Proof that:
$$ \lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda}$$
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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:
$$ \lim_{n\to +\infty}(1-\frac{\lambda}{n})^{n}=e^{-\lambda} $$