This is the question:
Let $\{N(t): t \ge0\}$ be a Poisson process, and $m(t)$ is its renewal function. Prove:
$$P\left[\lim_{t\to\infty}\left(1+ \frac{m(t)}{tN(t)}\right)^{N(t)} = e^{\lambda}\right] = 1$$
So, we know for a Poisson process, m(t) which is the renewal function is $\lambda t$. By substituting it, we'll have:
$$ P\left[\lim_{t\to\infty}\left(1+ \frac{\lambda}{N(t)}\right)^{N(t)} = e^{\lambda}\right] = 1$$
We also know that:
$$\lim_{t\to\infty}\left(1+ \frac{\lambda}{N(t)}\right)^{N(t)} = e^ {\lambda}$$
But what about the probability part? How should we approach that probability equals $1$ part? Does it have something to do with the Strong Law of Large Numbers?
Thanks in advance.
Best