Let $X~Bin(n, \lambda/n)$, $\lambda >0$. Show that for fixed $k \geq 0$, $P(X=k)\equiv \frac{e^-\lambda)\lambda^k}{k!}$

Question

Let $X - Bin(n, \lambda/n)$, $\lambda >0$. By using approximation $(1-\frac{x}{n})^n\approx e^{-x}$. Show that for fixed $k\geq 0$, $P(X=k)\approx \frac{e^{-\lambda}\lambda^k}{k!}$

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$P(X=k)=\frac{n!}{k!(n-k)!}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k}=\frac{n!}{k!(n-k)!}\frac{(\frac{\lambda}{n})^k}{(1-\frac{\lambda}{n})^k}(1-\frac{\lambda}{n})^n\approx\frac{n!}{k!(n-k)!}\frac{(\frac{\lambda}{n})^k}{(1-\frac{\lambda}{n})^k}e^{-\lambda}$

Where do I got from here?

Answer 1

Hint: Use $$\frac{n!}{(n-k)!} = n(n-1)\dots(n-k+1) = n^k(1+o(1))\approx n^k$$ and $$ \big(1-\frac{\lambda}{n}\big)^k = 1+o(1) \approx 1. $$