Proving $\lim_{n\to\infty} \frac{\left(1-\frac{m}{n}\right)^{n}}{\left(1-\frac{m}{n}\right)^{r}} = e^{-m}$

Question

I was going through the derivation of Poisson distribution in probability. And I stuck with following step:

$$\lim_{n\to\infty} \frac{\left(1-\frac{m}{n}\right)^{n}}{\left(1-\frac{m}{n}\right)^{r}} = e^{-m}$$ Can anyone explain the actual way to prove the above expression?

Answer 1

Consider $$A=\left(1-\frac{m}{n}\right)^{n-r}\implies \log(A)=(n-r)\log\left(1-\frac{m}{n}\right)$$ Using Taylor series $$\log(A)=(n-r)\left(-\frac{m}{n}-\frac{m^2}{2 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ Expand $$\log(A)=-m-\frac{m (m-2 r)}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$A=e^{\log(A)}=e^{-m}\left(1-\frac{m (m-2 r)}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$

Answer 2

Note that $\lim\limits_{n \to \infty} (1 - \frac mn)^r = 1$ while $\lim\limits_{n \to \infty}(1 - \frac mn)^n = e^{-m}$. Combining these observations gives the result. More generally, for any $\alpha \in \mathbb{R}$ it holds that $\lim\limits_{n\to\infty} (1 + \frac{\alpha}{n})^n = e^\alpha$. To see this, consider \begin{equation}\lim\limits_{n\to\infty}(1 + \frac{\alpha}{n})^n = \lim\limits_{n \to \infty} (1 + \frac{1}{n/\alpha})^{\frac{n}{\alpha} \alpha} = \left(\lim\limits_{n \to \infty}(1+\frac{1}{n})^n\right)^\alpha = e^\alpha. \end{equation} In the third equality above we just reparametrized the limit and used the continuity of the function $t \mapsto t^\alpha$ for $t > 0$.