I've seen the proof of $\lim_{N \to \infty} {N \choose k} p^k (1-p)^{N-k} = e^{-\lambda}\frac{\lambda^k}{k!}$ given that $\lambda = Np$ is a constant. However, if there's a situation which can be justified as a Binomial process with a very large N and a reasonably small p, then can we simplify the calculation with Poisson distribution? Note that we don't know if p is a function of N and gets smaller as N gets larger. More likely in real life, we can only say p is likely to be a small number.
An example might be the mutation on a (essentially discrete) DNA sequence is low at each possible position but it doesn't suggest the mutation rate gets even lower if we elongated the sequence. So I tried to prove
$$\lim_{N \to \infty} \frac{{N \choose k} p^k (1-p)^{N-k}}{e^{-Np}\frac{(Np)^k}{k!}}=1$$
but seems this is not true (correct me if I'm wrong).
So when can we use Poisson distribution to replace Binomial distribution without making impractical assumption like a varying p? I've seen many posts talking about replacing Binomial distribution with Poisson distribution to save calculation but I couldn't think of any real world case that would make sense.
If $p$ doesn't go to zero, then approximating Binomial($N,p$) by Poisson($Np$) will have a significant residual error as $N \to \infty$. This error can intuitively be understood because there is a discrepancy between the variances which remains significant as $N \to \infty$: the variance of the binomial is $1-p$ times the variance of the Poisson.
On the other hand, if $p$ is just fixed then you can use the normal approximation just fine, provided that $N \gg \max \{ 1/p,1/(1-p) \}$.