(\left|\frac{N}{A} - \lambda\right| > \frac{k\lambda}{A}\right) 
\le \frac{1}{k^2}
\quad , \quad
k \in \mathbb{R}^+
$$
Substitute $k = \frac{\epsilon A}{\lambda} \iff \epsilon = \frac{k\lambda}{A} \in \mathbb{R}^+$, 
$$
Pr\left(\left|\frac{N}{A} - \lambda\right| > \epsilon\right) 
\le \frac{\lambda^2}{\epsilon^2 A^2}
$$
$$
\implies \lim_{A \to \infty} Pr\left(\left|\frac{N}{A} - \lambda\right| > \epsilon\right) 
\le \lim_{A \to \infty} \frac{\lambda^2}{\epsilon^2 A^2} 
= 0
$$
$$
\implies \frac{N}{A} \overset{p}{\to} \lambda
$$
Since we have
$$
\lim_{A \to \infty} (1-\frac{\pi\omega^2}{A})^A
= e^{-\pi\omega^2}
\implies
\lim_{A \to \infty} \log\left((1-\frac{\pi\omega^2}{A})^A\right)
= -\pi\omega^2
$$
$$
\frac{N}{A} \overset{p}{\to} \lambda 
\implies \frac{N-1}{A} \overset{p}{\to} \lambda 
$$
Thus,
$$
\begin{align}
& \lim_{A \to \infty} \log\left((1-\frac{\pi\omega^2}{A})^{N-1}\right) \\
= & \lim_{A \to \infty} \left(\frac{N-1}{A}\log\left((1-\frac{\pi\omega^2}{A})^A\right)\right) \\
= & -\lambda\pi\omega^2
\end{align}
$$
$$
\lim_{A \to \infty} (1-\frac{\pi\omega^2}{A})^{N-1}
= e^{-\lambda\pi\omega^2}
$$
