Proof that: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda}$$ [hr] Firstly we establish this: $$(1-\frac{\lambda}{n})^{n} = e^{n*ln(1-\frac{\lambda}{n})} = e^{\frac{ln(1-\frac{\lambda}{n})}{\frac{1}{n}}}$$ Then we can do this: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n} = e^{\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}}$$ Then we can make use of the result we established above: $$= e^{\lim_{n\to +\infty}\frac{ln(1-\frac{\lambda}{n})}{\frac{1}{n}}}$$ This is in the form f(n)/g(n), so we can use L'Hopital's rule to evaluate the limit: $$= e^{\lim_{n\to +\infty}\frac{(\frac{\lambda}{n^{2}})\frac{1}{1-\frac{\lambda}{n}}}{-\frac{1}{n^{2}}}}$$ And then we can simplify: $$= e^{\lim_{n\to +\infty}-\lambda\frac{1}{1-\frac{\lambda}{n}}}$$ Then we see that anything divided by n, as n approaches positive infinity, tends to 0, so the fraction bit of the limit evaluates to 1. Then we have just a constant left, so the limit is gone, and so it evaluates to e^-[lambda], and hence: $$\lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda}$$