EDITED: Sorry. I made a mistake, I think I want to ask why this approximation is valid when $|D|$ is very small, rather than how to find its limit when $|D| \to 0^+$; also, I made a typo, it should be $(\lambda|D|)^k$, not $(-\lambda|D|)^k$ .
I am currently learning poisson point process, and see the following statement in my lecture notes:
When $|D|$ is very small, $$\frac{e^{-\lambda |D|}(\lambda |D|)^k}{k!}$$ where $k$ is a non-negative integer and $\lambda$ is a positive constant
$= (1-\lambda |D|)$ for $k=0$,
$= \lambda |D|$ for $k=1$ and
$=o(|D|)$ (negligible) for $k=2,3,4,...$
It means a Poisson distributed random variable with parameter $\lambda|D|$ will approximately become Bernoulli distributed with parameter $\lambda|D|$ when $|D|$ is very small.
I know that a Bernoulli distribution (sum of Bernoulli trials) can be approximated by a Poisson distribution under certain circumstances ($n \to \infty$, $p \to 0$ and $np \to \theta$, $\theta$ is fixed and preferably $>5$), but I just can't figure out how the reversion works.
How can I get this approximate? Thanks!
Let $f(x) =\frac{e^{-cx}(-cx)^k}{k!} $ where $c > 0$ and $k$ is a positive integer. (My notation is easier to use.)
We want to approximate $f(x)$ for small $x$.
For small $x$, $e^{cx} \approx 1+cx $, so $f(x) \approx \frac{(1-cx)(-cx)^k}{k!} = \frac{(1-cx)(-cx)^k}{k!} $, so $f(x) \approx 1-cx$ if $k=0$ and $f(x) = O(x^k)$ for $k \ge 1$.
Are you sure that this is what you want, because this isn't very interesting.