If X has Binomial distribution with parameters n and p such that np =λ, then \(\mathop {\lim }\limits_{n \to \infty } b\left( {x,n,p} \right);x = 0,1,2,.....\) is equal to:
With \(np=\lambda\), put \(p=\lambda/n\) in the Binomial PMF:
\[b(x,n,p)=\binom{n}{x}\left(\tfrac{\lambda}{n}\right)^x\left(1-\tfrac{\lambda}{n}\right)^{n-x}\]
Rearrange and take \(n\to\infty\): the factor \(\dfrac{n(n-1)\cdots(n-x+1)}{n^x}\to 1\), \(\left(1-\tfrac{\lambda}{n}\right)^{n}\to e^{-\lambda}\), and \(\left(1-\tfrac{\lambda}{n}\right)^{-x}\to 1\).
Therefore
\[\lim_{n\to\infty}b(x,n,p)=\dfrac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\dots\]
This is the Poisson PMF, matching option (1).
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