Were Devore's assumptions regarding Poisson processes incorrect? Page 134 of Devore's 9th Edition Probability and Statistics for Engineering and the Sciences lists the following two assumptions regarding Poisson processes:

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*There exists a parameter $α > 0$ such that for any short time interval of length $∆t$, the probability that exactly one event occurs is $α \cdot ∆t + o(∆t)$.

*The probability of more than one event occurring during $∆t$ is $o(∆t)$ [which, along with Assumption 1, implies that the probability of no events during $∆t$ is $1 - α \cdot ∆t - o(∆t)$].

...where a quantity is $o(∆t)$ if, as $∆t$ approaches $0$, so does $o(∆t)$/$∆t$.
I see two apparent issues with the above:

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*I don't see how these equations can yield probabilities. To illustrate, Example 3.41 gives $α = 6$ and $∆t = 0.5$, which would leave us with a probability greater than one.

*For the math to work, I believe "exactly one" in the first assumption should be "at least one" instead. Alternatively, the "+ $ o(∆t)$" term should be deleted from that sentence.

Or am I just missing something obvious here? To my knowledge, there is no published errata for this textbook.
Following is Example 3.41, for reference:
Suppose pulses arrive at a counter at an average rate of six per minute, so that $\alpha = 6$. To find the probability that in a $.5$-min interval at least one pulse is received, note that the number of pulses in such an interval has a Poisson distribution with parameter $\alpha t = 6(.5) = 3$ ($.5$ min is used because $\alpha$ is expressed as a rate per minute). Then with $X=$ the number of pulses received in the $30$-sec interval,
$P(1≤X) = 1-P(X=0) = 1-\frac{e^{-3}(3)^0}{0!}=.950$
 A: "Exactly one" and "at least one" both work equally well in the first statement.
If $\alpha$ is the rate of the Poisson process, then

*

*the probability of at least one event occurring in an interval of length $h$ is exactly $1 - e^{-\alpha h}$.

*the probability of exactly one event occurring in an interval of length $h$ is exactly $e^{-\alpha h} \cdot \alpha h$.

We have $\lim_{h \to 0} \frac{1 - e^{-\alpha h} - \alpha h}{h} = \lim_{h \to 0} \frac{\alpha e^{-\alpha h} - \alpha}{1} = 0$ by L'Hopital's rule, so $1 - e^{-\alpha h} = \alpha h + o(h)$.
We also have $\lim_{h \to 0} \frac{e^{-\alpha h} \cdot \alpha h - \alpha h}{h} = \lim_{h \to 0} (\alpha e^{-\alpha h} - \alpha) = 0$, so $e^{-\alpha h} \cdot \alpha h = \alpha h + o(h)$.
Note that this does not imply that these probabilities are the same! Two different expressions can both be $o(h)$.

In example 3.4.1, $\alpha = 6$ and $\Delta t = 0.5$ do not contradict the claim that $\alpha \Delta t + o(\Delta t)$ is a probability.
It just means that $o(\Delta t)$ stands in for a term which is asymptotically less than $\Delta t$ as $\Delta t \to 0$, but is negative (in fact, $-2$ or less) at the specific point you're looking at here.
A: Regarding issue 1, note that when we are using the o notations and all, we are talking about infinitesimal time intervals such that $\alpha \Delta t \ll 1$ which is certainly not the case in your example.
Regarding the 2nd issue, everything is correct; the probability for exactly one event to occur, also has some $o(\Delta t)$ terms. The exact expression is
$$e^{-\alpha \Delta t}\alpha \Delta t = \alpha \Delta t -\alpha^2 \Delta t^2 +\frac12\alpha^3\Delta t^3-\cdots = \alpha \Delta t + o(\Delta t)$$
