Probability of Poisson arrival in a vanishingly small interval

Question

Consider a homogeneous Poisson arrival process, and we look at the time interval $(t,t+\varepsilon]$ for a vanishingly small $\varepsilon>0$.

Why is it the case that the probability of more than 1 arrival in $(t,t+\varepsilon]$ is $o(\varepsilon)$?

Answer 1

The probability to have more than one arrival is then, assuming a rate $\lambda>0$ (seen as a constant), $$\begin{align} \mathbb{P}\{ N(t,t+\varepsilon] > 1 \} &= 1 - \mathbb{P}\{ N(t,t+\varepsilon] = 0 \} - \mathbb{P}\{ N(t,t+\varepsilon] = 1 \} \\ &= 1 - e^{-\lambda \varepsilon} - e^{-\lambda\varepsilon}(\lambda\varepsilon) \\ &= \frac{\lambda^2\varepsilon^2}{2} + o(\varepsilon^2) \end{align}$$ which is $o(\varepsilon)$.