How to prove that This limit was used in deriving the Poisson density

Question

From my understanding is to use l'hopital's rule is that right?
Please help me explain.
Thank you!

Answer 1

Suppose the rate of the Poisson process is $\lambda$. Then in small time interval $\Delta t$, the probability of one event occurring is $\lambda \Delta t$. The probability of no event occurring is $1-\lambda \Delta t$. Consider time $0$ to $t$ and cut it into a large number of $n$ intervals. Then the probability of $m$ events occurring in time $t$ is

$$C^n_m (1-\lambda \frac{t}{n})^{n-m}(\lambda \frac{t}{n})^m$$

$$=\frac{n!}{m!(n-m)!}(1-\lambda \frac{t}{n})^{n-m}(\lambda \frac{t}{n})^m$$

$$=\frac{n!}{m!(n-m)!n^m}(\lambda t)^m(1-\frac{\lambda t}{n})^{n-m}$$ $$=\frac{(\lambda t)^m}{m!}\frac{n(n-1)\cdots(n-m+1)}{n^m}(1-\frac{\lambda t}{n})^n(1-\frac{\lambda t}{n})^{-m}$$

Under the limit $n\rightarrow\infty$, we have

$$P=\frac{(\lambda t)^m}{m!}\lim_{n\rightarrow\infty}\frac{n(n-1)\cdots(n-m+1)}{n^m}\lim_{n\rightarrow\infty}(1-\frac{\lambda t}{n})^n\lim_{n\rightarrow\infty}(1-\frac{\lambda t}{n})^{-m}$$ $$=\frac{(\lambda t)^m}{m!}\times 1 \times e^{-\lambda t}\times 1$$ $$=\frac{(\lambda t)^m e^{-\lambda t}}{m!}$$

Answer 2

Recall in Poisson formulation $p \approx \frac{\lambda}{n}$, so when you have $(1-\frac{\lambda}{n})^n \to_n e^{-\lambda}$