In reading an old book I stumbled upon this expression
$\sum_{k=0}^n {n \choose k} q^k (1-q)^{n-k} f(n,k)=0$
With the trick of setting $\lambda=nq$ the author lets $n \to \infty$ and $q \to 0$, resulting in
$\sum_{k=0}^{\infty} e^{-\lambda} \frac{\lambda^k}{k!} \lim_{n,q \to \infty} f(n,k)=0.$
The limit of the summand is taken and n is summed up to infinity. But can those limits be taken separately? Is there a proper way to derive the expression given above?